chapter 4 study guide.pdf

Chapter 4

Definitions: Alkanes Alkenes Alkynes Cycloalkanes Cracking(thermal and catalytic) Polymerization Constitutional Isomers Haloalkanes Alcohols Cyclics Bicyclo Homologs/homologous series Solubility Acetylenic hydrogen/alkynide ion/acetylide ion Conformations/Conformational Analysis Newman Projections Eclipsed/Staggered/Gauche/Anti Chirality Stereochemistry Ring Strain Angle/Torsional Strain Chair/Boat Catenanes Pheromones Paraffins Alkylation Retrosynthesis

1. Oil Refining/Fractional Distillation Petroleum

source of alkanes

types of alkanes Petroleum refining fractional distillation

catalytic/thermal crackinGg polymerization

1

2. Alkane structure molecular formula/structural formula/condensed formula

line formula(zig-zag) geometry

alkyl substituents isopropyl isobutyl sec-butyl tert-butyl neo-pentyl 3. Nomenclature

A. Alkanes--chart p. 136(table 4.4)

B. IUPAC rules/common names 1. Locate longest continuous chain. Name it according to p. 136. a. if 2 chains have equal length. The chain with most substituents becomes the parent chain. 2. Number the chain a. if only one substituent, number to get to that substituent first b. if two substituents, number from side that first gets to a substituent c. if both substituents are equal distant from both sides and both are alkyl, use alphabet to determine the numbering d. if both substituents are equal go to next carbon with a substituent. Number from side that gives smaller total. I.E. 2,4 is better than 2,5 e. halides have same priority as alkyls. Use alphabet to determine priority.

2

f. alcohols have highest priority. If an alcohol is present number the chain to get to the alcohol first regarldless of alkyl and halide groups. Alcohols also take priority over alkenes and alkyenes. g. ethers have same priority as alkyls. Use alphabet to determine priority h. alkenes and alkynes are used to determine priority if there are no alcohols present in a molecule 3. Give each substituent a number and a name(3-bromo, 2-methyl) a. if two groups are on the same carbon it must be numbered twice i.e 2,2-dimethyl

4. Common substituents can be grouped together(2-methyl, 3-methyl can be called 2,3-dimethyl). 5. Numbers are separated by commas. Numbers and letters are separated by dashes.

6. Prefixes a. alkyl substituents are named alphabetically in the front(di, tri, sec,do not count for alphabetical, dimethyl comes after ethyl) b. Halides are named alphabetically in the front 1. Iiodo 2. Brbromo 3. Clchloro 4. FFluoro c. ethers are named alphabetically in the front 1. methyl ethermethoxy 2. ethyl etherethoxy 3. propyl etherpropoxyand so on

3

d. alkenyl side chains are named alphabetically in front 1. vinyl 2. allyl e. stereochemical designations come in front as prefixes 1. alkenescis/trans/E/Z 2. alkanesR/S

7. Suffixes a. alcohols are named by dropping e and adding ol b. alkenes are named by changing ending to ene. the location of the first carbon is placed in front of alkane name 1. butane to butene propane to propene if double bond is at 1st carbon1-octene

if double bond is at 2nd carbon2-octene if double bond is at 3rd carbon3-octene

c. if two double bondsdiene, three double bondstriene 1. 8 carbon unit with double bonds at 1st and 5th would be 1,5-octadiene 2. 9 carbon unit with double bonds at 1st, 4th and 7th

1,4,7-octatriene d. alkynes use the same rules as alkenes but change ending to yne. 8. Cyclic Alkanes/Alkenes a. use all previous rules add cyclo to root name(i.e octane becomes cyclooctane; 2-hexene becomes 2-cyclohexene) b. if a double bond is present it will be numbered 1 and 2. Numbering will precede to get to the substituent with lowest number.

4

1. if there is more than one substituent present use rules for alkanes(I.E. 2,2,3 is better than 2,2,5. if two equal substituents can be reached use alphabetical to determine which direction to precede. 2. if an alcohol is present it will be the 1st carbon and then precede to the next closest substituent in either clockwise or anticlockwise manner.

9. Other alkyl/alkene rules

a. when there is a cyclic chain attached to an alkyl group, the larger group becomes the base name. 1. cyclohexane ring with a pentyl group will be called 1-pentylcyclohexane 2. cyclopentyl ring with a hexane group will be called 1-cyclopentylhexane b. if an alkyl substituent is complex, use nomenclature rules to name it as its own molecule then change ending to yl. 1. 2,3-dimethy-4-propyloctane becomes 2,3-dimethyl- 4-propyloctyl 10. Bicyclics a. the total number of carbons in the ring becomes the parent name with the word bicyclo in front. b. the carbons where the rings are fused are called bridgehead carbons. c. the carbons between the bridgeheads are called bridges d. between bicyclo and the parent name is inserted in brackets the number of carbons of each bridge in order of decreasing length(bicyclo[3.2.1]octane). Periods are used to separate the numbers.

5

Br

OH

3-methyloctane

4-methyldecane

5-propylundecane

2,8-dimethyldecane

2-bromo-9-methyldecane

3,4,8-trimethyldecane

7,9-dimethyldec-4-ene

dec-8-en-3-ol or 8-decen-3-ol

O OHCl Br

6-bromo-10-chloro-8-ethyl-4-methoxydecan-2-ol

6

4-bromocyclohex-2-enol

OH

Br

4-bromo-2-ethyl-1-methylcyclohexane

Br

OH

2,5,7-trimethylcycloocta-1,3,5,7-tetraenol

OH

2-methylbicyclo[2.2.1]heptan-7-ol

2-methylbicyclo[2.2.1]hept-2-ene

bicyclo[3.3.2]decane

bicyclo[2.2.2]octane

7

More information on bicyclics... On bicyclics, the longest chain is always numbered 1st then the next longest... In naming bicyclics, the name is bicyclo[x.y.z]alkane, for example. Number to get to the substituent quickest in the longest chain. Substituents in shorter chains DO NOT take precedence over any substituent in a longer chain.

4

2

11

5

6

1

10

7

9

3

8

4

2

11

5

6

1

10

7

9

3

8

3

5

10

2

1

6

11

9

7

4

8Br

Br

3

5

10

2

1

6

11

9

7

4

8

Br

OH 11. Common names a. alkyl halides are often named by a common name(pentyl chloride instead of 1-chloropentane). b. alcohols are often named by common name(ethyl alcohol instead of 1-ethanol). c. other common names include acetic acid, formic acid, ethyl alcohol, ethylene glycol, glycerol, isopropyl alcohol, acetylene, dichloromethane, chloroform 12. Hydrogens a. Carbon with 3 Hsprimary hydrogens/carbon b. Carbon with 2 Hssecondary hydrogens/carbon c. Carbon with 1 Htertiary hydrogen/carbon d. Carbon with 0 Hquarternary carbon

8

RULES FOR NOMENCLATURE

Select the longest chain and name it methane, ethane,. If cyclic add prefix cyclo. If contains double bond change ending from -ane to ene. If there is an alcohol, number the chain to get to the alcohol first. If there is no alcohol but there is a triple bond, number the chain to get to the triple bond first. If there is no alcohol or a triple bond but there is a double bond, number the chain to get to the double bond first. If no alcohol or double or triple bonds, number the chain starting with the end with a functional group. If there is not a functional group on the first carbon go to the second If there are functional groups on both sides, priority is determined by alphabet. I. E. A bromine would get preference over a chlorine. If there are substituents hanging off the base chain name them. Iiodo, Brbromo, Clchloro, FFluoro If there are more than one substituent you can group them together with prefixes. 2di, 3tri, 4tetra, 5penta, 6hexa Put all substituents in alphabetical order with a number to indicate their position on the base chain. Separate numbers with commas. Separate numbers from letters with dashes. If there are two substituents on the same carbon you must use the number twice. The position of double and triple bonds are indicated by a number in front of the base name. If there are more than one double or triple bond you use the prefixes as above, i.e. di, tri, tetraFor example, hexane with two double bonds is 1,3-hexadiene. Cyclic compound are numbered with largest group being number 1. The numbering then proceeds in whichever direction gets to the next group quickest.

9

EXAMPLES First, the molecule is 10 carbons in length so it is a decane. Since there are double bonds it becomes decadiene. Since there are no alcohols, alkenes become highest priority. The two methyls are combined into one name. Therefore, the name of this molecule is

6,8-dimethyl-1,3-decadiene First, the molecule contains 6 carbons in a circle with a double bond. Thus, the base name is cyclohexene. A double bond in a carbon will always be the first carbon. We then head towards the bromine since bromine is alphabetically before methyl.

Br

3-bromo-6-methyl-1-cyclohexene First the base chain is heptane. We could number it from either side and get to a group at carbon #2. Since Br comes first alphabetically before Cl we number from the left to right. Therefore,

ClBr

2-bromo-6-chloro-4,4-dimethylheptane First the base chain is 9 carbons long with a double bond so it is nonene. If we number from the right we start with chlorine. From the left we start with double bond. Since double bonds take precedence over all non alcohol compounds we number from the left.

Cl

9-chloro-4,4,5,6,6,7-hexamethyl-1-nonene

10

OH

I Alcohols always are numbered to be as low as possible. We would name this compound 4-iodo-2-buten-1-ol.

Br

FOH OH We would name this compound 2-bromo-3-fluoro-2,4-octadiene-1,5-diol

Here the longest chain is 11 carbons long. See if you can find it. In either direction alcohol comes on the 6 carbon. If you number from the left you get to the double bond at the third carbon. From the right there is a methyl on the second carbon. Therefore, this is 2-methyl-4,6,8-undecatrien-6-ol.

At first glance you may try to number this from the left. There is an ethyl on the third carbon from the left and a methyl on the third carbon from the right. By rule ethyl comes alphabetically before methyl so you would assume that you would number from the left. But actually you number from the right because double bonds have the highest priority. 7-ethyl-3-methyl-4-nonene is the correct name.

OH

11

I. For the following structures give the name.

Br

Br

Cl

II. For the following names give the structure. 1. Octene 2. Cycloheptane 3. 4-methylnonane 4. 1,3-cyclooctadiene 5. 2-bromo-3-chlorobutane 6. 1-ethyl-2-methyl-3-pentylcyclopropane 7. 1,3,5-cyclohexatriene 8. 3-ethyl-4-fluoro-5,6,7-trimethyldecane 9. 3-bromo-1-cyclopentene 10. 7-methyl-5-propyl-3-octyne

12

I. For the following structures give the name.

Br

Br

Cl

4-ethyl-6-methylnonane

3-bromo-heptene

6,7,8-trimethyl-3-decene

4-propylcyclohexene

3-bromo-5-ethyl-1,5-octadiene

1,2,3-trimethylcyclopropane

4-methyl-1-pentyne

1-chlorocyclopentene

1,3,5,7-cyclooctatetraene

7-methyl-3,5-nonadiene

13

II. For the following names give the structure. 1. Octene

2. Cycloheptane

3. 4-methylnonane

4. 1,3-cyclooctadiene

5. 2-bromo-3-chlorobutane

Cl

Br

14

6. 1-ethyl-2-methyl-3-pentylcyclopropane

Technically, the name I gave is wrong. This should be named based on pentyl side chain. It should be called1-(2-ethyl-3-methylcyclopropanyl)pentane

7. 1,3,5-cyclohexatriene

8. 3-ethyl-4-fluoro-5,6,7-trimethyldecane

9. 3-bromo-1-cyclopentene

10. 7-methyl-5-propyl-3-octyne

Br

F

15

Practice ProblemsChapter 4Nomenclature

Cl

Br

OH

Cl

FBr

F

Br

Br

OHOH

F

Br

OH

F

Br

OH

AB

C

DE

F G

H

J

I

16

Draw the structures of the following: 1. 5-methyl-4-propyl-2,4,6,8,10-pentadecapenten-1-ol 2. Bicyclo[4.2.1]nonane 3. 3-chloro-4,5,6-trimethyl-1,4-cycloheptadien-1-ol 4. 5-sec-butyl-2-chloro-3-ethyl-8,9,9,11,12,13-hexamethyl-1,3,5-octadecatriene 5. 1-cyclopentyl-2-bromo-4-penten-2-ol 6. 1-bromo-2-chloro-3-ethyl-4-isobutyl-5- isopropyl-6-methyl-7-pentyl-1,3,5-

cycloheptatriene 7. 2-chloro-4,5-diethyl-7-iodo-3-vinyl-1,3-octadiene 8. 6-allyl-2-tert-butyl-3-chloro-5-ethyl-7,8,9-trimethyl-8-decen-6-ol

(after drawing this molecule rename it from the picture. You will come up with a different name because the name above has a mistake. What is the mistake?)

9. 1-bromo-3-chloro-4-(2-cyclopropyl-3-ethyl -4,5-diiodo-4-octenyl)-2-nonene (after drawing this molecule rename it from the picture. You will come up with a different name because the name above has a mistake. What is the mistake?) 10. 2,3-dichloro-4,6-diethylbicyclo[4.3.2]undecane

17

Answer Key to Practice Problems

A. 4-allyl-7-bromo-6-chloro-2,8-cyclononadien-1-ol B. 4-ethyl-3,7,8-trimethyl-5-decene C. 4-bromo-8-fluorobicyclo[10.2.2]hexadecane D. 5-ethyl-3,4,6,7-tetramethyl-2,4,6-nonatrien-1,9-diol E. 5-ethyl-6-(1-fluorobutanyl)-4-isopropyl-3,8-tridecadien-2-ol F. 8-bromo-5-sec-butyl-7-chloro-4-fluoro-6-methyl-2-nonene G. 5-(2-bromo-6,6-dimethyl-2,4-cyclohexadien-3-olyl)-4,4,6,8,9,9,10,10- octamethyl-1,5,7-undecatriene H. 4-bromo-1-(3-ethyl-2,4-dimethylcyclobutanyl)-2-pentene I. bicyclo[5.2.2]undecane J. 2-bromo-8-butyl-11-tert-butyl-17-fluoro-6-isopropyl-3,4,16-trimethyl-

4,8,14-octadecatriene 1.

2.

OH

18

3.

4.

5.

Cl

OH

C l

HO Br

19

6.

7.

8.

The correct name is 5-allyl-8-chloro-6-ethyl-2,3,4,9,10,10-hexamethyl- 2-undecen-5-ol

Br

Cl

Cl

I

Cl

HO

20

9.

The correct name is 9-(3-bromo-1-chloro-1-propenyl)-7-cyclopropyl-6-ethyl- 4,5-diiodo-4-tetradecene 10.

Br Cl

I

I

Cl

Cl

21

BrBr

Br Cl

Br

Br

ClCl

OH

Br

Br

OH

22

12

3

4

5

6

1

2

3

4

5

6

7

1

2

3

4

5

6

7

8

9

8

7

6

5

4

3

2

1

BrBr

Br Cl

Br

Br

ClCl

1

2

3

4

5

6

7

8

OH

Br

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

9

10

Br

OH

3-bromohexane

3-bromo-7-chlorononane

Br

2-bromo-5-methylheptane

2,7-dibromo-3,5-dichlorooctane

7-bromooctan-4-ol

9-bromo-7-methylnon-2-en-5-ol

9-bromo-7-methyl-2-nonen-5-ol

9-bromo-4-decene9-bromodec-4-ene

23

4. Physical Properties mp/bp

C 1-4gases; 5-17liquids; 18+--solids as molecular weight increasebp/mp increases as branching increasesbp decreases symmetry increases mp Density very low; halogens increase density considerably Solubility(alkanes) very low in water high solubility in low polar solvents 5. Conformational analysis a. Newman projections 1. eclipsed/staggered/anti/gauche anti staggered most favorable eclipsed least favorable gauche in between eclipsed and staggered gauche is the non favored staggered positions anti is the most favored staggered positions eclipsed have favored and non favored positions also

24

Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise.

Cl

H

sight down Carbons2-3 with 2 in front

tert-butyl is largest group, then phenyl,then pentyl chain, then Cl, then H.

25

Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise.

Cl

H

sight down Carbons2-3 with 2 in front

tert-butyl is largest group, then phenyl,then pentyl chain, then Cl, then H.

C 5H 12

H PhC H3

C 4H 9ClC5H 12

H Ph

Cl

CH3

C 4H 9

C5H 12

H PhC4H 9

ClH3C

C5H12

H Ph

H 3C

C4H9C l

C5H12

H PhCl

C4H9

C 5H12

H Ph

C H3

H3C

C 4H9

C l

26

Draw the Newman projections for the following molecule. In the first picture draw the molecule exactly as shown. Holding the first carbon fixed, rotate the back carbon to show all 6 conformations of this picture. Identify the gauche, anti and eclipsed pictures.

OHHSize Ph > Hexyl > F and CH3 > OH > H

FPh

OHHSize Ph > hexyl > F and CH3 > OH > H

Hexyl

Ph FCH3

OHH

Ph

F HexylCH3

OHH

F

Hexyl PhCH3

OHH

F

Hexyl

Ph

CH3

OHH

Hexyl

Ph

F

CH3

OHH

Ph

F

Hexyl

CH3

OHH

FPh

ECLIPSED

ECLIPSED ECLIPSEDANTI

GAUCHEGAUCHE

27

6. Cyclic Alkanes/Ring Strain a. ring strain 1. heats of combustion have been used to determine ring strain 2. table 4.6 page 155 cyclopropane has greatest ring strain. cyclononane,

cyclodecane and cyclooctane come next in ring strain. cyclopentane and heptane have very little ring strain cyclopentadecane and cyclohexane have almost no ring strain 3. angle strain makes up most of ring strain due to angles being less than the tetrahedral angles that are preferred cyclopropane is 60 degrees instead of 109.5 because of ring strain orbital overlap is low which weakens bonds. 4. Torsional strain due to locked ring structures leads to eclipsed newman projections cyclobutane actually is not planar(like a square) but bent. in bent conformation angles are 88 degrees which is even further from 109.5 than the 90 degrees of a square. However, the bent structure provides much less torsional strain which makes it a more favorable structure than the planar form. cyclopentane has angle of 108 vs 109.5 if planar. Therefore

little angle strain. However it doesnt take this form due to intense torsional strain. Almost all of the ring strain is due to torsional strain. Therefore it bends out of plane increasing its angle strain but relieving its torsional strain cyclohexane has a form that has almost no angle or torsional strain.

28

7. Cyclohexane a. boat vs. chair 1. cyclohexane actually has 7 conformations it can take. However, the boat and chair forms predominate. The chair form is free of ring and torsional strain. The boat form is free of ring strain but has torsional strain. Based on thermo- dynamical data, 99% of cyclohexane molecules exist in the

chair form. 2. Boat has torsional strain(due to eclipsed hydrogens2 and 3 eclipsed, 5 and 6 eclipsed). Boat has flagpole interactions(due to van der Waal interactions between H-1 and H-4).

3. half-twist forms has less torsional strain and less flagpole interactions

4. cycloheptane, cyclooctane, cyclodecanehave small to no ring strain. Most strain is due to eclipsed hydrogens(torsional strain) and due to van der waal interactions also called transannular strain.

b. axial vs equatorial

1. there are 6 equatorial and 6 axial positions on chair form 2. the equatorial positions are best for larger group as the equatorial positions avoid van der waal interactions. 3. As cyclohexane flips through all the axial positions become equatorial and vice versa 4. In cyclohexane these interactions are called 1,3 diaxial interactions. They are akin to gauche interactions in butane newman conformations. 5. as stated in #2, larger groups prefer equatorial position because larger groups will have larger diaxial interactions.

29

c. cis vs trans in disubstituted alkanes 1. 1,2 cis are e,a and a,e 2. 1,2 trans are e,e and a,a 3. 1,3 cis are e,e and a,a 4. 1,3 trans are e,a and a,e 5. 1,4 cis are e,a and a,e 6. 1,4 trans are e,e and a,a 7. 1,2 cis is meso; 1,2 trans is pair of enantiomers 8. 1,3 cis is meso; 1,3 trans is pair of enantionmers 9. 1,4 cis is meso and 1,4 trans is meso. 10. 1,4 cis is a diastereomer of 1,4 trans 11. 1,3 cis is a diastereomer of 1,3 trans 12, 1,2 cis is a diastereomer of 1,2 trans 8. Decalin/other bicyclics a. Know structures for decalin(cis and trans) b. Know adamantine, diamond, cubane, prismane, and bicyclo[1.1.0]butane, dodecahedrane 9. Pheremones a. undecane, 2-methylheptadecane, muscalure 10. Synthesis/Retrosynthesis a. hydrogenation of alkenes/alkynes b. reduction of alkyl halides c. deprotonation of alkynes/alkylation of alkynide ions d. retrosynthesis/identity of precursors/E. J. Corey

30

432

16

5

AB

B

A

A

B

B

A

B

A

B1 2

3

45

6

BA

B

A

B

B

A

A

B

A

B

A

A

43

2

16

5

A

B

1 23

45

6

BA

CHAIR RING FLIP OF CHAIR

CIS 1,2 and ring flip



43

2

16

5

A

B

1 23

45

6

B

A

432

16

5

B

B

1 23

45

6

BB

31

43

2

16

5

AB

B

A

A

B

B

B

B

A

B1 2

3

45

6

BA

B

A

B

B

A

A

B

A

B

A

A

CHAIRRING FLIP OF CHAIR

TRANS 1,2 and ring flip



432

16

5

A

A

1 23

45

6

A

A

432

16

5

B

A

1 23

45

6

B

A

43

2

16

5

A

A

1 23

45

6

AA

32

Draw the following cyclohexane in chair form. Draw the other chair form by flipping the chair. Pick which one is the more favored product. Phenyl > tert-butyl > pentyl > OH

4

56

1

23

3

45

6

12

6 1

2

34

5

O H

O H

Ph

d own

up

up

up

OH

be tte r pi cture all grou ps equ ato rial

O H

33

Homework #2Give the name for the structures below.

OH Br F

OH

OH

Br

F

Br

34

Redraw the following molecule in Newman Configuration in the first box. Put carbon 4 in front and carbon 5 behind. Then draw the other 5 main Newman Configurations by rotating the BACK carbon counterclockwise. Label the configurations, gauche, anti, eclipsed

1

2

3

4

5

6

HBr

PhCl

7

8

9

10

On carbon 4--Br is largest, next is propyl, and

hydrogen is the smallest.

On carbon 5--phenyl is largest, next is pentyl,

and hydrogen is the smallest.

35

Draw the following two molecules in chair configuration, then draw the ring flip of each.

Br

Cl

36

Homework #2Give the name for the structures below.

OH Br F 11-bromo-6-ethyl-13-fluoro-9-methyl-8-vinyl-2,6-tridecadien-5-ol 7-bromo-2-fluorobicyclo[4.2.2]decane 7,10-dimethylbicyclo[4.4.0]decan-3-ol

8-(3-bromo-2-methyl-2-pentenyl)-11-ethyl-10-isobutyl-4-isopropyl-3,7,11,13-tetradecatetraen-7-ol

OH

OH

Br

F

Br

37

Redraw the following molecule in Newman Configuration in the first box. Put carbon 4 in front and carbon 5 behind. Then draw the other 5 main Newman Configurations by rotating the BACK carbon counterclockwise. Label the configurations, gauche, anti, eclipsed

1

2

3

4

5

6

HBr

PhCl

7

8

9

10

On carbon 4--Br is largest, next is propyl, andhydrogen is the smallest.On carbon 5--phenyl is largest, next is pentyl,and hydrogen is the smallest.

Pentyl

Cl PhP

HBr Cl

Ph PentylP

HBr

PH

Pentyl ClP

HBr

PentylCl

PhP

HBr

Pentyl

ClP

PhBrCl

Ph

PentylP

HBr

ANTI ECLIPSED GAUCHE

ECLIPSED GAUCHE ECLIPSED

38

Draw the following two molecules in chair configuration, then draw the ring flip of each.

5

6

1

2

3

4

Br

6

1 2

3

45 5

6

1 2

3

4Br

Br

3

2

1

6

5

4

Cl

6

1 2

3

45

5

6

12

3

4Cl

Cl

39

4 3

2

16

5

OH

1

6 54

32

2

16 5

43

Pe

Ph

OHtBu

PePh

HO

tBu

OH

HO

40

4

3 21

65

BrCH3

OH

Cl

6

54

3

21

OH

Cl

Br

H3C

5

43 2

16 OH

ClBr

H3C

41

1

2 34

56

6

12 3

45

1 32

Cl

Br

CH3HO

up

down

4

HOCl

H3C

Br

OH

Cl

CH3

Br

42

IndexofHydrogenDeficiency

InordertocalculatetheIndexofHydrogenDeficiency(I.H.D.),youneedtoknowthefollowing:

1)FormulaofsaturatedalkaneisCnH2n+2.

2)Eachhalogenisreplacedbyaddingonehydrogen.3)Eachoxygenisreplacedbyaddingzerohydrogen.4)Eachnitrogenisreplacedbysubtractingonehydrogen.5)AfterconvertingformulatoCxHy,countthemissinghydrogenfromthesaturatedalkaneandthendividethatbytwotogettheI.H.D.EXAMPLESC10H20 First,converttoCxHy(alreadydone),C10H20

2ndfigureoutthesaturatedalkane;CnH2n+2,n=10,soC10H2(10)+2orC10H223rdfigureouthowmanyHaremissing;2220=24th,dividethenumberofmissingHby2;2/2=1I.H.D

C15H26 First,converttoCxHy(alreadydone),C15H26


C12H20Br2 First,converttoCxHy(eachhalogen=1H),C12H20Br2=C12H20H2=C12H22


44

C10H18ClF

First,converttoCxHy(eachhalogen=1H),C10H18ClF=C10H18HH=C10H20


C15H24Br2O2

First,converttoCxHy(eachhalogen=1H,oxygen=0H),C15H24Br2O2=C15H24H2=C15H26


C18H26O2N2

First,converttoCxHy(Halogen=1H,O=0H,N=1H),C18H26O2N2=C18H26H2=C18H24


C20H35Br3N2O

First,converttoCxHy(Halogen=1H,O=0H,N=1H),C20H35Br3N2O=C20H35H3H2=C20H362ndfigureoutthesaturatedalkane;CnH2n+2,n=20,soC20H2(20)+2orC20H423rdfigureouthowmanyHaremissing;4236=64th,dividethenumberofmissingHby2;6/2=3I.H.D

C24H40Cl2F2O3N2

First,converttoCxHy(Halogen=1H,O=0H,N=1H),C24H40Cl2F2O3N2=C24H40H4H2=C24H422ndfigureoutthesaturatedalkane;CnH2n+2,n=24,soC24H2(24)+2orC24H503rdfigureouthowmanyHaremissing;5042=84th,dividethenumberofmissingHby2;8/2=4I.H.D

45

Homework #2 Name: Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise. Label the anti, the 2 gauche and the 3 eclipsed. Circle the most energetically unfavorable drawing amongst the 6. Size of groups on the front carbon: Bromine > Methyl > Hydrogen Size of groups on the front carbon: t-butyl > Pentyl > Hydroxy (OH)

Br

t-Bu

H

HO

46

Name the following

Br

OH

47

Draw the following. 2-isopropyl-3,4,5-trimethyloctene 2,4-dibromocyclohept-2-4-dien-1-ol 2-bromobicyclo[4.2.2]decane Draw this product in the chair configuration. Draw the ring flip of what you just drew and then pick the best picture.

t-Bu

Cl

OH

Oh

48

Homework #2KEY Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise. Label the anti, the 2 gauche and the 3 eclipsed. Circle the most energetically unfavorable drawing amongst the 6. Size of groups on the front carbon: Bromine > Methyl > Hydrogen Size of groups on the front carbon: t-butyl > Pentyl > Hydroxy (OH)

Pe

tBuHO

Me

H Br

Pe

tBuHO

Br

Me H

HOtBu

Pe

Me

Br

H

GAUCHE WORST ECLIPSED GAUCHE

ECLIPSED ANTI ECLIPSED

Pe

tBuHO

H

Br Me

HOtBu

Pe

Br

H

Me

HOtBu

Pe

H

Me

Br

Br

t-Bu

H

HO

49

Name the following

Br

OH

3,4,6-trimethyloctane

4,8-dimethylnona-2,5-diene

6-bromo-1-methylcyclohex-1-ene

4,5-dimethylcycloocta-2,4-dienol

50

Draw the following.

1

2

3

4

5

6

7

8

2-isopropyl-3,4,5-trimethyloctene

1

2

34

5

67 OH

Br

Br

2,4-dibromocyclohept-2-4-dien-1-ol

12

3

4

56

8

7

Br Draw this product in the chair configuration. Draw the ring flip of what you just drew and then pick the best picture.

Cl

OHHO

Cl

OH

Br

t-Bu

Cl

OH

OH

t-Bu

t-Bu

51

CLCRectangle

Homework #2 Name: Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise. Label the anti, the 2 gauche and the 3 eclipsed. Circle the most energetically unfavorable drawing amongst the 6. Size of groups on the front carbon: Phenyl > Propyl > Hydrogen Size of groups on the front carbon: Isopropyl > Butyl > Hydroxy (OH)

HO

H

52

Name the following

OH

O

Br

Cl

OCH3

53

Draw the following. 2-bromo-4-methylhept-1,3-diene 4-chloro-3-isopropyl-2-methylcyclohexanol 3-methylbicyclo[4.2.1]noneane Draw this product in the chair configuration. Draw the ring flip of what you just drew and then pick the best picture.

OH

CH3

Ph

Br

54

Homework #2KEY Draw the Newman conformation for the following molecule exactly as drawn sighting down the indicated carbons. Then draw the other 5 conformations in order by rotating the front carbon clockwise. Label the anti, the 2 gauche and the 3 eclipsed. Circle the most energetically unfavorable drawing amongst the 6. Size of groups on the front carbon: Phenyl > Propyl > Hydrogen Size of groups on the front carbon: Isopropyl > Butyl > Hydroxy (OH)

HO

H

Bu

iPrHO

Pr

H Ph

Bu

IPrHO

Ph

Pr H

HO iPr

Bu

Pr

Ph

H

GAUCHE WORST ECLIPSED GAUCHE

ECLIPSED ANTI ECLIPSED

Bu

IPrHO

H

Ph Pr

HO iPr

Bu

Ph

H

Pr

HO iPr

Bu

H

Pr

Ph

55

Name the following

OH

O

Br

Cl

OCH3

2,3,4,6-tetramethylheptane

5-isopropyl-8-methylnona-6,8-dien-2-ol

2-bromo-3-chloro-6-ethoxy-3-methylheptane

3-methoxy-1,6-dimethylcyclohex-1-ene

56

Draw the following.

1

2

3

4

5

6

7

Br

2-bromo-4-methylhept-1,3-diene

1

2

3

4

5

6

HO

Cl

4-chloro-3-isopropyl-2-methylcyclohexanol

1

2

3

4

567

8

9

3-methylbicyclo[4.2.1]nonane Draw this product in the chair configuration. Draw the ring flip of what you just drew and then pick the best picture.

OH

CH3

Ph

Br

OH

CH3

PhBr

OH

CH3

Ph

Br

57

CLCRectangle

chapter.4.study.guidechapter.4.study.guidechapter.4.studyguide.new.pdfchapter.4.addendum.pdfchapter.4.study.guide.pdfchapter.4.study.guide.pdfuno.fall.07.studyguide.4.pdfpractice.nomenclature.pdfsupplemental.pdf

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Index of Hydrogen Deficiency

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