UNSATURATED HYDROCARBONS

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1 C20 09/16/ :47:28 Page 295 CHAPTER 20 UNSATURATED HYDROCARBONS SOLUTIONS TO REVIEW QUESTIONS 1. The sigma bond in the double bond of ethene is formed by the overlap of two sp 2 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 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 butene has a much lower melting point ( 185 C) than 2-methylpropene ( 14 C) although their boiling 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 2, but rather a hydrogen atom is replaced by a chlorine atom. C 6 H 6 þ 2! Fe C 6 H 5 þ H

2 C20 09/16/ :47:28 Page 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 benzene molecule must be planar. 11. According to the mechanism, the addition of an unsymmetrical module such as HX adds to a carboncarbon 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. H þ adds to C-1 forming the more stable cation, C þ HCH 2 A then adds to C-2 to form the product, CHCH chloropropene meets the requirement of two different groups on each carbon atom of the carboncarbon-double bond to have cis-trans isomers. 2-chloropropene does not meet this requirement. H H C C C C C C H H H H trans-1-chloropropene cis-1-chloropropene 2-chloropropene 15. They are the same compound. Both structures have the same name, 2,3-dimethylcyclohexene

3 C20 09/16/ :47:28 Page 297 SOLUTIONS TO EXERCISES 1. (a) 2. (a) 3. CH sp3 3 CH sp2 CH sp2 CH sp HC sp C sp sp 3 sp 3 sp 3 CH 2 CH 2 5. Isomeric iodobutenes, C 4 H 7 I

4 C20 09/16/ :47:29 Page (a) 7. (a) CH CCH CH 3-penten-1-yne (d) (e) (f) 8. (a) (d)

5 C20 09/16/ :47:29 Page 299 (e) (f) 9. (a) 23 sigma bonds, 1 pi bond; 17 sigma bonds, 1 pi bond; 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; 27 sigma bonds, 7 pi bonds; 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; 4,4-dimethyl-2-pentyne; 2-ethyl-l-pentene. 12. (a) 3-phenyl-l-butyne; 2-methyl-2-hexene; cis-3,4-dimethyl-3-hexene. 13. All the hexynes, C 6 H 10 CH 2 CH 2 CH 2 C CH CH 2 CH 2 C C CH 2 C CCH 2 1-hexyne 2-hexyne 3-hexyne 14. All the pentynes, C 5 H 8 CH 2 CH 2 C CH 1-pentyne CH 2 C C 2-pentyne 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.) 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-l-butyne (The parent chain is numbered from the end closest to the triple bond.). 17. Cis-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

6 C20 09/16/ :47:30 Page Cis-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) cis neither trans 20. (a) trans trans neither 21. (a) CH 2 CH 2 CH CH 2 þ Br 2! CH 2 CH 2 CHBrCH 2 Br (d) (e) 22. (a) CH 2 CH CH þ HBr! CH 2 CHBrCH 2 þ CH 2 CH 2 CHBr (CH 2 CH þ Br 2! CH 2 BrCHBr (d) (e) 23. (a) C C þ Br 2 ð1 moleþ! CBr CBr Two-step reaction: H CH CH þ H! CH 2 CH! CH 2 CH 2 CH 2 C CH þ H 2 ð1 moleþ! 3CH 25 2 CH 2 CH CH 2 C

7 C20 09/16/ :47:30 Page 301 Pt; (a) C CH þ H 2 ð1 molþ! C 1 atm CH CH 2 C C þ Br 2 ð2 molþ! CBr 2 CBr 2 Two-step reaction: C CH þ H! C ¼ CH 2! H C When cyclohexene, reacts with: (a) Br 2, the product is Br 1,2-dibromocyclohexane Br HI, the product is iodocyclohexane H 2 O, H þ, the product is cyclohexanol (d) KMnO 4 (aq), the product is cyclohexene glycol or 1,2-dihydroxycyclohexane 26. When cyclopentene, reacts with: (a) 2, the product is 1,2-dichlorocyclopentane HBr, the product is bromocyclopentane H 2, Pt, the product is cyclopentane 27. (d) H 2 O, H þ, the product is cyclopentanol (a)

8 C20 09/16/ :47:30 Page (a) 29. (a) meta meta ortho 30. (a) para meta para O 31. (a) CH (d) CH CH3 OH NO 2 (e) Br NH 2 Br CH 2 (f)

9 C20 09/16/ :47:31 Page 303 O 32. (a) COH (d) CH CH 2 Br (e) NO 2 NO 2 (f) 33. (a) bromodichlorobenzenes

10 C20 09/16/ :47:31 Page 304 The toluene derivatives of formula C 9 H 12 : 1,2,3-trimethylbenzene 1,2,4-trimethylbenzene H 3 C 1,3,5-trimethylbenzene CH2 o-ethyltoluene 34. (a) trichlorobenzenes CH 2 m-ethyltoluene CH 2 p-ethyltoluene 1,2,3-trichlorobenzene 1,2,4-trichlorobenzene The benzene derivatives of formula C 8 H 10 : 1,3,5-trichlorobenzene o-xylene or 1,2-dimethylbenzene m-xylene or 1,3-dimethylbenzene CH 2 H 3 C p-xylene or 1,4-dimethylbenzene ethylbenzene 35. (a) p-chloroethylbenzene (d) p-bromophenol propylbenzene (e) triphenylmethane m-nitroaniline 36. (a) styrene (d) isopropylbenzene m-nitrotoluene (e) 2,4,6-tribromophenol 2,4-dibromobenzoic acid

11 C20 09/16/ :47:31 Page (a) 2-chloro-1,4-diiodobenzene o-dibromobenzene m-chloroaniline 38. (a) 2,3-dibromophenol 1,3,5-trichlorobenzene m-iodotoluene 39. (a) 40. (a) 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 KMnO 4 solution; cyclohexane will not. (2) In the absence of sunlight, hexene will react with and decolorize bromine; cyclohexane will not

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13 C20 09/16/ :47:32 Page (Cont.) 44. (a) C þ H 2 primary carbocation (a positively charged carbon bonded to one other carbon) C þ H secondary carbocation (a positively charged carbon bonded to two other carbons) tertiary carbocation (a positively charged carbon bonded to three other carbons) 45. (a)

14 C20 09/16/ :47:33 Page 308 (d) 46. The reaction mechanism by which benzene is brominated in the presence of FeBr 3 : (a) FeBr 3 þ Br 2! FeBr 4 þ Brþ Formation of a bromonium ion (Br þ ), an electrophile. The bromonium ion adds to benzene forming a carbocation intermediate. A hydrogen ion is lost from the carbocation forming the product bromobenzene. H 47. (a) CHCH 2! CH 2 CHCH 2 þ CH CH H CH 2 CH 2 CH 2 CH 2! CH 2 CHCH 2 CH Yes, there will be a color change (loss of Br 2 color). The fact that there is no HBr formed indicates that the reaction is not substitution but addition. Therefore, C 4 H 8 must contain a carbon-carbon double bond. Three structures are possible. 49. Baeyer test: Add KMnO 4 solution to each sample. The KMnO 4 will lose its purple color with 1-heptene. There will be no reaction (no color change) with heptane

15 C20 09/16/ :47:33 Page I would not expect graphene to react easily with H in an addition reaction. Graphene is composed of sheets of aromatic rings that do not easily undergo addition reactions 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) HC sp C sp sp (a) Four isomers: CH 2 CHCH 2 C 4 H 8 1-butene CH 2 ¼ Cð Þ 2 2-methyl-1-propene One isomer C 5 H 8 CH ¼ CH 2-butene (cis and trans)

16 C20 09/16/ :47:34 Page 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 alkene alkane (d) alkyne or cycloalkene 58. Chemically distinguishing between benzene, 1-hexene, and 1-hexyne Step 1. Add KMnO 4 solution to a sample of each liquid. Benzene is the only one in which the KMnO 4 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) C CH! H C CH 2! HBr CBr C CH! Br 2 CBr CHBr! H CBrCH 2 Br C CH! 2 2 C 2 CH

3. What number would be used to indicate the double bond position in the IUPAC name for CH 3 CH 2 CH=CH CH 3 a. 1 b. 2 c. 3 d.

3. What number would be used to indicate the double bond position in the IUPAC name for CH 3 CH 2 CH=CH CH 3 a. 1 b. 2 c. 3 d. Chapter 2 Unsaturated Hydrocarbons MULTIPLE CHOICE 1. Name a difference between a saturated and an unsaturated hydrocarbon. a. Saturated hydrocarbons are composed of only carbon and hydrogen, and unsaturated