organic chemistry carbon - honourscheminnerweb.ukzn.ac.za/files/class slides with solutions...
TRANSCRIPT
10/4/2011
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CHEM 120 CHEMICAL REACTIVITY
ORGANIC CHEMISTRY
19 Lectures5 Tutorials
2 Quiz1 Test
Dr. Vincent O. Nyamori
http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx
worksheets
Textbooks
Recommended books
H. Hart, L.E. Craine, D.J. Hart and C.M. Hadad, Organic Chemistry - A
Short Course, 12th Edition, Houghton-Mifflin (students not intending to
continue with Chemistry).
P.Y. Bruice, Organic Chemistry, 6th Edition,
Pearson/Prentice Hall (Chemistry major students).
Any other relevant Organic textbook - Library
Prescribed book
T.L. Brown, H.E. LeMay Jr, B.E. Bursten, C.J. Murphy,
S. Langford and D. Sagatys, Chemistry: The Central
Science: A Broad Perspective, 2nd Edition, Pearson,
Australia, 2010.
Organic Chemistry• The study of carbon‐containing
compounds and their properties.
• The vast majority of organic
compounds contain chains or rings of
carbon atoms.
3
SurfactantC17H35COO
‐Ascorbic acidHC6H7O6
GlucoseC6H12O6
Carbon1s22s22p2
Periodic Table4
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Structure of Carbon Compounds
C C C C C C
Three hybridization states
sp3 sp2 sp
each satisfies the octet rule for each carbon!
1.54 Å1.20 Å
1.33 Å
“Chemistry: The Central Science: A Broad Perspective”, 2nd
Edition, by LeMay et al. Chapter 21.1 & Chapter 23.1 pg 852 5
“Organic Chemistry: A Short Course” 12th
Edition, by Hart et al. Chapter 1.14 ‐ 1.18
Geometries of carbon compounds
sp3
Tetrahedral
sp2
Trigonal planar
spLinear
C
C
6
Hydrogens are in a tetrahedral
arrangement around the sp3
hybridized carbon atom.
Hydrogens bond to the carbon sp3
orbitals with 1s orbitals.
Methane: CH4
Energy
sp3
2p
2s
1s
Hybridization
sp3 Hybridizationcarbon
7
Energy
sp2
2p
2s
1s
Hybridization
sp
2p
2s
1s
Hybridization
Energy
Ethene: C2H4
Ethyne: C2H2
sp2 Hybridization
sp Hybridizationcarbon
carbon
8
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HYDROCARBONS
• compounds composed of only carbon and hydrogen
• chain of carbon atoms bonded to enough hydrogen atoms
to satisfy the octet rule for each carbon
• chain is bent because of the 109.5° C–C–C tetrahedral angle
C
CC
CC
CC
H
H
H
H
HH
H
H
HH
CH3
CHHC
CH2
CC
H3C
e.g.
sp3
sp
sp2
Line notation
180°
109.5°
120°
9
Π = θ =
3 bonds16 bonds
How many bonds are 10
• They contain the maximum number of hydrogen atoms
• UNSATURATED hydrocarbons
HYDROCARBONS
• Hydrocarbons with all single carbon‐carbon bonds(no double or triple bonds)
• Alkanes are SATURATED
‐ they ARE NOT ALKANES
• contain carbon‐carbon multiple bonds
Alkanes
Alkenes, alkynes and aromatic compounds
“Chemistry: The Central Science: A Broad Perspective” ‐ Chapter 21
“Chemistry: The Central Science: A Broad Perspective”
‐ Chapter 23 & 24
Past exam Question
C‐1: _____ C‐3: _____
C‐2: _____ C‐4: _____
sp2 sp
sp3 sp3
11
Example
CC
C
CC
CC
C
O
CN
F H
O
ClH
HBr
H
H
H
12
3
46
78
910
1. Indicate the hybridization for carbons 1 – 10 and theirrespective geometry. Include bond angles in your answer.
Solution
Carbon 1, 2 and 6:
Carbon 3, 4 and 9: sp2
Carbon 7, 8 and 10:
sp3
sp
Hybridization Geometry Bond angles
Tetrahedral 109.5°
Trigonal planar 120°
Linear 180°12
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Hydrocarbons
• Four basic types:
‐ Alkenes
‐ Alkynes
‐ Aromatic hydrocarbons
‐ AlkanesCnH2n+2
C2H6 Ethane
CnH2n
C2H4 Ethene
C2H2 Ethyne
CnHn
CnH2n‐2
HH
H
H H
H
C6H6 Benzeneor
120°
Organic Nomenclature
• Three parts to a compound name:
1 2 3
1. Prefix
2. Base
3. Suffix
Chapter 2: “Organic Chemistry”, 5th Edition , Bruice P. Y.
Chapter 21: pg. 807‐9: “Chemistry: The Central Science: A
Broad Perspective”, 2nd Edition, LeMay et al.
Organic Nomenclature
Suffix:
Tells what type of compound it is.
suffix
What family?
15
Organic Nomenclature
Base/parent:
Tells how many carbons are in the longest continuous chain.
base suffix
How many carbons?
What family?
16
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Organic Nomenclature ‐ IUPAC Rules
Tells what substituent(s) are attached, if any.
prefix base suffix
What substituent? How many
carbons?
What family?
Prefix:
17
Alkanes
• Only van der Waals force: London force.
• Boiling point increases with length of chain.
• Combust to give mainly CO2 and H2O
• Nomenclature suffix “‐ane” E1
Properties
To Name a Compound…1. Determine what type of
compound it is. ‐ Suffix
2. Find the longest chain in the
molecule. ‐ Base
3. Number the chain from the
end nearest the first
substituent encountered.
4. List the substituents as a prefix
in alphabetical order along
with the number(s) of the
carbon(s) to which they are
attached.
CH2CH
CH2H2C
CH2
H3C
H3C
CH32‐Ethylhexane3‐Methylheptane Name?? 19
Example
If there is more than one type of
substituent in the molecule, list them
alphabetically i.e. name of substituent,
not prefix for frequency e.g. di, tri,
tetra, etc...are not considered.
CH3
CH2CH
CH CH
CHCH3
CH2 CH3
CH3
CH3
CH3
3‐Ethyl‐2,4,5‐trimethylheptane
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Cycloalkanes
• Carbon can also form ringed structures.
• Five‐ and six‐membered rings are most stable.
– Can take on conformation in which angles are
very close to tetrahedral angle.
– Smaller rings are quite strained.
cyclohexane cyclopentane cyclopropane
General formulaCnH2n
CYCLIC ALKANES
How do we name….CH3
CH2CH3
methyl
ethylEthylmethylcyclohexane
1
23
4
56
1‐Ethyl‐3‐methylcyclohexane
22
Unsaturated Hydrocarbons
Carbon can form multiple bonds with itself or other
atoms, e.g. N or O.
Multiple bonds affect, physical and chemical
properties (reactivity of organic molecules).
Hydrocarbons that contain one or more double or
triple bonds are called unsaturated hydrocarbons.
C
CC
C
CC
C C
C C
H
H
H
H
H H
H
H H
H
H
H
CH3C
O
CH3
CH3C
N
CH3
CH3C
S
CH3
H
Alkenes
• Contain at least one carbon–carbon double bond
• Unsaturated
– Have fewer than maximum number of hydrogens
– The C atoms on double bond are sp2 hybridized
VSEPR Theory
120°
Chapter 23: “Chemistry: The Central Science: A Broad Perspective”
2nd Edition, by LeMay et al.
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Structure of Alkenes
• Unlike alkanes, alkenes cannot rotate
freely about the double bond.
– Side‐to‐side overlap makes this
impossible without breaking ‐bond.
C C
25
Structure of Alkenes
This creates geometric isomers
difference in the spatial arrangement of groups about the double bond
Cis‐ isomer “Z”‐isomer
Trans‐ isomer “E”‐isomer
Z‐2‐Pentene
E‐2‐Pentene
26
27
NAMING ALKENES
1. Find the longest unbranched carbon chain containing the
double bond.
2. Number the carbon atoms in the main chain.
• Name chain according to number of carbon atoms.
add ‐ene as a suffix
• Start from the end of the chain that is closest to the
double bond.
location of the double bond is numbered with the
lowest‐numbered carbon in the double bond.28
1. The longest unbranched chain containing the double
bond is seven carbons long, so this is a heptene.
Carbon #2 is the lowest‐numbered carbon in the
double bond, so this is a 2‐heptene or hept‐2‐ene
1 23 4
5 6 7Name this alkene
Example
C
H
CHCH3 C
H
CH
CH2
CH2
CH
CH3
CH3
H2C
2. The chain numbering starts closest to the double bond.
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3. There are two substituent groups on this alkene:
C
H
CHCH3 C
H
CH
CH2
CH2
CH
CH3
CH3
H2C
1 23 4 5 6 7
5‐methyl
4‐cyclopropyl
hept the alkene is:
Compose the name…..
4‐cyclopropyl‐5‐methyl‐2‐ ene
• Add the substituent groups alphabetically to the alkene name
• Specify the position of each group on the main chain
29
Properties of Alkenes
Structure also affects physical properties of alkenes
2‐Methyl‐1‐propenebp. ‐7 ⁰C
1‐Butenebp ‐6 ⁰C
Cis‐2‐Butenebp +4 ⁰C
Trans‐2‐Butenebp +1 ⁰C
Example: C4H8
Can we have more than one double bond? 1,3‐butadiene
Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition
• Identical substituents on
opposite sides of the double
bond
CCH3
H
CH
CH3
C
CH3
HC
H
CH3
Alkenes exhibit cis‐trans isomerism.
GEOMETRICAL ISOMERS
• Identical substituents on
same side of the double
bondtrans‐ cis‐
Stick diagram
Trans‐ isomer “E‐” Cis‐isomer “Z‐”
31
Priorities are assigned by the atomic numbers of the atoms
bonded to the carbon in the double bond.
“Z‐”“E‐”
Geometric isomers of Alkenes
• Cis‐alkenes have similar higher priority elements or group in
the chain on the same side of the molecule (or Z‐isomer i.e.
have higher priority elements but not necessarily the same on
the same side of the molecule)
• Trans‐alkenes have similar higher priority elements or group in
the chain on opposite sides of the molecule (or E‐isomer i.e.
have higher priority elements but not necessarily the same on
opposite sides of the molecule).
2‐Pentene 32
2‐Pentene
1‐Pentene
(Z)‐
(E)‐
e.g.
C5H10
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Examples1. Name the following alkenes and determine whether there
are geometric isomers are either Trans‐ (E‐) or Cis‐ (Z‐)isomers if applicable.
H
a)
3‐Methylhex‐3‐ene
Br
Hb)
3‐Fluoro‐2‐methylpent‐2‐eneF
c)
(E)‐
1 2
1
1
1
2 2
2
Equal priority
No geometric isomers formed.
3‐Bromopent‐2‐ene
33
(Z)‐
34
Each functional group is specified by a suffix or prefix
Certain groups of atoms give a molecule a.... FUNCTION
Acidic, basic, alcohol, etc…
The GROUPS are called functional groups.
depicted on the nomenclature of the organic molecule
FUNCTIONAL GROUPS
34
Functional groups are given an order of priority to decide
on which is the suffix.
HOMEWORK!
• “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay et
al. Chapter 24.1, 25.1, 26.1 & 26.3
• “Organic Chemistry” 5th Edition, by Bruice
• Hart et al. “Organic Chemistry: A Short Course” 12th Edition
Hint: Please refer to your textbooks !!
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Alkane RH alkyl- -ane
Alkene R2C=CR2 alkenyl- -ene
Alkyne RC≡CR' alkynyl- -yne
Benzene derivative
RC6H5
RPhphenyl- -benzene
Ethane
2-phenylpropaneisopropylbenzene
Ethene
Ethyne
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Haloalkane RX halo-alkyl
halide
Fluoroalkane RF fluoro-alkyl
fluoride
Chloroalkane RCl chloro-alkyl
chloride
Bromoalkane RBr bromo-alkyl
bromide
Iodoalkane RI iodo-alkyl
iodide
Chloroethane
Ethyl chloride
Fluoromethane
Methyl fluoride
Chloromethane
Methyl chloride
Bromomethane
Methyl bromide
Iodomethane
Methyl iodide
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Group / Family
Group FormulaStructural Formula
Prefix Suffix Example
Amines
Primary amine
(1º)RNH2 amino- -amine
Secondary amine
(2º)R2NH amino- -amine
Tertiary amine
(3º)R3N amino- -amine
Quaternaryammonium
ionR4N+X-
ammonio--ammonium
H3C
N
H CH3
CH3
Cl
Methylamine
Dimethylamine
Trimethylamine
Trimethyl-ammonium
chloride
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Alcohol ROH hydroxy- -ol
Ketone RCOR' keto-, oxo- -one
Aldehyde RCHO aldo- -al
Methanol
2-ButanoneMethyl-
ethyl ketone
EthanalAcetaldehyde
Group / Family
FormulaStructural Formula
Prefix Suffix Example
Carboxylic acid
RCOOH carboxy- -oic acid
Acyl halide RCOX haloformyl- -oyl halide
Ether ROR' alkoxy-alkyl alkyl
ether
Ester RCOOR'alkyl
alkanoate
Ethanoic acidAcetic acid
Ethanoyl chlorideAcetyl chloride
Ethoxy ethaneDiethyl ether
Ethyl butanoateEthyl butyrate
Primary (1°) alcohols and amines
General structure
AlcoholR1 C
H
H
OH
Amine
Example
1° alcohols
1° amine
C
H
H
OHCH3CH2
N
H
H
CH3CH2R1 N
H
HEthylamine
Propan‐1‐ol
1‐Propanol
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Secondary (2°) alcohols and amines
General structure
AlcoholR1 C
H
R2
OH
Amine
Example
2° Alcohol
2° Amine
C
H
CH3
OHCH3CH2
N
H
CH3
CH3CH2R1 N
H
R2
Ethylmethylamine
Butan‐2‐ol
2‐Butanol
Tertiary (3°) alcohols and amines
General structure
AlcoholR1 C
R3
R2
OH
Amine
Example
3° alcohols
3° Amine
C
CH3
CH3
OHCH3CH2
N
CH3
CH3
CH3CH2R1 N
R3
R2
Ethyldimethylamine
2‐Methylbutan‐2‐ol
2‐Methyl‐2‐butanol
42
Quaternary amines
General structure
AmineR4 N
R2
R1
R3+ N
CH3
CH3
CH3CH3CH2
+
Example
Ethyltrimethylammonium ion
43
Exercise
1. Draw the structures of the following alcohols and amines
and classify them as either 1°, 2°, 3° or quaternary
a) Pentan‐1‐ol
b) Dimethylamine
c) 3‐Ethylhexan‐3‐ol
d) Diethylmethylamine
e) Butan‐2‐ol
f) Triethylmethyl ammonium ion
44
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CC
CC
CHO
H
HH
H
H
HH
H
H
HH
a) Pentan‐1‐ol
Solutions
N
H
CH3
CH3
b) Dimethylamine
OH
CH3
CH2 C
CH2
CH2
CH3CH2
CH3
c) 3‐Ethylhexan‐3‐ol
N
CH2CH3
CH2CH3
H3C
d) Diethylmethylamine
C
OH
CH3
H CH2CH3
e) Butan‐2‐ol
N
CH2CH3
CH2CH3
H3C
CH3CH2
f) Triethylmethylammonium ion
1° alcohol 2° amine3° alcohol
3° amine 2° alcoholQuaternary amine
A45 46
Naming Hydrocarbons with Functional Groups
Name the other substituent groups, using the
prefixes for alkyl groups and the prefixes for any
other functional groups
Specify the position of each group on the main chain.
Add the substituent groups alphabetically to the name
of the alkane (or alkene or alkyne) along with the
frequency of each group.
Example…..
47
Example: Name this organic molecule:
CH3 CHCH
OH
CH3
NO2
48
Example: Name this organic molecule:
1. This molecule contains a hydroxyl and a nitro group
CH3 CHCH
OH
CH3
NO2
Only the hydroxyl group has priority.
So this is an ALCOHOL…..
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49
Example: Name this organic molecule:
1. This molecule contains a hydroxyl group and a nitro
group
longest carbon chain containing the hydroxyl
group has four carbons.
Therefore its a “butanol”.
CH3 CHCH
OH
CH3
NO2
this is an ALCOHOL.
An alcohol suffix is ‐OL
50
3. Number the carbons, starting NEAREST the
functional group.
CH3 CHCH
OH
CH3
NO2 Butanol
1 2 3 4
The hydroxyl group is on position 2
…so this is a 2‐butanol or butan‐2‐ol
4. This molecule has one substituent
A nitro group on position 3
3‐nitro‐2‐butanol 3‐nitrobutan‐2‐olor
51
Example: Name this organic molecule
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
1. Identify functional groups
carboxyl group amino group
52
Example: Name this organic molecule:
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
1. Molecule contains a carboxyl group and an amino group.
a carboxylic acid
Highest priority
Now determine longest carbon chain
The carboxyl group has highest priority, hence
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53
Example: Name this organic molecule:
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
The longest carbon chain has 8 carbons.
BUT………………...
THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group
54
Example: Name this organic molecule
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
The longest carbon CONTAINING the ‐CO2H
chain has 7 carbons.
so this molecule is based on a heptane.
55
Highest priority functional group is a carboxyl group
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
Suffix ‐OIC ACIDHence… Heptanoic acid
Now number chain…….
3. Number the carbons starting with the functional group
123
4
5 6 7
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O
H2N
Heptanoic acid
The carboxyl group is on position 1,
do not include in the name because
the carboxyl group is always a terminal group.56
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57
Compose the name
2‐ethyl3‐isopropyl
5‐amino
5‐methyl
CH3 CH CH CH
CH3
C
CH2
CH2C
CH2
CH3
CH3
CH3
OH
O123
4
5 6 7H2N
heptanoic acid5‐methyl3‐isopropyl‐2‐ethyl‐5‐amino‐
ALPHABETICAL LIST
INTERPETING AN IUPAC NAME…...
Heptanoic acid
4. This molecule has four substituents
i.e. at carbons 2, 3 and two at 5
58
WHAT IS THE STRUCTURAL FORMULA OF
BUTANONE ?
4 CARBONS
KETONE GROUP
CH3 CH2C
O
CH3
CANNOT BE AT END!!!! WHY?????
C CH2CH2
O
CH3H
BUTANAL
Examples
Give the correct IUPAC name for the following compounds.
(Z)‐3‐methylhex‐2‐ene
2,3,7‐trimethyloctane
Br
H
6‐bromo‐4‐ethyl‐2,3,7‐trimethyloctane
3,3‐dimethylbutan‐2‐ol
OH
Alkane
BromoalkaneAlkylbromide
Alkene
Alcohol(2°)
a)
c)
b)
d)
59
O
H
O
CH3
O
Cl
OH
2‐chloropentanoic acid ethylbutanoate
3‐methylhexan‐2‐one2‐ethylpentanal
Examples
Give the correct IUPAC name for the following compounds.
f) g)
h) i) O
O
Aldehyde Ketone
Carboxylic acid Ester
60
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Examples
Draw structural formulae for the following compounds:
a) 2,3,5‐trimethylhexane
b) (Z)‐3‐chlorohept‐2‐ene
c) 3‐ethylnonanol
d) 2,3‐dimethylpentanoic acid
e) methyl hexanoate
f) 3‐iodohexanal
g) pentan‐2‐one
h) 3‐aminopentane 61
Solutions
2,3,5‐trimethylhexane
Cl
H
(Z)‐3‐chlorohept‐2‐ene
OH
3‐ethylnonanol
O
OH
2,3‐dimethylpentanoic acid
a) b)
c) d)
62
Solutions
methyl hexanoate 3‐iodohexanal
pentan‐2‐one
e) f)
g) h)
O
O
OI
H
OH2N
3‐aminopentane
63
ISOMERS
(a) Structural isomers
CONSTITUTIONAL ISOMERS
Constitutional isomers have different properties:
These are…..?
e.g. butane (C4H10) has 2 structural isomers
Molecules with the same chemical formula but different
bonds between the atoms
Now called...
CH3 CH2 CH2 CH3 CH3 CH CH3
CH3
n‐Butane: C4H10 2‐methylpropane: C4H10
bp = ‐12 °C mp = ‐159 °Cbp = 0 °C mp = ‐138 °C
Two types: (a) Structural isomers (b) Stereoisomers
64
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65
CH2CH3 CH2 CH2 CH3
n‐pentane
Solution:
2‐methylbutane
1) 2) CH3 CH CH2
CH3
CH3
CCH3 CH3
CH3
CH3
2,2‐dimethylpropane
3)
ExampleHow many constitutional isomers are formed from C5H12? Draw their structures.
THREE CONSTITUTIONAL ISOMERS
CONSTITUTIONAL (STRUCTURAL) ISOMERS
The general formula for ALKANES is…... 22 nnHC
The number of ISOMERS increases with n…..
n = 1, 2 and 3 1 ISOMER
n = 4 2 ISOMERS
n = 5 3 ISOMERS
n = 6 5 ISOMERS
n = 7 9 ISOMERS
n = 8 18 ISOMERS
n = 9 35 ISOMERS
n = 10 75 ISOMERS
n = 20 366,319 ISOMERS
n = 40 62,491,178,805,831 ISOMERS
HOMEWORK: DRAW THE ISOMERS OF C40H82 !!! 66
Constitutional isomers for multibonds
(E,E)‐2,4‐hexadiene
(E)‐1,3‐hexadiene
(Z,E)‐2,4‐hexadiene
(Z)‐1,3‐hexadiene
(Z,Z)‐2,4‐hexadiene
C6H10
Example: Alkene
1,5‐hexadiene67 68
Constitutional isomers for multibonds
Example: Alkyne
C6H6
H3C C C C C CH3
C C C C CH2
CH3
H
1,3‐hexadiyne
hexa‐1,3‐diyne
2,4‐hexadiyne
hexa‐2,4‐diyne
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69
OPTICAL ISOMERISM
Optical isomerism arises when molecules have a
structure such that the mirror image is not
superimposable on the original molecule.
occurs whenever there are four different groups
bound to the same tetrahedral carbon atom.
Some terminology…….
70
OPTICAL ACTIVITY IN ORGANIC COMPOUNDS
The carbon involved is called a chiral carbon or
stereogenic carbon and the molecule is known as a chiral
molecule.
Stereogenic centre has four different groups attached
to a tetrahedral carbon atom
Example: 2‐butanol
ZX
W
Y
C
Stereogenic centre
Chiral carbon atom
*
C
CH2CH3
OH
H
H3C
2‐ButanolDash shows bond going
backwards from the viewer
*
Solid wedge represents a bond extending out towards the viewer
Simple line for Bonds aligned to the asymmetric
center in the plane
Chiral centre
Perspective formula
71
Wedge‐dash notation
CH2CH3HO
H
CH3
72
2‐Butanol
*
Fischer projection
Bonds aligned to the asymmetric
center in the plane
Chiral centre
Bond going backwards from
the viewer
Bond extending out towards the
viewer
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(non‐superimposable mirror images of each other)
View in 3‐DView in 3‐D“mirror”
• 2‐butanol has two optical isomers.
• A pair of isomers called enantiomers ‐
C
CH2CH3
OH
H
H3C
C
OH
HCH3CH3CH2
ENANTIOMERISM in ORGANIC CHEMISTRY
Chapter 22.3: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition
E
Identify the chiral carbon (stereogenic centre) and draw the
structural formula for each of the following molecules:
Solution
(a) 1‐chloroethanol (b) 2,3,5‐trimethylhexane
(c) Methylcyclohexane (d) 1,3‐dimethlycyclopentane
C CH3
Cl
HO
H
(a) 1‐chloroethanol
*
*
* *No chiral carbon
Question
(b) 2,3,5‐trimethylhexane
(d) 1,3‐dimethlycyclopentane(c) methylcyclohexane
74
75
NAMING OPTICAL ISOMERS
Solution: Use R‐S nomenclature system for designating
the configuration
Stereogenic centre creates twomolecular optical isomers
We assign priorities as in the E, Z system……...
How do we name these isomers??
CH
Cl
OHH3C
CH
Cl
HO CH3
Two configurations
Enantiomers* *
1. Assign relative priorities to each of the four groups on
the stereogenic carbon to describe the configuration.
The priorities are given by rules:
• Higher atomic numbers are given higher priorities.
• If necessary, the second atom in each substituent is
used to determine the priorities.
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition
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20
2. Draw the molecule with the lowest‐priority group
pointing directly into the page….
and the other three groups pointing out of the page in an
arrangement like a steering wheel orMercedes‐Benz symbol
Example: 2‐Butanol…..
CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE
CH2CCH3
OH
CH3
H
Draw molecule as a wedge and dashed line diagram
C
78
H3CCH2CH3
C
OH
HH3CH2C
C
OH
CH3
H
2‐ butanol
DO NOT FORGET THE OTHER ISOMER……….
CH2CCH3
OH
CH3
H
THESE ARE THE TWO…. ENANTIOMERS...or ....OPTICAL ISOMERS
p
Chapter 22; pg. 829‐839. “Chemistry: The Central Science: A Broad Perspective” 2th Edition.
79
1. Assign priorities:
O > CMe = CEt > H
C,H,H > H,H,H
CEt > CMe
2‐ butanol
C
CH3
CH2CH3
OH
H
80
CH3
CH2CH3
C
OH
H
1
2
3
4
H3C CH2CH3
C
OH1
2
3
2. Redraw the molecule with the lowest priority group facing in.
Now what?????
O > CEt > CMe > H
1. Assign priorities to each group:
2‐Butanol
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81
3. Look at the direction in which the priorities decrease.
If they decrease in a clockwise direction, the stereogenic centre is called “R” or rectus
which is Latin for “right.”Or….
H3C CH2CH3
OH
1
2
3
C
2‐Butanol
(R)‐2‐Butanol
CH3CH3CH2
OH
1
2 3
If the priorities decrease in a counter‐clockwise
direction the stereogenic centre is called “S” or
sinister, which is Latin for “left.”
C
2‐Butanol
(S)‐2‐Butanol
82
A83
Example:
CH3
CH2CH3
C
NH2
HCH3C
CH2CH3H
NH2
(S)‐2‐butanamine(R)‐2‐butanamine
H3C CH2CH3
C
NH2
3 2
1CH3
CH2CH3
C
H2N
1 2
3
N > CEt > CMe > H
What are the configurations of the following chiral molecules?
84
Example: Give the configuration of the stereogenic
centre in each of the following molecules:
CH
CH2OH
ClC OH
O
CHCH2OH
Cl
HO C
O
THEY LOOK DIFFERENT BUT ARE THEY???
(A) (B)
2‐chloro‐3‐hydroxy‐propanoic acid
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85
A)
CH
CH2OH
ClC OH
O
1. Assign priorities to each group:
Cl > Calc = Ccarb > H
O,O,O > O,H,H
Go to next atoms on tied carbons
Ccarbox > Calc
Cl > Ccarb > Calc > H
86
A)
CH
CH2OH
ClC OH
O1
3
24
1. Assign priorities to each group:
Cl > Ccarb > Calc > H
86
ClC
1
2
3 CH2OH
C OH
O
The priorities decrease anti‐clockwise, so this centre is “S”
(S)‐2‐chloro‐3‐hydroxy‐propanoic acid
2. Redraw the molecule with the
lowest priority group facing in.
3. Look which way the priorities decrease.
1. Assign priorities to each group
Cl > Ccarb > Calc > H CHCH2OH
Cl
HO C
O
B)
2
1
4
3
2. Redraw the molecule with the
lowest priority group facing in.
C
32
Cl
CH2OHHO C
O
1
3. Direction of the priorities
decrease?
The priorities decrease anti‐clockwise, so this centre is “S”
(S)‐2‐chloro‐3‐hydroxy‐propanoic acid 88
Example: Give the configuration of the stereogenic
centre in each of the following molecules:
THEY LOOK DIFFERENT BUT ARE THEY???
2‐chloro‐3‐hydroxy‐propanoic acid
WE JUST FOUND OUT THAT THEY ARE IDENTICAL!!!!!
WHAT IF?????
CHCH2OH
Cl
HO C
O
CH
CH2OH
ClC OH
O
B)A)
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23
If we flip the COOH and H ???????
CH
CH2OH
ClC OH
O
A)
C H
CH2OH
Cl
C
O
HO
A’)
ClC
1
2
3 CH2OH
C OH
O
C
2
1
3 CH2OH
C
O
HO Cl
“S” configuration “R” configuration89 90
What is the structural formula of (R)‐2‐chloro‐2‐butanol?Hint: Make use of perspective formula or dash‐wedge notation.
DRAW MOLECULAR STRUCTURE
CH2CH3C
OH
CH3
Cl
ASSIGN PRIORITIES
1
3
2
4
EXAM QUESTION
Solution:
91
(R)‐2‐chloro‐2‐butanol
CH2CH3C
OH
CH3
Cl
DRAW MOLECULE…..
1
3
2
4
Draw steering wheel with lowest priority group pointing in…….
MAKE SURE PRIORITY GOES
CLOCKWISE FOR “R” configuration
Cl CH2CH3
OH
1
2
3
C
Assign prioritiesDRAW MOLECULE TO SEE
ALL GROUPS…..
Make sure priority goes clockwise for “R”
Cl
OH
H3CCH2CH3
C
Cl CH2CH3
OH
1
2
3
C
(R)‐2‐chloro‐2‐butanol
Remember lowest priority group has a “dashed” bond
92
Perspective formulaor
Dash‐wedge notation
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enantiomeric pairs enantiomeric pairs
[2R,3R] [2S,3S] [2R,3S] [2S,3R]
• Enantiomeric pair differ only in optical activity
a b c d
R
R
C
C
CH3
CH3
H
Cl
Br
H
S
S
C
C
CH3
CH3
Br
H
H
Cl
R
S
C
C
CH3
CH3
H
H
Br
ClR
SC
C
CH3
CH3
Br
Cl
H
H
CHCH3 CH
Br Cl
CH3* *
93Chapter 22.2, page 838‐9: “Chemistry: The Central Science: A Broad Perspective”
R
S
C
C
CH3
CH3
H
H
Br
ClR
SC
C
CH3
CH3
Br
Cl
H
H
R
R
C
C
CH3
CH3
H
Cl
Br
H
S
S
C
C
CH3
CH3
Br
H
H
Cl
a b c d
Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp., , solubility etc…
Diastereomers b & ca & c
b & da & d
molecule with n stereogenic centres may exist inmaximum of 2n stereisomeric forms, with maximum of2n/2 enantiomeric pairs 94
meso compound ‐ an achiral (optically inactive)
diastereomer of compound with stereogenic centres
arises because 4 different groups making each of C‐2 & C‐3
stereogenic are same 4 different groups…(!)
Meso compoundCHHOC CH COH
OH OH
OO* *
R
R
C
C
CO2H
CO2H
OH
H
H
HO
S
S
C
C
CO2H
CO2H
H
OH
HO
H
R
S
C
C
CO2H
CO2H
OH
OH
H
HR
SC
C
CO2H
CO2H
H
H
HO
HO
170 °C 170 °C 140 °C
+12° ‐12° 0°
95
Meso compounds
Enantiomeric pairs
possess plane of symmetry bisecting central C‐C bond
Meso compound
R
R
C
C
CO2H
CO2H
OH
H
H
HO
S
S
C
C
CO2H
CO2H
H
OH
HO
H
R
S
C
C
CO2H
CO2H
OH
OH
H
HR
SC
C
CO2H
CO2H
H
H
HO
HO-------------------- -------------------- -------------------- --------------------
Enantiomers, Chiral Identical, achiral, Meso form
96
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25
Shows arrangements in space
Newman projections
e.g. Ethane C2H6
H
H
H
HH
H
“dash‐wedge”
H H
HHH H
“dash‐wedge”
HH
HH
HH
H
HHH
H H
“sawhorse”
“sawhorse”
60°
0°
Newman
Newman
Staggeredconformation
Eclipsed conformation
H
H
H
H
H
H
HH
HHH
H
97
a
e
e
ae
a
aa
e
e
a
e
"flip"16
5
43
2
a
e
e
a e
a
aa
e
e
a
e
16
5
43
2
H
H H
HH
H
HH
H
HH
H
Chair conformations
Boat conformation
Cyclohexane conformations
Cyclohexane
98
Examples1,2‐Dimethylcyclopentane
CH3 CH3
H H
H
H
H
H
H H
CH3
CH3H
H H
H
H
H
H H
Cis‐1,2‐Dimethylcyclopentane Trans‐1,2‐Dimethylcyclopentane
99
Different bond pattern
isomers
Structural (constitutional)
isomer
Stereoisomer
Interconvertible by single bond rotation
Not interconvertibleby bond rotation
Same bond pattern
Conformers (rotamers)
Configurational isomers
Summary of isomerism
Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54
100
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26
Exercise
1. Draw the structures of
a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐heptadiene
b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐bromobutan‐2‐ol
2. Using the Newman projection draw the structure of a
staggered conformation of butane.
3. Using the Fischer projection draw a structure of
(S)‐2‐methylpentanoic acid.
4. Draw the structure for the cis‐ and trans‐isomers of
1‐bromo‐2‐chlorocyclopropane 101
Solutions
1. (a) (Z)‐3‐methyl‐2‐pentene
CH3
H2C CH3
H3C
H
(b) (S)‐2‐bromopropanol
CC
CH3
Br
OHH
H
H
C
C
CH3
Br
H3C
H
HHO
(c) (E,Z)‐2,4‐heptadiene (d) (2S,3R)‐3‐bromobutan‐2‐ol
102
CH 2C H2CH 3
COOH
H
H 3C
CH3
Staggered conformation of butane
Solutions
2.
60°Newman projectionH
H
CH3
H
H
Fischer projection
3.
*
(S)‐2‐methylpentanoic acid
103
(4) 1‐bromo‐2‐chlorocyclopropane
H
H
Br
H
Cl
H
H
H
Br
H
H
Cl
Cis‐1‐bromo‐2‐chlorocyclopropane
Trans‐1‐bromo‐2‐chlorocyclopropane
104
http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx
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27
Substitution reaction: Substitution of one atom or group of
atoms by another atom or group of atoms.
Addition reaction: Addition of one molecule or atom to another
to give a new molecule . normally on double or triple bond.
Elimination reaction: Elimination of two atoms or group of atoms
from a molecule (reverse of the addition reaction).
Oxidation and reduction reactions: Involve the loss or gain of
electron density by a carbon respectively.
Rearrangement reaction: Conversion of one structure to an
isomeric structure.
Five important organic reactions
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Uses
shows redistribution of valence electrons a molecule
defining resonance
chemical reaction direction
defining a reaction mechanism.
Two common types of electron redistribution:
From a bond to an adjacent atom, and
From an atom to an adjacent bond.
Arrow Notation
Homolytic bond breaking Heterolytic bond breaking
Bond forming
A B BA + A B B-A+ +
Homolytic bond forming Heterolytic bond forming
BA + A B
Bond breaking
A BB-A+ +
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Terminology
Nucleophile (Nu):- electron rich species and/or negatively
charged species
examples
-OH, Br-, -NH2,
Electrophile (E+):- electron poor species and/or positively
charged species
examples
H+, M+, H3O+
O
C
O
O
CC C
10/4/2011
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PowerPoint to accompany
Chapter 21
Alkanes: Scaffolds for
Organic Chemistry
Reactions of Alkanes
Because of its structural role, alkanes are relatively
unreactive.
They make excellent nonpolar solvents.
Forcing conditions are required for a reaction
Combustion: give mainly H2O and CO2
Free-radical reactions: chain reactions
In order to be able to predict the product of a reaction, it is
important to be able to distinguish between carbon (and
hydrogen) types in an alkane.
Primary, secondary, tertiary and quaternary carbon atoms.
Primary, secondary and tertiary hydrogens.
Figure 21.25
Classification
hydrogen and carbon atoms in alkanes
Combustion
Combustion of alkane gives CO2 and H2O.
Combustion reactions are exothermic.
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) ∆H = -2855 kJ
e.g. ethane
Write a balance equation for the combustion of butane
Question
Solution
C4H10(g) + O2(g) CO2(g) H2O(l)+6½ 4 5
2C4H10(g) + O2(g) CO2(g) H2O(l)13 8 10+
Homework Try combustion of pentane C5H12 ??
10/4/2011
29
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Combustion - Napalm
The commonly quoted composition is 21% benzene,
33% gasoline, and 46% polystyrene. This mixture is
difficult to ignite.
Movie – Vietnam.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Free-Radical Reactions
Consider the following halogenation reactions:
Reaction 1
Reaction 2
CH4 + Cl2 no reaction
CH4 + Cl2 CH3Cl + CH2Cl2 + CHCl3 + CCl4
dark
Light
λ
• This type of reaction is called a substitution reaction.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
The free-radical halogenation is also an example
of a radical chain reaction.
Such reactions are characterised by three steps
Initiation
Propagation
Termination
Initiation produces the first free-radicals by a
homolytic bond cleavage.
Cl Cl λ
Cl2
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Propagation steps - a series of new reactions
which generate other radicals
e.g.
radical
radical
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Termination steps: stop the formation of free-radicals
neutral molecular species formed
e.g.
radical pair
radical pair
Free-Radical Reactions
Given the opportunity, free-radical reactions are regiospecific.
More so for bromination than chlorination.
3°> 2° > 1°most stable least stable
R
C R = akly group - EDG
1°> 2° > 3°most stable least stable
X
C e.g X = halide - EWG
PowerPoint to accompany
Chapter 23
Alkenes and Alkynes
Electrophilic addition and Substitution reactions
The presence of carbon-carbon double or triple bonds in a
compound markedly increases its chemical reactivity.
Most characteristic reactions of alkene and alkynes are addition
reactions:
Halogenation
Hydrohalogenation
Hydration
The electron-rich nature of a multiple bond leads to enhanced
reactivity with electrophiles (electron-loving species).
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Alkenesreactions
Figure 23.20
Mechanism for HX Addition rxn
Two-steps mechanism
1. First step is slow, rate-determining step.
2. Second step is fast.
slow
H Br
Fast
carbocation
carbocation
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Addition of HX
In the first step, -bond breaks and new C—H bond
and cation form.
Figure 23.22
In the second step, a new bond forms between negative
bromide ion and positive carbon. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Regioselectivity
A occurs when a non-symmetric alkene is used.
The preference based on intermediate carbocation stability:
Primary < Secondary < Tertiary
(Less Stable) (Most Stable)
The regioselectivity is stated as Markovnikov’s Rule:
“In the addition of HX to an alkene, the hydrogen adds to the
carbon atom of the double bond bearing the greater number of
hydrogen atoms bonded directly to it.”
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Addition of HX
H Br
C C
H
H CH3
CH31 2
Major product
Minor product
Exercise
H 3CC
C H 3
C H 3
CH 2
C H 3
C H 2
+
+
+
+
(A ) (B )
(C ) (D )
1. Which of the following is the most stable carbocation and why?
Soln:
• It is a tertiary (3˚)
carbocation.
• Inductive effect from
electron rich groups
2. What will be the major product from the reaction by reflux in dark
of 2-methylbut-2-ene with hydrochloride in a carbon tetrachloride
solution? Name and draw the structure of the product.
(D)
C C
H3C CH3
H3C H
2-methylbut-2-ene
HCl
CCl4
C C
CH3 CH3
CH3 H
ClH (A)
C C
CH3 CH3
CH3 H
HCl (B)
Soln:
2-chloro-3-methylbutane
2-chloro-2-methylbutane
minor product
major product
Addition of H2O
The hydration of alkenes and alkynes requires the addition of a
strong acid catalyst e.g. H2SO4
The first step is rate determining.
Leads to the formation of a carbocation.
hydronium ion(electrophile)
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Second step is fast
Carbocation reaction with excess H2O (nucleophile)
leads to the formation of an oxonium ion.
Loss of H+ from the oxonium ion leads to formation of an alcohol
and regenerates the hydronium ion, hence it is catalytic reaction.
butan-2-ol2-butanol
hydronium ionregenerated
oxonium ion
Addition of H2O (Reversible)
The fact that the reaction is always described by equilibria
means that the overall reaction is reversible.
That is, in the presence of H2SO4, alcohols can be dehydrated to
yield alkenes.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Formation of Alkenes
Previously, we have discussed the formation of alcohols from alkenes.
The reverse process is also possible.
This dehydration reaction is also known as an elimination reaction.
Requires the elimination of a good leaving group.
I- > H2O > Br- > Cl- >> F- > CH3COO- > OH-
Formation of Alkenes
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Exercise
1. Draw and name the major product in the following reactions
(a)
(b)
H
OH
H+ / H2O
pentan-2-ol2-pentanol
OH H2SO4
Heatcyclohexene
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Halogenation
One of the easiest and most
dramatic reactions is the
addition of Br2 to an alkene.
Colour change
Useful test for alkenes
Figure 23.23
Reactions of Alkynes
Many similar reactions as alkenes.
Reactions involve the replacement of -bonds with -bonds.
C CH3C CH3
Example
Br2
CH2Cl2
C C
H3C
CH3Br
Br
Br2
CH2Cl2C C
CH3
CH3
Br Br
Br
Br
(E)-2,3-dibromobut-2-ene
2,2,3,3-tetrabromobutane
Problems1. What will be the product for a reaction between hex-2-yne and one
mole of chlorine in dichloromethane under anhydrous condition? Draw
the structure of the product and provide it IUPAC name.
2. How would you form 1,1,2,2-tetrabromo-1,2-difluoroethane using
bromine as one of your reagent?
Solutions
C C
F
F
Br Br
Br
Br
C C
F
FBr
BrBr2
CH2Cl2 C CF FBr2
CH2Cl2
1,2-difluoroethyne(E)-1,2-dibromo-1,2-difluoroethene
1,1,2,2-tetrabromo-1,2-difluoroethane
H3C C C CH2
H2C CH3(1)
(E)-2,3-dichlorohex-2-ene
Cl2
CH2Cl2
(2)
H3C
C C
H2C CH2
CH3
Cl
Cl
hex-2-yne
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Catalytic Hydrogenation
Reaction requires molecular hydrogen (H2) and a metal
catalyst
Ni, Pd, Pt, Ru typical metals used
Because a gas, liquid and solid metal are used, this
reaction is known as a heterogeneous catalysis
Problems
H2
Pd/C
3-ethylpent-1-ene
(1)
3-ethylpentane
(2)
(3)
Br2
CH2Cl2
Br2
H2O
Br
Br
1,2-dibromo-3-ethylpentane
HO
Br
1-bromo-3-ethylpentan-2-ol
Provide the major organic product(s) in the reactions below
3-ethylpent-1-ene
3-ethylpent-1-ene
PowerPoint to accompany
Chapter 24
Alcohols, Ethers and Haloalkanes
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Conventions in Organic Chemistry
Figure 24.1
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36
Alcohol Properties
The difference in electronegativity between oxygen and
hydrogen allows alcohols to hydrogen bond. Compared to
alkanes they:
Are more soluble in water
Have higher boiling and melting points
Figure 24.4
Alcohol Solubility
There are differences in solubility:
Like dissolves like Table 24.2
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ether Properties
Ethers are more polar than alkanes but are unable to hydrogen bond.
They are chemically inert.
Both these properties mean that ethers are excellent solvents!
Ether Nomenclature
Figure 24.10
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Reactions of Alcohols
Figure 24.12
Fill in the reagents andname type of reaction !!Fill in the reagents andname type of reaction !!
Homework
Reactions of Alcohols
Alcohols react with sodium metal to form
alkoxides and H2(g).
Alkoxides are excellent nucleophiles and
strong bases.
Figure 24.13
2 CH3CH2O H + 2 Na 2 CH3CH2O-Na+ + H2(g)
Sodium ethoxide
2 CH3CH2O H + 2 Na 2 CH3CH2O-Na+ + H2(g)
Sodium ethoxide
Basicity of Alcohols
An alcohol can act as a weak base, generating an oxonium ion
under acidic conditions, which then loses water to form
carbocation intermediate.
The stability of the carbocation is an enormous driving force for
the nature of the reactivity of alcohols.
CH3CH2 H + H+O CH3CH2 O
H
H
+
CH3CH2+ + H2O
carbocation
oxonium ion
10/4/2011
38
Alcohols to Haloalkanes
Haloalkanes can be prepared from alcohols.
Haloalkanes are important for introducing alkyl groups in
chemical reactions.
Two reactions are common
Depend on the type of alcohol
Dehydration of Alcohols Dehydration of alcohols is an acid catalysed process. The rate-
determining step is the elimination of water to form the corresponding
carbocation. Removal of a small group from a larger molecule is called
elimination reaction.
Water is classed as a good leaving group because of its stability and
relatively unreactive nature. Hydroxide ion, on the other hand, does not
eliminate easily and so is classed as a poor leaving group.
Elimination reactionoxonium ion
Order of leaving group:
I- > H2O > Br- > Cl- >> F- >
CH3COO- > OH- > NH2-
Nucleophilic Substitution (Haloalkanes)
Nucleophilic substitution reactions of haloalkanes proceed by attack of the
nucleophile at the carbon bearing the halogen.
Because of the difference in electronegativity between carbon and any of
the halides, this bond is polarised such that the carbon becomes mildly
electrophilic.
This is a SN2 or biomolecular (2), nucleophilic (N), substitution (S)
reaction.
Figure 24.15
Transition state
SN2Mechanism
Example
1. Show the reaction mechanism between sodium hydroxide
and 1-bromoethane.
CCH3
BrH
H
NaOH
Soln
Transition state1-bromoethane. ethanol
mechanism
+
10/4/2011
39
2. Use reaction mechanisms to explain the why the inversion of the
carbon reaction centre is observed when (S)-2-bromobutane is
reacted with NaOH.
C
H3C Br
H
CH2CH3
Soln
HO-
HO C Br
CH2CH3
CH3
_
(S)-2-bromobutan
C
CH3
H
CH2CH3
HO
(R)-butan-2-ol
Inversion:- the (S)-configuration now becomes (R)-configuration.
Transition state
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
University of KwaZulu-Natal, Westville Campus, Durban
Examinations : November 2010
CHEMICAL REACTIVITY - CHEM120
Duration: 3 hours Total Marks for Examination: 100
Internal Examiners: V.O. Nyamori, H.G. Kruger, M.D. Bala, B.S. Martincigh,
P.G. Ndungu and G.D. Dawson
QUESTION 6Primary organic halides react with nucleophiles through a SN2mechanism. In the mechanism inversion of the carbon reaction centre isobserved. Provide the mechanism of the SN2 mechanism below.
(4)
C
H
H2CH3C
H
BrHO
SN2
(a) (b) (c)
starting material transition state product(s)
Four Factors affecting the SN2 reaction rate
Substrate
The nucleophile attacks from the back of the substrate and hence to
maximise the rate of the SN2 reaction, the back of the substrate must
be as unhindered as possible.
Order of reactivity: Primary > secondary > tertiary substrates
(Most reactive) (not reactive)
Nucleophile
Steric hindrance affects the nucleophile's strength e.g. methoxide
anion is better nucleophile compared to tert-butoxide.
Nucleophile strength is also affected by charge and electronegativity:
nucleophilicity increases with increasing negative charge and
decreasing electronegativity, e.g. nucleophile OH- > H2O and I- > Br-
(in polar protic solvents).
Solvent.
Solvents may or may not surround a nucleophile, thus hindering or
not hindering its approach to the carbon atom.
Polar aprotic solvents because polar protic solvents will be solvated
by the solvent hydrogen bonding to the nucleophile and thus
hindering it from attacking the carbon with the leaving group.
Leaving group
The more stable the leaving group is, the more likely that it will take
the two electrons of its carbon-leaving group bond with it when the
nucleophile attacks the carbon.
Therefore, the weaker the leaving group is as a conjugate base,
and thus the stronger its corresponding acid, the better the leaving
group. Examples of good leaving groups are therefore the halides
(except fluoride) and tosylate, whereas HO- and H2N- are not
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University of KwaZulu-Natal, Westville Campus, DurbanExaminations : November 2010
CHEM120 - CHEMICAL REACTIVITYSection B
QUESTION 6Primary organic halides react with nucleophiles through a SN2 mechanism.In the mechanism inversion of the carbon reaction centre is observed.Provide the mechanism of the SN2 reaction below. (4)
C
H
H2CH3C
H
BrHO
SN2
(a) (b) (c)
starting material transition state product(s)
Soln
Example
Nucleophilic Substitution (Haloalkanes)
Table 24.4
SN1 reaction mechanism
C R2
LG
R1
R3Nu + CR2
R1
R3
Nu
CNu
R2
R1
R3C
Nu
R2
R1
R3
and
Racemic mixture: Both (R) and (S)-configurationenantiomeric mixtures
carbocation
or
This allows two different avenues forthe nucleophilic attack, one on eitherside of the planar molecule.
If the reaction takes place at astereocenter yielding a racemicmix of enantiomers.
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Example
Step 1
Step 2
Step 3
Separation of the leaving group
Nucleophilic attack
Deprotonation
Carbocation
oxonium ion
Product : 2-methylpropan-2-ol
The SN1 mechanism tends to dominate when the central carbon
atom is surrounded by bulky groups. SN2 reaction hindered by steric
factor.
Bulky substituents on the central carbon favour carbocation
formation because of the relief of steric strain that occurs.
The resultant carbocation is also stabilized by both inductive
stabilization and hyperconjugation from attached alkyl groups.
The SN1 mechanism dominates in reactions at tertiary alkyl centers
and is further observed at secondary alkyl centers in the presence
of weak nucleophiles.
The carbocation intermediate formed in the reaction's rate limiting
step is an sp2 hybridized carbon with trigonal planar molecular
geometry.
Scope of the reaction
Haloalkanes to Alkenes: -Elimination
The process of elimination of a good leaving group.
Haloalkanes undergo elimination under basic conditions. This
process is called -elimination reaction. It involves the
dehydrohalogenation of the haloalkane.
For the reaction to proceed, a hydrogen atom on the carbon atom
adjacent to the carbon bearing the halide must exist and be
removed.
The -elimination reaction is a E2 or biomolecular (2) elimination
(E) reaction.
When there are two sets of -hydrogens present in the
haloalkane, there is only one product formed and it is determined
by the Zaitsev’s Rule:
The major alkene formed by dehydrohalogenation is the more
substituted one.
majorminor
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Figure 24.16
monosubstituted
disubstituteddisubstituted disubstituted
trisubstituted
tetrasubstituted
ethene
C C
H
H
H
X
HH
X = Cl, Br, I
C2H5O-Na+
C2H5O-Na+
C2H5O-Na+
C2H5O-Na+
C C
H
H
H
X
RR
C C
R
H
R
X
HR
C C
R
H
R
X
RR
C C
R
H
H
X
HR
HX +
HX +
HX +
HX +
C C
H
H
H
X
HR
C2H5O-Na+Increasing degree of
substitution
+HX
Substitution vs Elimination
Both Substitution reactions (SN1 & SN2) and elimination are important,
however, each of these reactions does not work in isolation.
E.g. the addition of strong base to an alkyl halide yields an alkene:
This elimination reaction competes with a substitution reaction which,
in this case, forms ethers:
Substitution vs Elimination
Mechanism shows a trigonal planar intermediate formed by
concerted bond breaking and bond making processes
There are usually stereochemical consequences
Elimination vs Substitution
Thus, there is a competition between elimination and substitution reactions and the result is dependant on:
Size of the alkoxide
The accessibility of the carbon bearing the halide
The accessibility of the -hydrogen
Figure 24.17
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Elimination vs Substitution
Figure 24.19
We looked at the
hydration of alkenes
The reverse reaction
is also possible!
The presence of other
nucleophiles add another
dimension to the reaction.
PowerPoint to accompany
Chapter 25
Aldehydes and Ketones
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Carbonyl Compounds
Contain C=O double bond.
Include many classes of compounds:
aldehydes, ketones, carboxylic acids, esters, amides.
C
O
CR H
O
CR R'
O
CR O
O
HC
R O
O
R'
CR N
O
R'
R''Aldehyde Ketone Carboxylic acid Ester Amide
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Aldehydes
Characterised by the functional group CHO.
Contains a carbonyl group Both carbon and oxygen are sp2 hybridised.
Figure 25.1
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Aldehyde Nomenclature Name the longest chain containing the carbonyl group
and replace ane with al.
Must, by definition, occur at one of the ends of a
molecule. Remember that aldehyde group takes priority
over both methyl and hydroxyl groups.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ketones
Name the longest chain containing the carbonyl
group and replace ane with one.
Carbonyl group is bonded to alkyl groups on either
side.
O
O
H
H
H
progesterone
O
camphor
O
OH
H
H
H
testosterone
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Ketone Nomenclature
Longest chain containing the carbonyl group.
Replace ane with one.
Constitutional isomers occur for C5 or greater.
Preparation of Aldehydes and Ketones
Oxidation of alcohols using a range of reagents
• Most are chromium(VI) compounds (CrO3).
Again, different types of alcohols yield different products
• Primary alcohols can lead to aldehydes and carboxylic acids.
• Secondary alcohols lead to ketones.
• Tertiary alcohols do not oxidise.
Figure 25.2
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Preparation of Aldehydes and Ketones Preparation of Aldehydes and Ketones
Using pyridinium chlorochromate (PCC)
Preparation of Aldehydes and Ketones
Secondary alcohols can be oxidised by PCC, Jones
Reagent (CrO3) or K2Cr2O7 to yield the corresponding
ketone.
Aldehydes and ketones can also be prepared in good yields by the
reaction of alkenes with ozone, followed by a mild reduction using
dimethyl sulfide. This reaction is called ozonolysis
ozonolysis
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Reactions Aldehydes and Ketones
In general
Reduction reaction
Reduction Reactions
Aldehydes and ketones can be reduced to give alcohols
Catalytic hydrogenation is one way
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Examples
H
O
O
O
O
H
H2
H2
H2
H2
Pd/C or Pt
Pd/C or Pt
Pd/C or Pt
Pd/C or Pt
(A)
(C)
(B)
(D)
H
OH
OH
OH
OH
H
QuestionProvide the structure of the reaction products from the followingequations.
Reduction Reactions
Hydrides:
Figure 25.5
NaBH4 LiAlH4
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
ExamplesQuestion
Provide the structure of the reaction products from the followingequations.
H
O
O
O
O
H
LiAlH4(A)
(C)
(B)
(D)
LiAlH4
LiAlH4
LiAlH4
H
OH
OH
OH
OH
H
Tautomerism in Aldehydes and Ketones
Aldehydes and ketones bearing -hydrogen are able to undergo
tautomerism to form enols (i.e. compounds bearing an alkene and an
alcohol). Enols are constitutional isomers of their respective keto form
(i.e. aldehyde or ketone).
Tautomerism is the process by which two isomers are interconverted by
a formal movement of an atom or group bearing in mind valency rules.
Hence, an enol and its keto form are tautomers of the same structure.
Aldehydes and ketones are in equilibrium with their enol forms and this
equilibrium is acid catalysed. The process involves two separate proton
transfers:
The extent of tautomerisation is dependent on the aldehyde or ketone:
Halogenation of Aldehydes and Ketones
Aldehydes and ketones undergo a substitution reaction at the α-carbon in the presence of halogen in good yield.
This reaction is regiospecific in that only the α-hydrogen issubstituted under these conditions. The rate determining step inthis reaction is the conversion of the ketone to its enol isomer,e.g.
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PowerPoint to accompany
Chapter 26
Carboxylic Acids and their
Derivatives
Carboxylic Acids Have hydroxyl group bonded to carbonyl group.
Carboxylic acids are weak acids.
Figure 26.3
Carboxylic Acids
Figure 26.4
Figure 26.5
Carboxylic Acids - Acidity
Table 26.2
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Question
Which of the following organic acids is expected to be the
strongest and why? Provide an explanation
O
CC O H
ethanoic acid
H
H
H
O
CC O H
F
F
F
O
CC O H
CH3
H3C
CH3
2,2,2-trifluoroacetic acid
2,2,2-trifluoroethanoic acid
acetic acid pivalic acid
2,2-dimethylpropanoic acid
Soln
(A) (B) (C)
(B)
Inductive effect from the substituents on the α-carbon.
F atom is the most electronegative, hence, stabilizes thecarboxylate anion generated
O
CC O-
F
F
F
+ H+carboxylate anion
Carboxylic Acids - Acidity
Stronger acid than alcohols because of resonance stabilisation.
pKa = 4.8
Figure 26.7
Carboxylic Acids - Reactions with Bases
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
ExamplesQuestion
Provide the structure of the reaction products from the followingequations.
HO
O
O
OH
NaOH(A)
(B)
NaOH
+Na-O
O
O
O-Na+
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Esters and Esterification
Products of reaction between carboxylic acids and alcohols.
Found in many fruits, perfumes and many other everyday
products:
Figure 26.9
O
CR O
+ HO R'H
O
CR O R'
+ OHH
Esters
Table 26.3
Examples
A
Ester Hydrolysis
Saponification = hydrolysis of ester using aqueous base.
Figure 26.12
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Carboxylic Acid Derivatives
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Acid Chlorides
Formed by treatment of a carboxylic acid with thionyl chloride.
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Acid Anhydrides
Formed by the reaction of an acid chloride and carboxylic acid or by dehydration.
Questions
(a) Provide the structures for the reaction products from the followingequations.
(b) Provide appropriate starting compounds or reagents used to form theproducts in the following equations.
CH2
CH3
O
Cl
C CH2
CH3O
H3CC
O
O
SOCl3
? ?
O
OH
+ HO
O
O
HO
C CH2
CH3OH3C
C
O
OH
(i)
(ii)
O
O F Dilute NaOH
H2OHO F
O
OH +
PowerPoint to accompany
Chapter 28
Nitrogen-Containing
Compounds of Biological
Relevance
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Biological Chemistry
Research at the interface of biology and
chemistry.
The principles of chemistry are applied to
understand the molecular basis of biological
processes.
Figure 28.1
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Biological Chemistry
We will investigate: Amines and amides
Amino acids
Peptides, proteins and enzymes
Vitamins
DNA
Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Amines
General formula:
Can be aliphatic, heterocyclic or aromatic:• Next slide shows a number of simple
amines.
• Classified as 1o, 2o, 3o
• Quaternary (4o) ammonium salts are also likely.
Figure 28.3
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Figure 28.4
Simple amines are named using the same guidelines for alcohols.
In many instances, the amino group is named as a substituent.
Amine Nomenclature
Amine Nomenclature Reactivity of Amines
Amines react with acids to form ammonium salts.
Amines react with alkyl halides to form ammonium salts.
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Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia
Amides
Amides are derivatives of carboxylic acids.
They are classified as 1o, 2o and 3o
O
CH3C N
CH3
H
O
CH3C N
H
H
O
CH3C N
CH3
CH3
acetamide N-methylacetamide N,N-dimethylacetamide
2°1° 3°
Formation of Amides
Amides are formed by the reaction of amines with “activated”
carboxylic acids.
C
O
H3C Cl NCH2CH3
H
H
+ C
O
H3C N
H
CH2CH3
+ HCl
CHEM 120 CHEMICAL REACTIVITY
ORGANIC CHEMISTRY
19 Lectures5 Tutorials
2 Quiz1 Test
Dr. Vincent O. Nyamori
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