有机化学课后习题答案 南开大学5章 羧酸衍生物.docx

有机化学课后习题答案 南开大学5章 羧酸衍生物第十五章 羧酸衍生物 1给下列化合物命名或写出它们的结构式。 对甲基-N，N-二甲基苯甲酰胺 3-溴丁腈 苯甲酰氧基甲酸乙酯 3-甲基戊二酸二异丙酯 O(5)COClO(6)CH3CH2CCH2CO(CH2)7CH32完成下列反应式： (1)CH3CH2CH2COOH(NH4)2CO3CH3CH2CH2CONH2NH3P2O5CH3CH2CH2CN(2)HO(CH2)3COOHH+CH2=CHCOOHCH2=CHCONH2(3)CH2ClNaCNCH2CNLiAlH4CH2CH2NH2COOH(4)CH2NH2COOC2H5(5)COCl(CH3CH2)2CdONHCOOC2H5COCH2CH3O(6)CH3CH2OCCl+NH3O(7)NH2CCl+CH3O-OCH3CH2OCNH2OCH3OCNH2(8)NCCONHCH3H2/NiNH2CH2CONHCH3(9)CONHCH3LiAlH4CH2NHCH3O(10)2CH3CH2CH2COOC2H5Na/C6H6H+OHCHCH2CH2CH3CH3CH2CH2C(11)BrCH2Cl+CH3COONaCH3COOCH2Cl CH3(12)CH3CHCOOHPCl3CH3CH3CHCOClOCH3COONaOOCCH3CH3CHC3写出下列反应的机理： O(1)C6H5OCH3ONaOO-O-BrOOCH3CH2CH2BrC6H5CH2CH2BrC6H5COOCH3C6H5COOCH3O(2)ONH2NH2C2H5O-OOC2H5ONH2HNOOC2H5-OC2H5OHNH2NOH5C2OOCHOH5C2OOC_H2OH5C2OOC4.下列酯用不同浓度的碱处理可得到不同产物，分别写出它们的生成过程 C6H5CHCH=CHCH3OCCH3O5mol/LNaOHC6H5CHCH=CHCH3O-C6H5CHCH=CHCH3OHCOCH3OH构型保持稀NaOH+C6H5CHCH=CHCH3OCCH3O+C6H5CHCH=CHCH3C6H5CH=CHCHCH3HO-C6H5CH=CHCHOHCH3外消旋5下列化合物A水解生成中间体B，进一步反应生成C，写出各步反应历程： OCH=CHNA-OOCH2CF3CH3HO-OOCH2CF3CH3CH=CNCCOCH2CF3O-CCOCH=CNCCOCH3OO-OOCH2CF3OCHNOCH3BO-OOHOCHHO-NCH3CHNCH3CH=CHNCCCOCH36比较下列酯碱性水解的速度： (1)CH3COOXX=NO2>Br>H>OCH3(2)RCOOC2H5R=CH3>CH3CH2CH2(CH3)2CH>(CH3)3CCH3NCOOCH37.OOCPh可卡因OHNCOOH芽子碱CH38分别用两种方法完成下列转化： (1)OHCHOOHPBr3BrMgEt2OHCHOH3O+CH2OHCrO3/ 吡啶CHOOHPBr3BrMgEt2O1)CO22)H3O+COOHPCl3H2/Pd/BaSO4S喹啉CHO(2)OHCH2NH2OHPBr3MgEt2O1)HCHO2)H3O+CH2OHCrO3/ 吡啶CHOH2/Pd/NH3CH2NH2OHPBr3MgEt2O1)CO22)H3O+COOHPCl3COClNH3LiAlH4CH2NH29用简单化学方法鉴别下列各组化合物 (1)CH3CH2OCH3CH3CH2COOHCH3CH2COCH3NaHCO3CO2CH3CH2COOH黄色沉淀CH3CH2COCH3不反应2，4-二硝基苯肼不反应CH3CH2OCH3用红外光谱鉴别： CH3CH2CH2CN在22602240cm-1有特征吸收； CH3CH2CH2CONH2在35003100cm-1有两个特征吸收峰； CH3CH2CH2CON(CH3)2在上述两处无吸收峰； 10Lifbrate是控制体内胆甾醇的药物。由不超过6个碳的有机原料起始，设计一条合成路线： ClOHNaOHClONaCl2CHCOONaH+(ClO)2CHCOOHHONCH3O(ClO)2CHCONCH311写出下列反应中AE的结构式： O+OAlCl3OOACOOHZnHgCOOHPCl3浓HClBCOClAlCl3C2H5MgBrH+OCDEC2H512治疗高血压、心衰等疾病的药物Carvedilol可由下列方法合成，写出中间体AD和Carvedilol的结构式： OOOONCH2CH2OOCOCH2CHOD+OH2NCH2CH2OHNCH2CH2OHOABPBr3ONCH2CH2BrOOHOCH3OH-NH2NH2H2NCH2CH2OCH3CODCH3COCH2OCH2CHOHCH2NHCH2CH2OCH3CONHNHCarvedilol13.甲CH3CH2COOH乙HCOOC2H5丙CH3COOCH314由苯、乙酰氯及其他必要的无机试剂合成下列化合物： (1)CH3COClCOCH3CH3CHOOH-COCH=CHCH3(其中 CH3COClBr2FeLiAlHOC(CH3)33CH3CHO）PCl3(2)MgEt2OMgBrCO2H3O+COOHCOClO苯胺CNH(其中 HNO3H2SO4NO2FeHClNH2）O(3)CH3COClOH-OOCCH3Ag2OCH3CO-CH3COClCH3COH-(4)CH3COClLiAlHOC(CH3)33CH3CHOCH3CH=CHCHOCH3CH=CHCOOH(5)CH3COClC2H5OH吡啶CH3COOC2H5C2H5MgBrH3O+CH3CH3CH2COHNaC2H5BrCH2CH3CH3COCl(其中 LiAlH4CH3CH3CH2COCH2CH3CH3CH2OHPBr3MgEt2OC2H5MgBr）CH2CH3O15.A.HOOCCH2CHCH2COOHCH3B.CH3OO Br16.CH3CH2CHCOOCH(CH3)2O17.A.CH2OCCH3B.CH2CH2COOHO18.A.CCH3B.OOCCH3