wade powerpoint/lecture for chapter 5

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7/25/2019 Wade Powerpoint/Lecture for Chapter 5 http://slidepdf.com/reader/full/wade-powerpointlecture-for-chapter-5 1/72 Chapter 5 Stereochemistry: the study of the 3-D the study of the 3-D structure of molecules structure of molecules Organic Chemistry , 8 th  Edition L. G. Wade, Jr.

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Page 1: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 172

Chapter 5

Stereochemistrythe study of the 3-Dthe study of the 3-D

structure of moleculesstructure of molecules

Organic Chemistry 8th Edition

L G Wade Jr

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 272

Stereochemistry

bull For pharmaceuticals slight

differences in the 3D spatial

arrangement can make the difference

between targeted treatment and

undesired side-effects

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372

ypes of somers

Structural or Constitutional sameformula different bonding se$uence

CH3CH3

CH3CH3

CH3

CH3

CH3CH2OHH3C-O-CH3

and

and

C amp (

C ) amp

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472

ypes of somers

Stereoisomers same formulasame bonding se$uence but

different spatial arrangement of the

atoms +eometric or cis-trans

) nantiomers mirror image isomers

3 Diastereomers not mirror images ofeach other

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572

Stereoisomersbull Same bonding se$uence

bull Different arrangement in space

bull ample ampC-CampCamp-Camp

has two geometric cis-trans0 isomers

OHOH

O

O

OHOH

H H

O O

fumaric acid mp )12C maleic acid mp 31C

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

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Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

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he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

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Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

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Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

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hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

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+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

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Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

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Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

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C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

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nd of Chapter 5

Page 2: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

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Stereochemistry

bull For pharmaceuticals slight

differences in the 3D spatial

arrangement can make the difference

between targeted treatment and

undesired side-effects

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372

ypes of somers

Structural or Constitutional sameformula different bonding se$uence

CH3CH3

CH3CH3

CH3

CH3

CH3CH2OHH3C-O-CH3

and

and

C amp (

C ) amp

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472

ypes of somers

Stereoisomers same formulasame bonding se$uence but

different spatial arrangement of the

atoms +eometric or cis-trans

) nantiomers mirror image isomers

3 Diastereomers not mirror images ofeach other

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572

Stereoisomersbull Same bonding se$uence

bull Different arrangement in space

bull ample ampC-CampCamp-Camp

has two geometric cis-trans0 isomers

OHOH

O

O

OHOH

H H

O O

fumaric acid mp )12C maleic acid mp 31C

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 3: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 372

ypes of somers

Structural or Constitutional sameformula different bonding se$uence

CH3CH3

CH3CH3

CH3

CH3

CH3CH2OHH3C-O-CH3

and

and

C amp (

C ) amp

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472

ypes of somers

Stereoisomers same formulasame bonding se$uence but

different spatial arrangement of the

atoms +eometric or cis-trans

) nantiomers mirror image isomers

3 Diastereomers not mirror images ofeach other

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572

Stereoisomersbull Same bonding se$uence

bull Different arrangement in space

bull ample ampC-CampCamp-Camp

has two geometric cis-trans0 isomers

OHOH

O

O

OHOH

H H

O O

fumaric acid mp )12C maleic acid mp 31C

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

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nd of Chapter 5

Page 4: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 472

ypes of somers

Stereoisomers same formulasame bonding se$uence but

different spatial arrangement of the

atoms +eometric or cis-trans

) nantiomers mirror image isomers

3 Diastereomers not mirror images ofeach other

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572

Stereoisomersbull Same bonding se$uence

bull Different arrangement in space

bull ample ampC-CampCamp-Camp

has two geometric cis-trans0 isomers

OHOH

O

O

OHOH

H H

O O

fumaric acid mp )12C maleic acid mp 31C

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 5: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 572

Stereoisomersbull Same bonding se$uence

bull Different arrangement in space

bull ample ampC-CampCamp-Camp

has two geometric cis-trans0 isomers

OHOH

O

O

OHOH

H H

O O

fumaric acid mp )12C maleic acid mp 31C

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 6: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 672

dentify the following pairs as

either constitutional isomers

stereoisomers or identical

)

3

(

5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 7: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 772

Chirality

bull 4ampandedness right glo6e doesn7t fit the left handbull 8irror-image ob9ect is different from

4 nonsuperimposable0 the original ob9ect

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 8: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 872

hich ob9ects are chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 9: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 972

Chirality in 8olecules

bull he cis isomer is achiral ie not handed0bull he trans isomer is chiral

bull nantiomers nonsuperimposable mirrorimages different molecules

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 10: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1072

ampelpful 4rulelt

bull if both hal6es of a molecule are

symmetric about an 4internal

mirror plane then the molecule is

achiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 11: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1172

=rob 5-) Draw the mirror image of the original

structure gt determine whether the mirror image is

the same cmpd abel each structure as chiral

or achiral and label each pair of enantiomers

c )-bromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 12: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1272

=rob 5-) Continued

CH3

H

H

CH3

CH3

CH3

H

H

O

a cis-)-dimethylcyclobutane b trans-)-dimethylcyclobutane

e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 13: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1372

Chiral Carbonsbull etrahedral carbons with ( different attached groups

are chiral bull f there7s only one chiral carbon in a molecule its

mirror image will be a different compound enantiomer0

bull Chiral carbons are designated with a

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 14: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1472

H Cl

CH3

OH

H

H

Sol6ed =rob 5- Star each

asymmetric carbon atom

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 15: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1572

=rob 5-3 Draw structures and star

asymmetric carbon atoms

a0 )-pentanol b0 3-pentanol c0 alinine

g0 h0

CH3 CH3

OH

CH3 CH3

OH

CH3 OH

O

NH2

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1672

8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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7252019 Wade PowerpointLecture for Chapter 5

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Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

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nd of Chapter 5

Page 16: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

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8irror =lanes of Symmetry

bull A molecule containing ) or more chiral carbons willbe achiral if the molecule has an 4internal mirror

plane of symmetry

bull A molecule with an internal mirror plane cannot be

chiral

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 17: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1772

=rob 5-5 Determine whether

each compound has an internal

mirror plane of symmetry

CH3

CH3CH3

h0

Bo

asymmetric

carbons

g0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 18: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1872

Baming nantiomersusing the Cahn-ngold-=relog con6ention

bull nantiomers are different moleculestherefore they must ha6e different

names

bull sually only one enantiomer will bebiologically acti6e

bull Configuration around the

chiral carbon is specifiedwith E0 or S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 19: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 1972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 20: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2072

Assigning =riorities

CH3

H

NH2

OHOCH3 Br

Br

CH2CH2Br

Br CH3

Halinine

1

2

3

4

3-dibromobutane

1

2

3

4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 21: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2172

Assigning E0 or S0

bull orking in 3-D rotate molecule so that lowest priority group faces away from you is behind theC 0

bull Draw an arrow from highest to lowest prioritygroup

bull Clockwise E0 Counterclockwise S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 22: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2272

he steering wheel analogylt

bull urn the steering wheel clockwiseand the car turns to the right E0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 23: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2372

Sol6ed =rob 5-) Draw the

enantiomers of 3-dibromobutane

and label them as E0 or S0

H2CH2C CH3

Br Br

HC

3-dibromobutane

CBr Br

CH3 H3C

HH

CH2CH2Br CH2CH2Br

1 1

2 2

3 3

4 4

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 24: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2472

Sol6ed =rob 5-3 he structure of one of the

enantiomers of car6one is shown below Find

the asymmetric carbon atom determine

whether it has the E0 or S0 configuration

car6one

1

23

4

(E ) S0

O

C

CH2H3C

H

s it or

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 25: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2572

hat to do when

group ( is in front

bull Eotate the

molecule in

space so that

group (

points away

from you E

S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 26: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2672

hat to do when

group ( is in front

bull Draw the arrow

but then apply the

rule backwards

eg now clock-

wise is S and

counterclockwise

is E Elt S0

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 27: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2772

hat to do when

group ( is in front

bull Switch group ( with whate6er group is

pointed away from you Bow determine

configuration as normal he originalcompound is the new compound7s

enantiomer

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 28: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2872

ample Amphetamine

E0S0 E0

NH2

CH3

H

NH2

H

CH3

NH2

CH3 H

mirror

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 29: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 2972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 30: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3072

Answer From Solutions

8anual to =rob 5-

Cl H

ClH

HCl

ClH

(d)

HH

ClCl

(e)

H H

ClCl

S0

S0

S0

E0

Assume lower groups

are back and away

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 31: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3172

=roblem Are these pairs of

compounds enantiomers or

diastereomers

CH3CH3

CH3

CH3

CH3 CH3

CH3

CH3

S

S

E E

CH3

CH3

OH

Cl

CH3Cl

OH

CH3

E E E S

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 32: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3272

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 33: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3372

+ood and Had nantiomers

he errible Case of halidomide

bull A +erman drug company launched thalidomide

in I52bull t was found to act as a

bull analgesic bull tran$uiliGer bull antiemetic for treating morning sickness

bull Current knowledge held that drugs could not

pass the placental barrier and harm a fetusbull FDA ne6er appro6ed of use in the SA but

thalidomide came in from o6erseas and for use in

clinical trials

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 34: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3472

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 35: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3572

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 36: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 37: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3772

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 38: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3872

Eemember

bull E0 and S0 are names

nomenclature terms0

bull d0 and l0 or K0 and -0 are physical properties and must

be determined bymeasurement

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 39: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 3972

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 40: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4072

Eacemic 8itures

bull $ual $uantities of d- and l- enantiomers

a 5J5J mi0

bull Botation dl0 or plusmn 0

bull Bo optical acti6ity

bull he miture may ha6e different bp and

mp from the enantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 41: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4172

Eacemic =roducts

f optically inacti6e reagents combineto form a chiral molecule a racemic

miture of enantiomers is formed

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 42: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4272

ptical =urity

bull Also called enantiomeric ecess

bull Amount of pure enantiomer in ecess ofthe racemic miture

bull f op 5JL then the obser6ed rotationwill be only 5JL of the rotation of the pure enantiomer

bull 8iture composition would be 25-)5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 43: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4372

Chirality of Conformers

bull f e$uilibrium eists between two chiral

conformers molecule is not chiral

bull Mudge chirality by looking at the most

symmetrical conformer

bull Cycloheane can be considered to be

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 44: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4472

8obile Conformers

Bonsuperimposable mirror

images but e$ual energyand intercon6ertible

So consider it

planar on a6erage

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 45: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4572

ip earn to Draw Cylics in

=lanar 4Approimations

H

H

HH

HH

H

HHH

H H

O

H

H

HH

HH

HH

H H

H

H

H

HH

H

H

H

H

H O

H

H

H

HH

H

H

H

H

H

H

H H

H

H

H

H

HH

H

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 46: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4672

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 47: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4772

=rob 5-( Draw each compound in its most

symmetric conformation and determine

whether it is capable of showing optical acti6ity

c )-dichloroproane

Br

H

H

Br

d cis-3-

dibromocycloheane

Br

H

H

Br

f trans-(-

dibromocycloheane

Cl

Cl

H3C H

HH

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 48: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4872

Bonmobile Conformers

f the conformer is sterically hinderedit may eist as enantiomers

Allenes

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 49: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 4972

Allenes

bull Chiral compounds with no chiral carbonbull Contains sp hybridiGed carbon with

ad9acent double bonds -CCC-

Allene is achiral

MAKE THE MODELS

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 50: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5072

Allenes ill He Chiral f heir

nd +roups Are Different

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 51: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5172

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 52: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5272

Draw the

carbon chain

6ertically

Fi h = 9 ti E l

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 53: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5372

Fischer =ro9ection Eules

bull Carbon chain is on the 6ertical linebull ampighest oidiGed carbon at top

bull Eotation of 1J deg

in plane doesn7t

change molecule

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 54: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5472

Fischer =ro9ection Eules

bull Do not rotate IJ deg

bull Do not turn o6er out of plane ie noflipping like a pancake

Fi h 8i

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 55: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5572

Fischer 8irror magesbull asy to draw easy to find enantiomers

easy to find internal mirror planes

bull ample )S3E0-)3-dibromobutane

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 56: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5672

Fischer E0 and S0

bull owest priority usually amp0 comes forwardso assignment rules are backwards

bull Clockwise -)-3 is S0 andcounterclockwise -)-3 is E0

bull ample

(S )

(S)

Br

Br

CH3

CH3

H

H

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 57: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5772

ip Draw the Fischer=ro9ection and then

determine whether it is E0or S0 f it is not the one

you want 9ust switch the

horriGontal groups

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 58: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5872

=rob 5-2 Draw a Fischer

=ro9ection for each compound

CH3

OH

OH

CH3

Br

OH

a S0-propane-)-diol

b E0-)-bromobutan--ol

= b 5 1 8 k Fi h

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 59: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 5972

=rob 5-1 8ake a Fischer

=ro9ection Draw the mirror

image Determine whether themirror image is the same as or

different from the original

H

CH2OH

CHO

OH

CH2OH

Br H

CH2OH

Br

CH2Br

Br

CH3

a b c

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 60: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6072

=rob 5-I abel each asymmetric

carbon atom as E0 or S0 in each

Fischer =ro9ection

H

H

CH2OH

CH2

OH

OH

OH

CH3

COOH

HH2N

d h i

Br

CH2OH

CH3

Cl

C i f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 61: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6172

bull Are cisOtrans isomers enantiomers or

diastereomers

Categories of somers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 62: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6272

Diastereomers

bull Stereoisomers that are not mirror imagesbull +eometric isomers cis-trans0

bull 8olecules with ) or more chiral carbons

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 63: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6372

Diastereomer Alkenes

Cis-trans isomers are not mirrorimages so they are diastereomers

CH3

H

CH3

H

CH3

CH3

H

H

Diastereomeric Eing

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 64: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6472

Diastereomeric Eing

Compoundsbull Cis-trans isomers possible

bull 8ay also ha6e enantiomers

bull ample trans-)-dimethylcyclopentane

8 C d

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 65: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6572

8eso Compoundsbull An achiral compound containing two or

more chiral carbonsOchirality centers

bull Also contains an internal mirror plane of

symmetry

H

H

CH2OH

CH2OH

OH

OH

S t f ti f

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 66: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6672

wo Sets of nantiomers of

the Same Compound are

Diastereomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 67: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6772

=rob 5-)J For each pair gi6e

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 68: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6872

=rob 5 )J For each pair gi6e

the relationship between the

two compoundsCl

Cl

ande OHH

OHH

CH2OH

CHO

HHO

OHH

CH2OH

CHO

andf

Br Br

and

d

= ti f Di t

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 69: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 6972

=roperties of Diastereomersbull Diastereomers ha6e different physical

properties mp bpbull hey can be separated easily by

distillation

crystalliGation

chromatography bull nantiomers differ only in reaction with other

chiral molecules and the direction in which polariGed light is rotated

bull nantiomers are difficult to separate

E l ti f ti

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 70: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7072

Eesolution of nantiomers

Eeact a racemic miture with a chiral compoundto form diastereomers which can be separated

Chromatographic Eesolution

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 71: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7172

C o atog ap c eso ut o

of nantiomers

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5

Page 72: Wade Powerpoint/Lecture for Chapter 5

7252019 Wade PowerpointLecture for Chapter 5

httpslidepdfcomreaderfullwade-powerpointlecture-for-chapter-5 7272

nd of Chapter 5