C20 09/16/2013 15:47:28 Page 295

C H A P T E R 2 0

UNSATURATED HYDROCARBONS

SOLUTIONS TO REVIEW QUESTIONS

1. The sigma bond in the double bond of ethene is formed by the overlap of two sp2 electron orbitals and is

symmetrical about a line drawn between the nuclei of the two carbon atoms. The pi bond is formed by the

sidewise overlap of two p orbitals which are perpendicular to the carbon-carbon sigma bond. The pi bond

consists of two electron clouds, one above and one below the plane of the carbon-carbon sigma bond.

2. The restricted rotation around the carbon-carbon double bond in 1,2-dichloroethene means two different

structural isomers exist, one with two chlorines on the same side of the double bond (cis) and when the

chlorines are on opposite sides of the double bond (trans). For 1,2-dichloroethyne, the triple bond also

has restricted rotation. However, as all atoms lie in a line, there is only one way to arrange the atoms for

1,2-dichloroethyne.

3. 1-pentene is a liquid at room temperature (25�C) because its boiling point is above room temperature

(30�C). 2-methylpropene is a gas at room temperature because its boiling point (�7�C) is much lower

than 25�C.

4. 1-butene has a much lower melting point (�185�C) than 2-methylpropene (�14�C) although theirboiling points are almost the same (1-butene, �6�C; 2-methylpropene, �7�C). 1-butene has a much

different molecular shape than 2-methylpropene. This shape difference has a great effect on melting

points because the molecules fit closely together in a solid. In a liquid the molecules are rapidly moving

and molecular shape differences are much less important. Thus, the boiling points for these two

molecules are almost the same.

5. Trans fats contain double bonds in the trans isomer form. In contrast, most naturally occurring unsaturated

fats have double bonds in the cis form. Our bodies can’t metabolize the trans fats because they are the

wrong shape. The trans fats accumulate over time and increase the risk of cardiovascular disease.

6. During the 10-year period from 1935 to 1945, the major source of aromatic hydrocarbons shifted from

coal tar to petroleum due to the rapid growth of several industries which used aromatic hydrocarbons as

raw material. These industries include drugs, dyes, detergents, explosives, insecticides, plastics, and

synthetic rubber. Since the raw material needs far exceeded the aromatics available from coal tar,

another source had to be found, and processes were developed to make aromatic compounds from

alkanes in petroleum. World War II, which occurred during this period, put high demands on many of

these industries, particularly explosives.

7. Benzene does not undergo the typical reactions of an alkene. Benzene does not decolorize bromine

rapidly and it does not destroy the purple color of permanganate ions. The reactions of benzene are

more like those of an alkane. Reaction of benzene with chlorine requires a catalyst. Benzene does not

readily add Cl2, but rather a hydrogen atom is replaced by a chlorine atom.

C6H6 þ Cl2�!Fe C6H5Clþ HCl

- 295 -

C20 09/16/2013 15:47:28 Page 296

8. The 11-cis isomer of retinal combines with a protein (opsin) to form the visual pigment rhodopsin.

When light is absorbed, the 11-cis double bond is converted to a trans-double bond. This process

initiates the mechanism of visual excitation which our brains perceive as light or light forms.

9. Polycyclic aromatic hydrocarbons are potent carcinogens.

10. Cyclohexane carbons form bond angles of about 109�in a tetrahedron; this causes the ring to be either a“boat” or a “chair” shape. Benzene carbons form bond angles of 120�in a plane. Therefore, the benzenemolecule must be planar.

11. According to the mechanism, the addition of an unsymmetrical module such as HX adds to a carbon-

carbon double bond. In the first step, the H adds to the carbon of the carbon-carbon double bond that has

the most hydrogen atoms on it, according to Markovnikoff’s rule. The second step completes the

addition by adding the more negative element, X of the HX.

12. “Cracking” means breaking into pieces, which, according to the pyrolysis products, is what occurs in

the reaction.

13. Hþ adds to C-1 forming the more stable cation,

CH3CþHCH2CH3

A Cl� then adds to C-2 to form the product,

CH3CHClCH2CH3

14. 1-chloropropene meets the requirement of two different groups on each carbon atom of the carbon-

carbon-double bond to have cis-trans isomers. 2-chloropropene does not meet this requirement.

C C

CH3

H

H

Cl

trans-1-chloropropene

C C

CH3

H

Cl

H

cis-1-chloropropene

C C

CH3

Cl

H

H

2-chloropropene

15. They are the same compound. Both structures have the same name, 2,3-dimethylcyclohexene.

- 296 -

- Chapter 20 -

C20 09/16/2013 15:47:28 Page 297

SOLUTIONS TO EXERCISES

1. (a) (b) (c)

2. (a) (b) (c)

3. CHsp3

3��CHsp2 ����CH

sp2

��CHsp3

3

4. HCsp ������C

sp

��CH2

sp3

CH2

sp3

CH3

sp3

5. Isomeric iodobutenes, C4H7I

- 297 -

- Chapter 20 -

C20 09/16/2013 15:47:29 Page 298

6. (a)

(b)

7. (a) (b)

(c) CH������CCH����CHCH3

3-penten-1-yne

(d)

(e) (f)

8. (a) (b)

(c) (d)

- 298 -

- Chapter 20 -

C20 09/16/2013 15:47:29 Page 299

(e) (f)

9. (a) 23 sigma bonds, 1 pi bond; (b) 17 sigma bonds, 1 pi bond; (c) 10 sigma bonds, 3 pi bonds;

(d) 17 sigma bonds, 1 pi bond; (e) 15 sigma bonds, 2 pi bonds; (f) 33 sigma bonds, 3 pi bonds.

10. (a) 23 sigma bonds, 1 pi bond; (b) 27 sigma bonds, 7 pi bonds; (c) 20 sigma bonds, 5 pi bonds;

(d) 13 sigma bonds, 1 pi bond; (e) 19 sigma bonds, 1 pi bond; (f) 21 sigma bonds, 1 pi bond.

11. (a) trans-6-chloro-3-heptene; (b) 4,4-dimethyl-2-pentyne; (c) 2-ethyl-l-pentene.

12. (a) 3-phenyl-l-butyne; (b) 2-methyl-2-hexene; (c) cis-3,4-dimethyl-3-hexene.

13. All the hexynes, C6H10

CH3CH2CH2CH2C������CH CH3CH2CH2C������CCH3 CH3CH2C������CCH2CH3

1-hexyne 2-hexyne 3-hexyne

14. All the pentynes, C5H8

CH3CH2CH2C������CH

1-pentyne

CH3CH2C������CCH3

2-pentyne

15. (a) 1-butyne (The smallest numbered carbon that is involved in the triple bond is used to number the

triple bond.); (b) 2-hexyne (The parent chain is the longest continuous carbon chain that contains the

triple bond.); (c) propyne (The triple bond has only one possible location in propyne and no number is

needed.)

16. (a) 4-methyl-2-pentyne (The parent chain is numbered from the end closest to the triple bond.);

(b) 4-methyl-2-hexyne (The parent chain is the longest continuous carbon chain that contains the triple

bond.); (c) 3-bromo-l-butyne (The parent chain is numbered from the end closest to the triple bond.).

17. Cis-trans isomers exist for (c) only. Alkynes do not have cis-trans isomerism. Alkenes have cis-trans

isomerism only when each double-bonded carbon is attached to two different groups.

- 299 -

- Chapter 20 -

C20 09/16/2013 15:47:30 Page 300

18. Cis-trans isomers exist for (b) only. Alkynes do not have cis-trans isomerism. Alkenes have cis-trans

isomerism only when each double-bonded carbon is attached to two different groups.

19. (a) cis

(b) neither

(c) trans

20. (a) trans

(b) trans

(c) neither

21. (a) CH3CH2CH2CH����CH2 þ Br2 ���! CH3CH2CH2CHBrCH2Br

(b)

(c)

(d)

(e)

22. (a)

(b) CH3CH2CH����CHCH3 þ HBr���! CH3CH2CHBrCH2CH3 þ CH3CH2CH2CHBrCH3

(c) (CH2����CHClþ Br2 ���! CH2BrCHClBr

(d)

(e)

23. (a) CH3C������CCH3 þ Br2 ð1 moleÞ ���! CH3CBr����CBrCH3

(b) Two-step reaction:

CH������CHþ HCl���! CH2����CHCl���������!HClCH3CHCl2

(c) CH3CH2CH2C������CHþ H2 ð1 moleÞ ��!Pt25�C

CH3CH2CH2CH����CH2

- 300 -

- Chapter 20 -

C20 09/16/2013 15:47:30 Page 301

24. (a) CH3C������CHþ H2 ð1 molÞ �����!Pt;25�C1 atm CH3CH����CH2

(b) CH3C������CCH3 þ Br2 ð2 molÞ ���! CH3CBr2CBr2CH3

(c) Two-step reaction:

CH3C������CHþ HCl���! CH3CCl ¼ CH2�!HCl CH3CCl2CH3

25. When cyclohexene, reacts with:

(a) Br2, the product is

Br

Br1,2-dibromocyclohexane

(b) HI, the product is iodocyclohexane

(c) H2O, Hþ, the product is cyclohexanol

(d) KMnO4(aq), the product is cyclohexene glycol or

1,2-dihydroxycyclohexane

26. When cyclopentene, reacts with:

(a) Cl2, the product is 1,2-dichlorocyclopentane

(b) HBr, the product is bromocyclopentane

(c) H2, Pt, the product is cyclopentane

(d) H2O, Hþ, the product is cyclopentanol

27.

(a)

(b)

(c)

- 301 -

- Chapter 20 -

C20 09/16/2013 15:47:30 Page 302

28. (a)

(b)

(c)

29. (a) meta

(b) meta

(c) ortho

30. (a) para

(b) meta

(c) para

31. (a) CH

O

(d) CH

CH3

CH3

(b)OH

NO2

(e)NH2

BrBr

(c)CH2 (f)

CH3

CH3

- 302 -

- Chapter 20 -

C20 09/16/2013 15:47:31 Page 303

32. (a) COH

O

Br

(d)CH CH2

(b)

Cl

ClCl (e)

(c)

CH3

NO2

(f)

NO2

33. (a) bromodichlorobenzenes

- 303 -

- Chapter 20 -

C20 09/16/2013 15:47:31 Page 304

(b) The toluene derivatives of formula C9H12:

p-ethyltoluene

CH3

CH2CH3

m-ethyltoluene

CH3

CH2CH3

o-ethyltoluene

CH3CH2CH3

1,3,5-trimethylbenzene

CH3H3C

CH3

1,2,4-trimethylbenzene

CH3

CH3

CH3

1,2,3-trimethylbenzene

CH3

CH3

CH3

34. (a) trichlorobenzenes

1,3,5-trichlorobenzene

ClCl

Cl

1,2,4-trichlorobenzene

ClCl

Cl

1,2,3-trichlorobenzene

ClCl

Cl

(b) The benzene derivatives of formula C8H10:

CH2CH3

ethylbenzenep-xylene or 1,4-dimethylbenzene

CH3H3C

m-xylene or 1,3-dimethylbenzene

CH3

CH3

o-xylene or 1,2-dimethylbenzene

CH3

CH3

35. (a) p-chloroethylbenzene (d) p-bromophenol

(b) propylbenzene (e) triphenylmethane

(c) m-nitroaniline

36. (a) styrene (d) isopropylbenzene

(b) m-nitrotoluene (e) 2,4,6-tribromophenol

(c) 2,4-dibromobenzoic acid

- 304 -

- Chapter 20 -

C20 09/16/2013 15:47:31 Page 305

37. (a) 2-chloro-1,4-diiodobenzene

(b) o-dibromobenzene

(c) m-chloroaniline

38. (a) 2,3-dibromophenol

(b) 1,3,5-trichlorobenzene

(c) m-iodotoluene

39. (a)

(b)

40. (a)

(b)

41. When reacts with HBr, two products are possible:

The first will strongly predominate. This is the product according to Markovnikov’s rule and forms

because the tertiary carbocation intermediate formed is more stable than a primary carbocation.

42. Two tests can be used. (1) Baeyer test—hexene will decolorize KMnO4 solution; cyclohexane will

not. (2) In the absence of sunlight, hexene will react with and decolorize bromine; cyclohexane

will not.

- 305 -

- Chapter 20 -

C20 09/16/2013 15:47:32 Page 306

43.

- 306 -

- Chapter 20 -

C20 09/16/2013 15:47:32 Page 307

43. (Cont.)

44. (a) CH3 CþH2 primary carbocation (a positively charged carbon bonded to one other carbon)

(b) CH3 CþHCH3 secondary carbocation (a positively charged carbon bonded to two other

carbons)

(c)tertiary carbocation (a positively charged carbon bonded to three other

carbons)

45. (a)

(b)

- 307 -

- Chapter 20 -

C20 09/16/2013 15:47:33 Page 308

(c)

(d)

46. The reaction mechanism by which benzene is brominated in the presence of FeBr3:

(a) FeBr3 þ Br2 ���! FeBr�4 þ Brþ

Formation of a bromonium ion (Brþ), an electrophile.

(b)

The bromonium ion adds to benzene forming a carbocation intermediate.

(c)

A hydrogen ion is lost from the carbocation forming the product bromobenzene.

47. (a) CH3CHClCH2CH3���!��HClCH2����CHCH2CH3 þ CH3CH����CHCH3

(b) CH2ClCH2CH2CH2CH3 ���!�HClCH2����CHCH2CH2CH3

(c)

48. Yes, there will be a color change (loss of Br2 color). The fact that there is no HBr formed indicates that

the reaction is not substitution but addition. Therefore, C4H8 must contain a carbon-carbon double bond.

Three structures are possible.

49. Baeyer test: Add KMnO4 solution to each sample. The KMnO4 will lose its purple color with 1-heptene.

There will be no reaction (no color change) with heptane.

- 308 -

- Chapter 20 -

C20 09/16/2013 15:47:33 Page 309

50. I would not expect graphene to react easily with HCl in an addition reaction. Graphene is composed of

sheets of aromatic rings that do not easily undergo addition reactions.

51.

52. The double bonds in fatty acids are oxidized. This reaction adds oxygen and the double bond converts to

a single bond. This reaction is thought to start the process that eventually results in “hardening of the

arteries.” This is an oxidation reaction.

53. (a)

(b) HCsp

������Csp��CH3

sp3

(c)

54. (a) Four isomers:

C4H8

CH2����CHCH2CH3

1-butene

CH2 ¼ CðCH3Þ22-methyl-1-propene

CH3CH ¼ CHCH3

2-butene (cis and trans)

(b) One isomer

C5H8

- 309 -

- Chapter 20 -

C20 09/16/2013 15:47:34 Page 310

55. (b) is the correct structure, (a) is an incorrect structure because the carbons where the two rings are

fused have five bonds, not four bonds to each carbon.

56. Carbon-2 has two methyl groups on it. The configuration of two of the same groups on a carbon of a

carbon-carbon double bond does not show cis-trans isomerism.

57. (a) alkyne or cycloalkene

(b) alkene

(c) alkane

(d) alkyne or cycloalkene

58. Chemically distinguishing between benzene, 1-hexene, and 1-hexyne

Step 1. Add KMnO4 solution to a sample of each liquid. Benzene is the only one in which the KMnO4

does not lose its purple color.

Step 2. To 0.5 mL samples of 1-hexene and 1-hexyne add bromine solution dropwise until there is no

more color change of the bromine (from reddish-brown to colorless). 1-hexyne (with a triple bond)

will decolor about twice as many drops of bromine as 1-hexene. Thus the three liquids are identified.

59. (a) CH3C������CH���!HClCH3CCl����CH2���!HBr

CH3CClBrCH3

(b) CH3C������CH���!Br2CH3CBr����CHBr���!HCl

CH3CClBrCH2Br

(c) CH3C������CH���!2Cl2CH3CCl2CHCl2

- 310 -

- Chapter 20 -