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1、_ _ _ 209 CHAPTER 9 ALKYNES SOLUTIONS TO TEXT PROBLEMS 9.1The reaction is an acidbase process; water is the proton donor. Two separate proton-transfer steps are involved. 9.2 Atriple bond may connect C-1 and C-2 or C-2 and C-3 in an unbranched chain of fi ve carbons. One of the C5H8isomers has a bra
2、nched carbon chain. 3-Methyl-1-butyne CH3CHCCH CH3 1-Pentyne CH3CH2CH2CCH 2-Pentyne CH3CH2CCCH3 ?CCHH Acetylene HO ? Hydroxide ion ?C ? CH Acetylide ionWater HHO ? Carbide ion CH Acetylide ion HO ? Hydroxide ionWater HHOC ? C ? C ? BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website 210ALK
3、YNES 9.3The bonds become shorter and stronger in the series as the electronegativity increases. NH3H2OHF Electronegativity:N (3.0)O (3.5)F (4.0) Bond distance (pm):NH (101)OH (95)FH (92) Bond dissociation energy (kJ/mol):NH (435)OH (497)FH (568) Bond dissociation energy (kcal/mol):NH (104)OH (119)FH
4、 (136) 9.4(b)Aproton is transferred from acetylene to ethyl anion. The position of equilibrium lies to the right. Ethyl anion is a very powerful base and depro- tonates acetylene quantitatively. (c)Amide ion is not a strong enough base to remove a proton from ethylene. The equilibrium lies to the le
5、ft. (d)Alcohols are stronger acids than ammonia; the position of equilibrium lies to the right. 9.5(b)The desired alkyne has a methyl group and a butyl group attached to a CC unit. Two alkylations of acetylene are therefore required: one with a methyl halide, the other with a butyl halide. It does n
6、ot matter whether the methyl group or the butyl group is introduced fi rst; the order of steps shown in this synthetic scheme may be inverted. (c)An ethyl group and a propyl group need to be introduced as substituents on a CC unit. As in part (b), it does not matter which of the two is introduced fi
7、 rst. 1. NaNH2, NH3 2. CH3CH2CH2Br Acetylene1-Pentyne3-Heptyne CHHC 1. NaNH2, NH3 2. CH3CH2Br CH3CH2CH2CCHCH3CH2CH2CCCH2CH3 1. NaNH2, NH3 2. CH3Br AcetylenePropyne2-Heptyne CHHC 1. NaNH2, NH3 2. CH3CH2CH2CH2Br CH3CCHCH3CCCH2CH2CH2CH3 ? Amide ion (stronger base) NH2 ? Ammonia (weaker acid) Ka10?36 (p
8、Ka 36) NH3 2-Butyn-1-olate anion (weaker base) CH3CCCH2O ? ? 2-Butyn-1-ol (stronger acid) Ka ? 10?16?10?20 (pKa ? 16?20) HCH3CCCH2O ? Ammonia (stronger acid) Ka 10?36 (pKa 36) NH3 Vinyl anion (stronger base) CH2CH? Amide ion (weaker base) NH2 ? Ethylene (weaker acid) Ka ? 10?45 (pKa ? 45) HCH2CH CH3
9、CH3 Ethane (weaker acid) ? Acetylide ion (weaker base) HCC ? Ethyl anion (stronger base) CH2CH3 Acetylene (stronger acid) HCCH Ka 10?26 (pKa 26) Ka ? 10?62 (pKa ? 62) BackForwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website 9.6Both 1-pentyne and 2-pentyne can be prepared by alkylating acetylene
10、. All the alkylation steps involve nucleophilic substitution of a methyl or primary alkyl halide. A third isomer, 3-methyl-1-butyne, cannot be prepared by alkylation of acetylene, because it re- quires a secondary alkyl halide as the alkylating agent. The reaction that takes place is elimination, no
11、t substitution. 9.7Each of the dibromides shown yields 3,3-dimethyl-1-butyne when subjected to double dehydro- halogenation with strong base. 9.8(b) The fi rst task is to convert 1-propanol to propene: After propene is available, it is converted to 1,2-dibromopropane and then to propyne as described
12、 in the sample solution for part (a). (c)Treat isopropyl bromide with a base to effect dehydrohalogenation. Next, convert propene to propyne as in parts (a) and (b). (d)The starting material contains only two carbon atoms, and so an alkylation step is needed at some point. Propyne arises by alkylati
13、on of acetylene, and so the last step in the synthesis is The designated starting material, 1,1-dichloroethane, is a geminal dihalide and can be used to prepare acetylene by a double dehydrohalogenation. 1. NaNH2, NH3 2. H2O CH3CHCl2 1,1-Dichloroethane HCCH Acetylene 1. NaNH2, NH3 2. CH3Br HCCH Acet
14、ylene CH3CCH Propyne (CH3)2CHBr NaOCH2CH3 PropeneIsopropyl bromide CH3CHCH2 CH3CH2CH2OH H2SO4 heat Propene1-Propanol CH3CHCH2 1. 3NaNH2 2. H2O oror(CH3)3CCCH3 Br Br 2,2-Dibromo-3,3- dimethylbutane (CH3)3CCH2CHBr2 1,1-Dibromo-3,3- dimethylbutane Br (CH3)3CCHCH2Br 1,2-Dibromo-3,3- dimethylbutane (CH3)
15、3CCCH 3,3-Dimethyl-1-butyne ? E2 CH3CHCH3 Br Isopropyl bromide HCCH Acetylene CH2CHCH3 Propene HC Acetylide ion C ? 1. NaNH2, NH3 2. CH3CH2Br Acetylene1-Butyne2-Pentyne CHHC 1. NaNH2, NH3 2. CH3Br CH3CH2CCHCH3CH2CCCH3 Acetylene1-Pentyne CHHCCHCH3CH2CH2C 1. NaNH2, NH3 2. CH3CH2CH2Br ALKYNES211 BackFo
16、rwardMain MenuTOCStudy Guide TOCStudent OLCMHHE Website (e) The fi rst task is to convert ethyl alcohol to acetylene. Once acetylene is prepared it can be alkylated with a methyl halide. 9.9 The fi rst task is to assemble a carbon chain containing eight carbons. Acetylene has two carbon atoms and can be alkylated via its sodium salt to 1-octyne. Hydrogenation over platinum converts 1-octyne to octane. Alternatively, two successive alkylations of acetylene with CH3CH2CH2Br could be car
