有机化学答案之有机十四章

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(2)CH3CO-OH3CCO-O(3)CH2=CHCH=CH2CH2CH=CHCH2(4)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+()(1)C2H5ONa(2)CH3COOH(1)分析：CH3CHCH3COCH32222CCCCHHHH3333IIII解：CH3CCHCOC2H5CH3OOCH3CCH2COC2H5OOCH3IC2H5O-Na+CH3IC2H5O-Na+CH3CHCH3COCH3CH3CCCOC2H5OOCH3CH3(1)OH-,(2)H+,(3)∆CH3OHCH3IHI(2)分析：CH3CHCH2CH2CH2CH3OHH2NiCH3CCH2CH2CH2CH3OBBBBrrrrCCCCHHHH2222CCCCHHHH2222CCCCHHHH3333解：CH3CCH2COC2H5OOCH3CCHCOC2H5CH2CH2CH3OOC2H5O-Na+CH3CH2CH2Br(1)5%NaOH(2)H+(3)∆,-CO2CH3CCH2CH2CH2CH3OCH3CHCH2CH2CH2CH3OHH2Ni()正丙基溴可由所给的甲醇和乙醇变化而得：CH3OHCrO3H2SO4H2C=OCH3CH2OHCH3CH2BrCH3CH2MgBrCH2OH+MgHBrCH3CH2CH2OHCH3CH2CH2BrHBr(3)分析：CH3CHCHCOOHCH3CH3CCCCHHHH3333CCCCHHHHBBBBrrrrCCCCHHHH3333CCCCHHHH3333IIII解：CH3CCH2COC2H5OOC2H5ONaCH3CCHCOC2H5OO-Na+CH3CHCH3BrCH3CCHCOC2H5CH(CH3)2OOC2H5O-Na+-CH3CCCOC2H5Na+OOCH(CH3)2CH3CCCOC2H5CHCH3CH3CH3OO()CH3CHCHCOOHCH3CH3(1)40%NaOH(2)H+(3)∆,-CO2CH3I(4)分析：CH3CCH2CH2COOHOBBBBrrrrCCCCHHHH2222CCCCOOOOOOOOCCCCHHHH2222CCCCHHHH3333解：CH3CCH2COC2H5OOC2H5ONaCH3CCHCOC2H5Na+OO-BrCH2COOC2H5CH3CCHCOC2H5CH2COC2H5OOO(1)5%NaOH(2)H+(3)∆,-CO2()CH3CCH2CH2COOHO(5)分析：CH3CCH2OCH2CCH3OIIII2222解：2CH3CCH2COC2H5OO2C2H5ONa2CH3CCHCOC2H5Na+OO-CH3CCH2CH2CCH3OOCHCC2H5CH3COOCHCC2H5CH3COOI2(1)5%NaOH(2)H+(3)∆,-CO2()(6)分析：CH3CCH2OCCH3OCCCCllllCCCCCCCCHHHH3333OOOO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)+(1)C2H5ONa(2)H+CHCOOC2H5CH3CCC2H5OOO(2)+(1)C2H5ONa(2)H+HCCH2COOC2H5O(3)+(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：CH3CHCHCOOHCH3CH3CCCCHHHH3333IIIICCCCHHHH3333CCCCHHHHBBBBrrrrCCCCHHHH3333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：CHCH3CH2COOHHOOCCH211112222333344445555CCCC2222HHHH5555OOOOCCCCCCCCHHHH====CCCCHHHHOOOOCCCCHHHH3333解：CH3CH=CHCOOC2H5CH2(COOC2H5)2C2H5O-Na+CH3CHCH2COOC2H5CH(COOC2H5)2(1)NaOH/H2O(2)H+(3)∆,-CO2CH3CHCH2COOHCH2COOHHOOCCH2CHCH2COOHCH3(3)分析：HOOCCH2CHCH2CH2COOHCH3““““””””““““””””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：CH3CHCH2CH2COOHOHCCCCHHHH2222CCCCHHHHOOOOCCCCHHHH3333OCH3OH+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：OCH3OCH3CCH2OCH2COOHCH3CHCH2CH2COOHOHγ-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：CCCCOOOOOOOOHHHHCCCCHHHH2222CCCCHHHH2222CCCCllllCCCCHHHH2222CCCCHHHH2222CCCCllll解：CH2(COOC2H5)22C2H5O-Na+CH2CH2CH2CH2BrBrCOOC2H5COOC2H5COOH(1)NaOH/H2O(2)H+(3)∆,-CO2(6)分析：HOOCCOOHBBBBrrrrCCCCHHHH2222CCCCHHHH2222BBBBrrrrCCCCHHHH2222BBBBrrrr2222解：2CH2(COOC2H5)22C2H5O-Na+2CH-(COOC2H5)2Na+CH2CH2BrBr(C2H5OOC)2CHCH2CH2CH(COOC2H5)22C2H5O-Na+CH2BrBrCOOHHOOC(1)NaOH/H2O(2)H+(3)∆,-CO2COOC2H5COOC2H5C2H5OOCC2H5OOC或者：HOOCCOOH11112222333344445555BBBBrrrrCCCCHHHH2222CCCCHHHH2222BBBBrrrrCCCCHHHH2222====CCCCHHHHCCCCNNNNoooorrrrCCCCHHHH2222====CCCCHHHHCCCCOOOOOOOOCCCC2222HHHH5555CH2(COOC2H5)2C2H5O-Na+CH2=CHCOOC2H5CH2CH2COOC2H5CH(COOC2H5)22C2H5O-Na+MichealCH2CH2BrBr(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)OCH3CHOOCH