发布时间 : 星期日 文章有机化学 高鸿宾 第四版 答案 第十四章 二羰基更新完毕开始阅读
(1) 丙酸乙酯＋乙二酸二乙酯 (1) C2H5ONa(2) H+OOC2H5OCCCHCOOC2H5
CH3O
(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+CCHCH2COOC2H5COOC2H5OOCCH3
OO(4) CHC(CH)COCH32425(1) C2H5ONa(2) H+
OOO(1) C2H5ONa(2) H+(5) CHC(CH)COCH32325
O(6) CH2 CH2CH2COOC2H5 CH2CH2COOC2H5(1) C2H5ONa(2) H+OCOOC2H5OCOOC2H5C2H5O2C
COOC2H5(7) CH3CH2OOCCH2CH2CHCHCH3(1) C2H5ONa(2) H+CH3COOC2H5(七) 用丙二酸二乙酯法合成下列化合物：
(1)
CH3CHCHCOOHCH3CH3 (2)
HOOCCH2CHCH2COOHCH3O COOH
来自丙二
(3)
HOOCCH2CHCH2CH2COOHCH3 (4) CH3O(5)
COOH (6) HOOC上CH3CH(1) 分析：
BrCH3CHCHCOOHCH3C2H5O-Na+CH3CH3(CH3)2CHCl
上CH3IC2H5O-Na+
解：
CH2(COOC2H5)2(CH3)2CHCH(COOC2H5)2(1) 5%NaOHCH3I(CH3)2CHC(COOC2H5)2CH3(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：
O上C2H5OCCH=CHCH3解： CH2(COOC2H5)2(1) NaOH/H2O(2) H+12345HOOCCH2CHCH2COOHCH3来自丙二
C2H5O-Na+CH3CH=CHCOOC2H5CH3CHCH2COOC2H5CH(COOC2H5)2CH3CHCH2COOHCH2COOHCHCH2CH32C2H5O-Na+HOOCCH2CHCH2COOHCH3
(3) ??? -CO2(3) 分析：
HOOCCH2来自“丙二”CH2COOH来自“丙二”
BrBr+CH3CHCH2解：
2CH2(COOC2H5)22CH-(COOC2H5)2Na
(1) NaOH/H2O(C2H5OOC)2CHCHCH2CH3HOOCCH2CHCH2CH3CH(COOC2H5)2(2) H+(3) ??? -CO2 CH2COOH
(4) 目标化合物是内酯，应首先制备4-羟基戊酸：
CH3OOH+CH3CHCH2OHCH2COOH来自丙二
上CH3CHCH2OCH3解1：CH2(COOC2H5)2C2H5ONa-+CH-(COOC2H5)2OCH3CHCH2CH(COOC2H5)2
O-(1) NaOH/H2O(2) H+(3) ??? -CO2CH3CHCH2CH2COOHOHH+CH3OO
或者，将4-戊酮酸还原得到4-羟基戊酸：
CH3O内酯OH+OHCH3CHCH2CH2COOH?-羟基酸OH2NiCH3CCH2上溴代丙酮CH2COOH
来自丙二解2：
CH2(COOC2H5)2C2H5O-Na+OCH3COCH2BrCH3CCH2CH(COOC2H5)2
(1) NaOH/H2O(2) H+(3) ??? -CO2H+OCH3CCH2CH2COOHH2/NiOHCH3CHCH2CH2COOH
CH3OO
(5) 目标分子可看作乙酸的二烷基取代物：
CH2CH2Cl上CH2CH2ClBrBrCOOH
来自\丙二\解： CH2(COOC2H5)22C2H5O-Na+CH2CH2CH2CH2COOC2H5COOC2H5
(1) NaOH/H2O+ (2) H(3) ??? -CO2COOH
上 CH2Br2(6) 分析： 来自丙二HOOCCOOH来自丙二
上BrCH2CH2BrBrBrCH2解：
2CH2(COOC2H5)22C2H5O-Na+2CH-(COOC2H5)2Na+2C2H5O-Na+CH2
(C2H5OOC)2CHCH2CH2CH(COOC2H5)2C2H5OOCC2H5OOCCOOC2H5(2) H+(1) NaOH/H2OBrCH2Br COOC2H5(3) ??? -CO2324HOOCCOOH
或者：来自丙二HOOC15COOH上 CH2=CHCNor CH2=CHCOOC2H5
上BrCH2CH2BrCH2(COOC2H5)2BrCH2BrCH2C2H5O-Na+CH2=CHCOOC2H5Micheal加成CH2CH2COOC2H52C2H5O-Na+CH(COOC2H5)2(1) NaOH/H2O(2) H+(3) ??? -CO2
C2H5OOCC2H5OOCCOOC2H5HOOCCOOH
(八) 写出下列反应的机理：
OCH3COOCH3NaOCH3CH3OHH3O+OCH3COOCH3
O解：
CH3O-+OCH3CH3COOCH3O-CH3OCH3O-OCH3OCH3CH3COOCH3
COOCH3O
OCH3COOCH3
CH3OCCHCH2CH2CHCOCH3O
CH3COOCH3
(九) 完成下列转变：
O(1)
COOC2H5COOC2H5CH3CH2
(2)
CH2COOC2H5CH2COOC2H5OO
O(3)
COOEtCH2COOEtO
OHOHC2H5CH3(4) CH3COCH2COOC2H5CH3CHCHCH2CCH3
OO(5)
COOEtCH2CH2CNO
(6)
CH3COCH2COOC2H5(CH3C)2CHCH2Ph