CHAPTER 20 UNSATURATED HYDROCARBONS SOLUTIONS TO REVIEW QUESTIONS

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1 APTER 20 UNSATURATED YDROARBONS SOLUTIONS TO REVIEW QUESTIONS 1. The sigma bond in the double bond of ethene is formed by the overlap of two 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. owever, as all atoms lie in a line, there is only one way to arrange the atoms for 1,2-dichloroethyne pentene is a liquid at room temperature (25 ) because its boiling point is above room temperature (30 ). 2-methylpropene is a gas at room temperature because its boiling point ( 7 ) is much lower than butene has a much lower melting point ( 185 ) than 2-methylpropene ( 14 ) although their boiling points are almost the same (1-butene, 6 ; 2-methylpropene, 7 ). 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 metabolized 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

2 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 but rather a hydrogen atom is replaced by a chlorine atom Fe " The 11-cis isomer or 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. yclohexane 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 benzene molecule must be planar. 11. According to the mechanism, the addition of an unsymmetrical module such as X adds to a carbon-carbon double bond. In the first step, the adds to the carbon of the carboncarbon 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 X. 12. racking means breaking into pieces, which, according to the pyrolysis products, is what occurs in the reaction

3 SOLUTIONS TO EXERISES 1. (a) ethane ethene ethyne 2. (a) propane propene propyne 3. Isomeric iodobutenes, 4 7 I 2 I cis-1-iodo-1-butene 2 I 2 2-iodo-1-butene 2 I iodo-1-butene I trans-2-iodo-2-butene 2 I trans-1-iodo-2-butene I 2 trans-1-iodo-1-butene I 2 3-iodo-1-butene 3 cis-2-iodo-2-butene 2 I cis-1-iodo-2-butene I 1-iodo-2-methylpropene I 2 I 2 3-iodo-2-methylpropene

4 4. (a) 3 5 trans-1-chloropropene 2 2-chloropropene chlorocyclopropane cis-1-chloropropene chloropropene 5. (a) 2,5-dimethyl-3-hexene cis-4-methyl-2-pentene 3-penten-1-yne trans-3-hexene (e) 2 3 (f) methyl-1-pentyne 2 6. (a) ethyl-3-methyl-1-pentene 3-methyl-2-phenylhexane cis, 1,2-diphenylethene

5 cyclopentene 3-phenyl-1-butyne (e) (f) ( ) 2 1-methylcyclohexene 3-isopropylcyclopentene 7. (a) 23 sigma bonds, 1 pi bond; 17 sigma bonds, 1 pi bond; 10 sigma bonds, 3 pi bonds; 17 sigma bonds, 1 pi bond; (e) 15 sigma bonds, 2 pi bonds; (f) 33 sigma bonds, 3 pi bonds. 8. (a) 23 sigma bonds, 1 pi bond; 27 sigma bonds, 7 pi bonds; 20 sigma bonds, 5 pi bonds; 13 sigma bonds, 1 pi bond; (e) 19 sigma bonds, 1 pi bond; (f) 21 sigma bonds, 1 pi bond. 9. (a) 10. (a) 2 2 Numbering was started from the wrong end of the structure. orrect name: 2-methyl-1-butene 2 Numbering was started from the wrong end of the structure. orrect name: cis-2-pentene 2 This compound does not have cis-trans isomers. orrect name: 2-methyl-2-pentene 2 2 Longest chain contains five carbon atoms. orrect name: 3-methyl-1-pentene

6 The bond in cyclohexene is numbered so that substituted groups have the smallest numbers. orrect name: 1-chlorocyclohexene 2 2 Numbering was started from the wrong end of the structure. orrect name: trans-2-hexene 11. (a) trans-6-chloro-3-heptene; 4,4-dimethyl-2-pentyne; 2-ethyl-1-pentene. 12. (a) 3-phenyl-1-butyne; 2-methyl-2-hexene; cis-3,4-dimethyl-3-hexene. 13. All the hexynes, hexyne 2-hexyne 3-hexyne 2 3-methyl-1-pentyne 4-methyl-2-pentyne 2 4-methyl-1-pentyne 3,3-dimethyl-1-butyne 14. All the pentynes, pentyne 2-pentyne 3-methyl-1-butyne 15. (a) 1-butyne (The smallest numbered carbon that is involved in the triple bond is used to number the triple bond.); 2-hexyne (The parent chain is the longest continuous carbon chain that contains the triple bond.); propyne (The triple bond has only one possible location in propyne and no number is needed.)

7 16. (a) 4-methyl-2-pentyne (The parent chain is numbered from the end closest to the triple bond.); 4-methyl-2-hexyne (The parent chain is the longest continuous carbon chain that contains the triple bond.); 3-bromo-1-butyne (The parent chain is numbered from the end closest to the triple bond.). 17. is-trans isomers exist for 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. 18. is-trans isomers exist for 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) I 2 I O + " 2 O 2 Pt, 25 2 ± 2 1 atm (e) + KMnO 4 2 O 99: cold O O 20. (a) O + " O ±

8 (e) KMnO 4 2 O 99: cold 2 2 O O 21. (a) + 2 (1 mole) Two-step reaction: + 2 " (1 mole) Pt 25 " (a) Pt, (1 mol) " 1 atm (2 mol) 2 2 Two step reaction: + 2 " When cyclohexene, reacts with: (a) the product is 1,2-dibromocyclohexane I I, the product is iodocyclohexane + the product is O cyclohexanol O the product is cyclohexene glycol or O 1,2-dihydroxycyclohexane 24. When cyclopentene, reacts with: (a) the product is 1,2-dichlorcyclopentane, the product is bromocyclopentane

9 the product is cyclopentane + O the product is cyclopentanol 25. (a) Two possible alkenes will yield the same product, 2 ( ) 2 (2-methyl-1-butene); and (2-methyl-2-butene) (3,3-dimethyl-1-butene) (2-methyl-1-propene) 3 (3,6-dimethylcyclohexene) 26. (a) 2-hexyne 3,3-dimethyl-1-butyne 4,4-dimethyl-1-pentyne 2-butyne 27. (a) (a)

10 O 29. (a) NO 2 O (a) NO 2 2 NO (a) I N

11 32. (a) 2 OO NO 2 NO 2 NO (a) bromodichlorobenzenes 3-bromo-1,2-dichlorobenzene 4-bromo-1,2-dichlorobenzene 2-bromo-1,3-dichlorobenzene 4-bromo-1,3-dichlorobenzene 5-bromo-1,3-dichlorobenzene 2-bromo-1,4-dichlorobenzene

12 The toluene derivatives of formula 9 12 : 1,2,3-trimethylbenzene 1,2,4-trimethylbenzene 3 1,3,5-trimethylbenzene 2 o-ethyltoluene 2 m-ethyltoluene 2 p-ethyltoluene 34. (a) trichlorobenzenes 1,2,3-trichlorobenzene 1,2,4-trichlorobenzene 1,3,5-trichlorobenzene The benzene derivatives of formula 8 10 : o-xylene or 1,2-dimethylbenzene m-xylene or 1,3-dimethylbenzene 2 3 p-xylene or 1,4-dimethylbenzene ethylbenzene 35. (a) p-chloroethylbenzene p-bromophenol propylbenzene (e) triphenylmethane m-nitroaniline

13 36. (a) styrene isopropylbenzene m-nitrotoluene (e) 2,4,6-tribromophenol 2,4-dibromobenzoic acid 37. (a) 2-chloro-1,4-diiodobenzene o-dibromobenzene m-chloroaniline 38. (a) 2,3-dibromophenol 1,3,5-trichlorobenzene m-iodotoluene 39. (a) Fe 3 ± 2 ± bromobenzene NO 2 SO 2 4 ± NO 3 ± 2 O 1,4-dimethyl-2-nitrobenzene 40. (a) ± +KMnO 4 2 O Al 3 isopropylbenzene OO + benzoic acid

14 41. When reacts with, two products are possible: 2 2 and 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 solution; cyclohexane will not. (2) In the absence of sunlight, hexene will react with and decolorize bromine; cyclohexane will not trans -3,4-dimethyl-2-hexene cis-3,4-dimethyl-2-hexene trans-3,5-dimethyl-2-hexene cis-3,5-dimethyl-2-hexene 3 trans -4,5-dimethyl-2-hexene cis-4,5-dimethyl-2-hexene

15 43. (cont.) 2 2 trans-4,4-dimethyl-2-hexene cis-4,4-dimethyl-2-hexene 2 2 trans-3,5,5-trimethyl-2-hexene cis-3,5,5-timethyl-2-hexene trans -3,4-dimethyl-3-hexene cis-3,4-dimethyl-3-hexene 2 2 trans -2,4-dimethyl-3-hexene cis-2,4-dimethyl-3-hexene trans-2,5-dimethyl-3-hexene cis-2,5-dimethyl-3-hexene

16 43. (cont.) 2 trans -2,2-dimethyl-3-hexene 2 cis-2,2-dimethyl-3-hexene 44. (a) 2 primary carbocation (a positively charged carbon bonded to one other carbon) secondary carbocation (a positively charged carbon bonded to two other carbons) tertiary carbocation (a positively charged carbon bonded to three other carbons) 45. (a) + 2 O O + 2 O O + 2 O O

17 + 2 O O 46. The reaction mechanism by which benzene is brominated in the presence of Fe 3 : (a) Fe Fe Formation of a bromonium ion an electrophile. ± ± ± The bromonium ion adds to benzene forming a carbocation intermediate. ± ± Fe 4 ± Fe 3 ± A hydrogen ion is lost from the carbocation forming the product bromobenzene. 47. (a) 2 " " Yes, there will be a color change (loss of color). The fact that there is no formed indicates that the reaction is not substitution but addition. Therefore, 4 8 must contain a carbon-carbon double bond. Three structures are possible Baeyer test: Add solution to each sample. The will lose its purple color with 1-heptene. There will be no reaction (no color change) with heptane. 50. I would not expect graphene to react easily with in an addition reaction. Graphene is composed of sheets of aromatic rings that do not easily undergo addition reactions

18 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 (a) 2 2 sp 2 sp 2 3 all sp sp sp sp 3 sp 3 3 sp 3 all sp (a) Four isomers: butene 2-butene (cis and trans) 2 ( ) 2 2-methyl-1-propene

19 One isomer 5 8 cyclopentene 55. 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. arbon-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 alkene alkane alkyne or cycloalkene 58. hemically distinguishing between benzene, 1-hexene, and 1-hexyne Step 1. Add solution to a sample of each liquid. Benzene is the only one in which the 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) " 3 2 " 3 2 " " "