19 09/16/2013 13:54:37 Page 283 APTER 19 ORGANI EMISTRY: SATURATED YDROARBONS SOLUTIONS TO REVIEW QUESTIONS 1. Two of the major reasons for the large number of organic compounds is the ability of carbon to form short or very long chains of atoms covalently bonded together and isomerism. 2. The carbon atom has only two unshared electrons, making two covalent bonds logical, but in 4, carbon forms four equivalent bonds. Promoting one 2s electron to the empty 2p orbital would make four bonds possible, but without hybridization, we could not explain the fact that all four bonds in 4 are identical, and the bond angles are equal (109.5 ). 3. The first ten normal alkanes: methane 4 hexane 6 14 ethane 2 6 heptane 7 16 propane 3 8 octane 8 18 butane 4 10 nonane 9 20 pentane 5 12 decane 10 22 4. Aromatic hydrocarbons contain benzene rings. Aliphatic refers to all other hydrocarbons including alkanes, alkenes, alkynes, and cycloalkanes. 5. Advantages to alkyl halide anesthetics include nonflammability, a pleasant odor, and a much reduced hepatotoxicity 6. yclopropane is more reactive than cyclohexane because cyclopropane s carbon carbon bond angles are substantially smaller than the normal tetrahedral angle. 7. (a) A substitution reaction allows an exchange of atoms or groups of atoms between reactants while in an elimination reaction a single reactant is split into two products. Two reactants combine together in an addition reaction while one reactant is split into two products in an elimination reaction. 8. Alkanes are hydrocarbons containing only the elements hydrogen and carbon. Alkyl halides also contain halogens and, thus, cannot be classified as alkanes. 9. E85 is a gasoline that contains 85% ethanol and 15% petroleum. This mixture reduces the use of petroleum, a nonrenewable resource. E85 is the cleanest burning gasoline now available. 10. Renewable gasoline differs in several ways from the gasoline we buy at the pump today: (1) Renewable gasoline is produced by bacteria instead of being pumped from the ground; (2) Renewable gasoline differs in composition depending on which bacteria are used. - 283 -
19 09/16/2013 13:54:37 Page 284 - hapter 19 - SOLUTIONS TO EXERISES 1. Lewis structures: l (a) 4 l 2 2. Lewis structures: (a) 2 l 6 l l l l l l 4 3 8 5 12 3. The formulas (a),, (f), and (g) represent isomers. 4. The same compound is represented by formulas (a),, and (f). 5. The number of methyl groups in each formula in Exercise #3 is as follows: (a) 2 2 3 (d) 4 (e) 1 (f) 3 (g) 4 (h) 3 6. The number of methyl groups in each formula in Exercise #4 is as follows: (a) 3 2 3 (d) 4 (e) 4 (f) 3 7. Isomers of heptane 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2-284 -
19 09/16/2013 13:54:37 Page 285 - hapter 19-8. Isomers of hexane 2 2 2 2 2 2 2 2 2 9. (a) 2 l 2 ; one 2 l 2 3 7 Br; two 2 2 Br Br 3 6 l 2 ; four 2 l 2 l 2 l 2 l 2 2 l l 2 (d) 4 8 2 nine 2 2 l 2 2 l 2 l l 2 2 l 2 l 2 2 2 l 2 l 2 ll 2 l 2 l 2 l 2 2 l l 10. (a) Br, one Br 2 5 l, one 2 l 4 9, four 2 2 2 I 2 I I 2 I (d) 3 6 Brl; five 2 Brl l 2 Br Br 2 l 2 l 2 2 Br Brl 11. (a) 5 6 5 12. (a) 7 7 6 13. IUPA names (a) 1-chloropropane 2-chloropropane 2-chloro-2-methylpropane (d) 2-methylbutane (e) 2,3-dimethylhexane 14. IUPA names (a) chloroethane l-chloro-2-methylpropane 2-chlorobutane (d) methylcyclopropane (e) 2,4-dimethylpentane - 285 -
19 09/16/2013 13:54:37 Page 286 - hapter 19-15. The tertiary carbons are circled in the following structures from Exercise #11: (a) 2 2 2 16. The tertiary carbons are circled in the following structures from Exercise #12: 2 (a) 2 2 2 2 2 2 2 3 2-286 -
19 09/16/2013 13:54:37 Page 287 2 - hapter 19-17. Structural formulas: 3 3 (a) 2,4-dimethylpentane 2 2, 2-dimethylpentane 2 2 3-isopropyloctane 2 2 2 2 2 ( ) 2 3 (d) (e) 5,6-diethyl-2,7-dimethyl-5-propylnonane 3 2 3 2 2 2 2 2 2 2-ethyl-1,3-dimethylcyclohexane 18. (a) 4-ethyl-2-methylhexane 2 2 2 4-t-butylheptane 2 2 2 2 ( ) 3 4-ethyl-7-isopropyl-2,4,8-trimethyldecane 2 2 2 2 2-287 -
19 09/16/2013 13:54:37 Page 288 - hapter 19-2 (d) 3-ethyl-2,2-dimethyloctane 2 2 2 2 (e) 1,3-diethylcyclohexane 2 2 19. (a) 3-methylbutane 2 Numbering was done from the wrong end of the molecule. The correct name is 2-methylbutane. 2-ethylbutane 2 2 The name is not based on the longest carbon chain (5 carbons). The correct name is 3-methylpentane. 2-dimethylpentane. Each methyl group needs to be numbered. Depending on the structure, the correct name is 2,2-dimethylpentane; 2,3-dimethylpentane; or 2,4-dimethylpentane. (d) 1,4-dimethylcyclopentane The ring was numbered in the wrong direction. The correct name is 1,3-dimethylcyclopentane. 20. (a) 3-methyl-5-ethyloctane 2 2 2 2 2 Ethyl should be named before methyl (alphabetical order). The numbering is correct. The correct name is 5-ethyl-3-methyloctane. 2 3,5,5-triethylhexane 2 2 2 2 The name is not based on the longest carbon chain (7 carbons). The correct name is 3,5-diethyl- 3-methylheptane. - 288 -
19 09/16/2013 13:54:38 Page 289 - hapter 19-4,4-dimethyl-3-ethylheptane (d) 2 2 2 2 Ethyl should be named before dimethyl (alphabetical order). The correct name is 3-ethyl-4, 4-dimethylheptane. 1,6-dimethylcyclohexane The ring was numbered in the wrong direction. The correct name is 1,2-dimethylcyclohexane. 21. (a) butane, 2 2 2-methylpropane, 22. (a) 2-methylbutane, 2 2,2-dimethylbutane, 2 23. An isomer is shown in. 24. Isomers are shown in (a) and. 25. (a) 2-chlorobutane 2-chloro-2-methylpropane 26. (a) 2-bromo-2,3-dimethylbutane 2-bromo-2-methylbutane 27. (a) ( ) 2 2 ( ), 2,5-dimethylhexane 1,2-dimethylcyclohexane 28. (a) ( ) 2, 2,2-dimethylpropane 2 2 2, pentane - 289 -
19 09/16/2013 13:54:38 Page 290 - hapter 19-29. (a) Five isomers: 2 l 2-chloro-4-methylpentane 2 2 2l 1-chloro-4-methylpentane 2 l 2 2 1-chloro-2-methylpentane 3 l 2 2 2-chloro-2-methylpentane l 2 3-chloro-2-methylpentane Five isomers: 2 l l l chloromethylcyclohexane 1-chloro-1-methylcyclohexane 2-chloro-1-methylcyclohexane l 3-chloro-1-methylcyclohexane l 4-chloro-1-methylcyclohexane 30. (a) Three isomers: Br Br 1,1-dibromocyclopentane Br Br 1,2-dibromocyclopentane Br Br 1,3-dibromocyclopentane - 290 -
19 09/16/2013 13:54:38 Page 291 - hapter 19 - Seven isomers: Br 2 2 1,1-dibromo-2,2-dimethylbutane 2 Brr 2 1,1-dibromo-3,3-dimethylbutane 2 Br Br 1,3-dibromo-2,2-dimethylbutane Br 2 Br 1,2-dibromo-3,3-dimethylbutane Br 2 2,2-dibromo-3,3-dimethylbutane 2 Br 2 Br 2 1-bromo-2-bromomethyl-2-methylbutane 2 Br 2 2 Br 1,4-dibromo-2,2-dimethylbutane 31. Names: (a) l-chloro-2-ethylcyclohexane l-chloro-3-ethyl-l-methylcyclohexane 1,4-diisopropylcyclohexane 32. Structural formulas: (a) 2 2 l 2 2 l - 291 -
19 09/16/2013 13:54:38 Page 292 - hapter 19-33. (a) chair chair 34. (a) chair boat 35. The formula for dodecane is 12 26. 36. F 2 2 F þ l 2!F 2 1F þ 1 37. Data: 1 gal 60 mi ; 60 mi traveled; 19 mol 8 18 ;T¼293K gal 2 8 18 þ 25 O 2!16 O 2 þ 18 2 O 1 gal 60 mi ¼ 1 gal gasoline used 60 mi 19 mol 8 18 16 mol O 2 2 mol 8 18 ¼ 1:5 10 2 mol O 2 PV ¼ nrt V ¼ nrt P V ¼ ð1:5 102 mol O 2 Þð0:0821 L atmþð293 KÞ 1 atm mol-k ¼ 3:6 10 3 LO 2 38. (a) elimination substitution addition 39. It is not possible to distinguish hexane from 3-methylheptane based on solubility in water because both compounds are nonpolar and, thus, insoluble in water. 40. (a) The compounds are isomers. The compounds are not the same and are not isomers. The compounds are not the same and are not isomers. (d) The compounds are not the same and are not isomers. 41. (a) Initiation: l 2! uv 2l Propagation: l þ l! 2 l þ l 2 l þ l 2! 2 l 2 þ l Termination: l þ l! l 2 2 l þ 2 l! 2 l 2 l 2 l þ l! 2 l 2-292 -
19 09/16/2013 13:54:38 Page 293 - hapter 19 - Initiation: l 2 uv 2 l Propagation: l + ( ) 3 + l Termination: ( ) 3 + l 2 l + l l 2 ( ) 3 l + l ( ) 3 + ( ) 3 ( ) 3 ( ) 3 ( ) 3 + l ( ) 3 l Initiation: l 2 uv 2 l Propagation: l + + l l + l 2 + l Termination: l + l l 2 + + l l (d) Initiation: l 2 uv 2 l Propagation: l + 3 + l + l 2 l + l Termination: l + l l 2 + + l l 42. Using a high mole ratio of methane to chlorine will allow a chlorine free radical to react with a methane molecule rather than a chloromethane molecule and minimize the formation of di-, tri-, and tetrachloromethane. 43. (a) 2 2 2 2 2 2 2 2 2 undecane 2 2 2 2 2 2 2 2 2 2 2 tridecane Both compounds are alkanes. They are composed of only carbon and hydrogen and have only single bonds between the carbon atoms. Both formulas agree with the general formula for alkanes, n 2nþ2. - 293 -
19 09/16/2013 13:54:38 Page 294 - hapter 19-44. yclopentane and cycloheptane are in the same homologous series (cycloalkanes) and share the same general formula, n 2n. The formula for cycloheptane is 7 14. F F F F 45. (a) R-32 F R-125 F R-134a F F F F F R-32, four sigma bonds; R-125, seven sigma bonds; R-134a, seven sigma bonds. 46. (a) 2-methylpentane; 2,4-dimethylheptane; 4-ethyl-3-methyloctane. 47. (a) 48. A mixture. A hydrocarbon of the formula, 4 10, can have two possible structures: (I) 2 2 and (II) ( ). Structure I can form only two monobromo compounds: 2 Br 2 2 and Br 2 Structure II can form only two monobromo compounds: 2 Br( ). and Br( ). A mixture of the two structures gives four monobromo compounds. - 294 -