有机化学-高鸿宾-第四版-答案-第十四章-二羰基

第十四章β-二羰基化合物(一)命名下列化合物：(1)HOCH2CHCH2COOHCH3(2)(CH3)2CHCCH2COOCH3O(3)CH3CH2COCH2CHO(4)(CH3)2C=CHCH2CHCH3OH(5)ClCOCH2COOH(6)CHOOCH3OH(7)OCH3NO2(8)CH2CH2OHCl解：(1)3-甲基-4-羟基丁酸(2)4-甲基-3-戊酮酸甲酯(3)3-氧代戊醛or3-戊酮醛(4)5-甲基-4-己烯-2-醇(5)丙二酸单酰氯or氯甲酰基乙酸(6)4-羟基-3-甲氧基苯甲醛(7)2-硝基苯甲醚(8)2-间氯苯乙醇or2-(3-氯苯基)乙醇(二)下列羧酸酯中，那些能进行酯缩合反应？写出其反应式。(1)甲酸乙酯(2)乙酸正丁酯(3)丙酸乙酯(4)2,2-二甲基丙酸乙酯(5)苯甲酸乙酯(6)苯乙酸乙酯解：(2)、(3)、(6)能进行酯缩合反应。反应式如下：(2)2CH3COO(CH2)3CH3CH3CCH2CO(CH2)3CH3OO(1)C2H5ONa(2)H+(3)2CH3CH2COOCH2CH3CH3CH2CCHCOC2H5OOCH3(1)C2H5ONa(2)H+(6)(1)C2H5ONa(2)H+CH2COOC2H5CH2CCHCOOC2H5O(三)下列各对化合物，那些是互变异构体？那些是共振杂化体？(1)CH3CCHCCH3OHOCH3CCHCH3COHO和(2)CH3CO-OH3CCO-O和(3)和CH2=CHCH=CH2CH2CH=CHCH2(4)和OOH解：(1)、(4)两对有氢原子核相对位置的移动，是互变异构体，(2)、(3)两对中只存在电子对的转移，而没有原子核相对位置的移动，是共振杂化体。(四)写出下列化合物分别与乙酰乙酸乙酯钠衍生物作用后的产物。(1)烯丙基溴(2)溴乙酸甲酯(3)溴丙酮(4)丙酰氯(5)1,2-二溴乙烷(6)α-溴代丁二酸二甲酯解：(1)CH3CCHCOC2H5OOCH2CH=CH2(2)CH3CCHCOC2H5OOCH2COOCH3(3)CH3CCHCOC2H5OOCH2COCH3(4)CH3CCHCOC2H5OOCOC2H5(5)CH3CCHCOC2H5OOH2CCH2CH3CCHCOC2H5OO(6)CH3CCHCOC2H5OOCHCH2COOCH3COOCH3(五)以甲醇、乙醇及无机试剂为原料，经乙酰乙酸乙酯合成下列化合物。(1)3-甲基-2-丁酮(2)2-己醇(3)α,β-二甲基丁酸(4)γ-戊酮酸(5)2,5-己二酮(6)2,4-戊二酮解：先合成乙酰乙酸乙酯CH3CH2OHKMnO4CH3COOHC2H5OHCH3COC2H5O2CH3COC2H5OCH3CCH2COC2H5OOH+（以下同）(1)C2H5ONa(2)CH3COOH(1)分析：CH3CHCH3COCH3来自三乙上2个CH3I解：CH3CCHCOC2H5CH3OOCH3CCH2COC2H5OOCH3IC2H5O-Na+CH3IC2H5O-Na+CH3CHCH3COCH3CH3CCCOC2H5OOCH3CH3(1)稀OH-,(2)H+,(3)酮式分解CH3OHCH3IHI(2)分析：CH3CHCH2CH2CH2CH3OHH2NiCH3CCH2CH2CH2CH3O来自三乙上BrCH2CH2CH3解：CH3CCH2COC2H5OOCH3CCHCOC2H5CH2CH2CH3OOC2H5O-Na+CH3CH2CH2Br(1)5%NaOH(2)H+(3)-CO2CH3CCH2CH2CH2CH3OCH3CHCH2CH2CH2CH3OHH2Ni（成酮分解）正丙基溴可由所给的甲醇和乙醇变化而得：CH3OHCrO3H2SO4H2C=OCH3CH2OHCH3CH2BrCH3CH2MgBrCH2O绝对乙醚H+Mg乙醚HBrCH3CH2CH2OHCH3CH2CH2BrHBr(3)分析：CH3CHCHCOOHCH3CH3来自三乙成酸分解上CH3CHBrCH3上CH3I解：CH3CCH2COC2H5OOC2H5ONaCH3CCHCOC2H5OO-Na+CH3CHCH3BrCH3CCHCOC2H5CH(CH3)2OOC2H5O-Na+-CH3CCCOC2H5Na+OOCH(CH3)2CH3CCCOC2H5CHCH3CH3CH3OO（成酸分解）CH3CHCHCOOHCH3CH3(1)40%NaOH(2)H+(3)-CO2CH3I(4)分析：CH3CCH2CH2COOHO上BrCH2COOCH2CH3来自三乙解：CH3CCH2COC2H5OOC2H5ONaCH3CCHCOC2H5Na+OO-BrCH2COOC2H5CH3CCHCOC2H5CH2COC2H5OOO(1)5%NaOH(2)H+(3)-CO2（成酮分解）CH3CCH2CH2COOHO(5)分析：CH3CCH2OCH2CCH3O来自三乙来自三乙用I2偶联解：2CH3CCH2COC2H5OO2C2H5ONa2CH3CCHCOC2H5Na+OO-CH3CCH2CH2CCH3OOCHCC2H5CH3COOCHCC2H5CH3COOI2(1)5%NaOH(2)H+(3)-CO2（成酮分解）(6)分析：CH3CCH2OCCH3O来自三乙上ClCCH3O解：CH3CCH2COC2H5OO+NaHCH3CCHCOC2H5Na+OO-+H2(此处用NaH取代醇钠，是为了避免反应中生成的醇与酰氯作用)CH3CClOCH3CCHCOC2H5OOCOCH3CH3CCH2CCH3OO(六)完成下列反应：解：红色括号中为各小题所要求填充的内容。(1)丙酸乙酯＋乙二酸二乙酯(1)C2H5ONa(2)H+CHCOOC2H5CH3CCC2H5OOO(2)乙酸乙酯＋甲酸乙酯(1)C2H5ONa(2)H+HCCH2COOC2H5O(3)苯甲酸乙酯＋丁二酸二乙酯　(1)C2H5ONa(2)H+CCHCH2COOC2H5COOC2H5O(4)(1)C2H5ONa(2)H+CH3C(CH2)4COC2H5OOOCCH3O(5)(1)C2H5ONa(2)H+CH3C(CH2)3COC2H5OOOO(6)(1)C2H5ONa(2)H+CH2CH2CH2COOC2H5CH2CH2COOC2H5OCOOC2H5(7)(1)C2H5ONa(2)H+CH3CH2OOCCH2CH2CHCHCH3COOC2H5COOC2H5CH3C2H5O2COCOOC2H5(七)用丙二酸二乙酯法合成下列化合物：(1)CH3CHCHCOOHCH3CH3(2)HOOCCH2CHCH2COOHCH3(3)HOOCCH2CHCH2CH2COOHCH3(4)OCH3O(5)COOH(6)COOHHOOC(1)分析：CH3CHCHCOOHCH3CH3来自丙二上CH3I上CH3CHBrCH3解：CH2(COOC2H5)2(CH3)2CHCH(COOC2H5)2C2H5O-Na+C2H5O-Na+(CH3)2CHClCH3I(CH3)2CHC(COOC2H5)2CH3(1)5%NaOH(2)H+(3)-CO2(CH3)2CHCHCOOHCH3(2)目标化合物为1,5-二羰基化合物，可经Michael加成制得：CHCH3CH2COOHHOOCCH212345来自丙二C2H5OCCH=CHOCH3上解：CH3CH=CHCOOC2H5CH2(COOC2H5)2C2H5O-Na+CH3CHCH2COOC2H5CH(COOC2H5)2(1)NaOH/H2O(2)H+(3)-CO2CH3CHCH2COOHCH2COOHHOOCCH2CHCH2COOHCH3(3)分析：HOOCCH2CHCH2CH2COOHCH3来自“丙二”来自“丙二”解：2CH2(COOC2H5)22C2H5O-Na+2CH-(COOC2H5)2Na+CH3CHCH2BrBr(C2H5OOC)2CHCHCH2CH(COOC2H5)2CH3(1)NaOH/H2O(2)H+(3)-CO2HOOCCH2CHCH2CH2COOHCH3(4)目标化合物是内酯，应首先制备4-羟基戊酸：CH3CHCH2CH2COOHOH来自丙二上CH2CHOCH3OCH3OH+解1：CH2(COOC2H5)2C2H5O-Na+OCH3CH-(COOC2H5)2CH3CHCH2CH(COOC2H5)2O-(1)NaOH/H2O(2)H+(3)-CO2CH3CHCH2CH2COOHOHH+OCH3O或者，将4-戊酮酸还原得到4-羟基戊酸：来自丙二OCH3OCH3CCH2OCH2COOHCH3CHCH2CH2COOHOH内酯-羟基酸上溴代丙酮H2NiH+解2：CH2(COOC2H5)2CH3CCH2CH(COOC2H5)2OC2H5O-Na+CH3COCH2Br(1)NaOH/H2O(2)H+(3)-CO2CH3CCH2CH2COOHOH2/NiCH3CHCH2CH2COOHOHH+OCH3O(5)目标分子可看作乙酸的二烷基取代物：来自丙二COOHCH2CH2ClCH2CH2Cl上解：CH2(COOC2H5)22C2H5O-Na+CH2CH2CH2CH2BrBrCOOC2H5COOC2H5COOH(1)NaOH/H2O(2)H+(3)-CO2(6)分析：HOOCCOOH上BrCH2CH2Br来自丙二来自丙二上CH2Br2解：2CH2(COOC2H5)22C2H5O-Na+2CH-(COOC2H5)2Na+CH2CH2BrBr(C2H5OOC)2CHCH2CH2CH(COOC2H5)22C2H5O-Na+CH2BrBrCOOHHOOC(1)NaOH/H2O(2)H+(3)-CO2COOC2H5COOC2H5C2H5OOCC2H5OOC或者：HOOCCOOH12345来自丙二上BrCH2CH2Br上CH2=CHCNorCH2=CHCOOC2H5CH2(COOC2H5)2C2H5O-Na+CH2=CHCOOC2H5CH2CH2COOC2H5CH(COOC2H5)22C2H5O-Na+Micheal加成CH2CH2BrBr(1)NaOH/H2O(2)H+(3)-CO2COOC2H5C2H5OOCC2H5OOCCOOHHOOC(八)写出下列反应的机理：OCH3COOCH3NaOCH3CH3OHH3O+OCOOCH3CH3解：OCH3COOCH3+CH3O-O-OCH3CH3COOCH3OOCH3CH3COOCH3CH3OCCHCH2CH2CHCOCH3OOCH3O-OCH3COOCH3CH3COOCH3CH3O(九)完成下列转变：(1)OCOOC2H5COOC2H5CH2CH3(2)CH2COOC2H5CH2COOC2H5OO(3)CH2COOEtCOOEtOO(4)CH3COCH2COOC2H5CH3CHCHCH2CCH3OHC2H5OHCH3(5)COOEtOCH2CH2CNO(6)CH3COCH2COOC2H5(CH3C)2CHCH2PhO(7)OCH3CHOOCH3OO解：(1)分析：CH2=PPh3COOC2H5CH3O不饱和酮，由羟醛缩合来OH-COOC2H5CH3OO123451,5-二羰基化合物，由Michael加成来CH3OCOOC2H5OH+EtO-由Witting反应引入COOC2H5CH3CH2解：C2H5OHOCOOC2H5+OC2H5O-Na+Micheal加成OCOOC2H5CH2CH2CCH2CH3ONaOHCOOC2H5CH3OPh3P=CH2Witting反应COOC2H5CH3CH2Robinson增环(2)分析：COOC2H5COOC2H5OOOOCOOC2H5COOC2H5HCOC2H5OC2H5OOCHCOOC2H5OOC2H5OOCC2H5O-C2H5O-成酮分解成酮分解CCOOOC2H5OC2H5COOC2H5COOC2H5HH解：CH2COOC2H5CH2COOC2H54C2H5ONaOCOO