高等有机化学期末复习题+答案.pdf

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1、H3CBrCH3OH1.CH3CH3C6H5C=O(1)p-ClC6H4MgBr2.+HCH3(2)H2O/HC6H5COCH33.+HCHO+NHOH4.PhCO3HC2H55.(CH3)3CCHO+Ph3P=CHCOCH3COCH3K2CO36.吡啶O=CC6H5OHOCH2NH2HNO27.OCCH2CH2CHCOC2H5C2H5ONa8.BrOC6H5CH3H2SO49.C6H5CCCH3OHOH1CH3O10.+CH=CHCCHC2H5ONa23OOOHOCCH1.CHCHONa,CHOHCOOC25252H5OC2H5ONa2.COOC2H5ClH2O3

2、.CH2CH2OTsHOOHNOCH3NH2Br2,NaOCH34.CH3OHOOO5.C2H5ONaCOCH3O1.(S)-3-溴-3-甲基己烷在H2O-丙酮中得外消旋的3-甲基-3-己醇。2.试解释化合物A的乙酸解反应的结果为构型保持产物B。OOO-C-CH3O-C-CH3CH3COONaCH3COOHOTsO-C-CH3OAB23.试说明为什么内消旋（赤式）-1，2-二溴-1，2-二苯乙烷和碱作用主要生成顺式-1-溴-1，2-二苯乙烯？而（苏式）-1，2-二溴-1，2-二苯乙烷却主要得到反式产物？4.请预测下列反应的主要产物为（A）还是（B）？并分析其原因。C

3、H3CO3HO+O(A)(B)5.当具有立体化学结构的3-溴-2-丁醇（A）与浓HBr反应时生成内消旋的2，3-二溴丁烷，而3-溴-2-丁醇（B）的类似反应则生成外消旋2，3-二溴丁烷，请从反应历程的角度说明这一反应的立体化学。BrCH3BrHHCH3HHH3COHH3COH(A)(B)3H3CBrH3COCH3H3COCH31.CH3OH+CH3CH3CH3CH3CH3CH3C6H5C6H5Cl-pC=O(1)p-ClC6H4MgBrHOC6H52.+HCH3HCH3(2)H2O/HC6H5C6H5COCH3COCH2CH2N3.+HCHO+NHOOH4.PhCO

4、3HOHC2H5C2H5H(H3C)3C5.(CH3)3CCHO+Ph3=CHCOCH3C=CHCOCH3COCH3COCH2COC6H5K2CO36.吡啶O=CC6H5OHOOHOCH2NH2HNO27.OCC6H5OCH2CH2CHCOC2H5C2H5ONaCOOC2H58.BrOC2H5OC6H5CH3CHO659.H2SO4C6H5CCCH3C6H5CCCH3OHOHCH34CH3CH3O10.+CH2=CHCCH3C2H5ONaOOOOHOCCHCH1.C2H5ONa,C2H5OHCOOC2H5OOOCCOC2H5CHC2H5O-解答：HO__OHoOCH

5、cC2H5OHCOOC2H5COC2H5__HC2H5OOC2H5ONa2.COOC2H5ClH_OO___Cl解答：C2H5O_C2H5OHClCl_OO_C2H5OOCHCOOC2H525_C2H5OH_COOC2H5C2H5O5H2O3.CH2CH2OTsHO+.._H2O解答：_OTs_+HOTsHOHOOHNOCH3NH2Br2,NaOCH34.CH3OHOO..HN__HBr解答：+BrBrHOO_H....OC2H5NBrN__BrCH3OH重排_N=C=O_BrCH3OH..HOCH3+OCH3+H转移N=CN=CO_OHHNOCH3异构化O6OO5

6、.C2H5ONaCOCH3O_OO_OCCOOC2H52H5解答：C2H5O_CH2OOOOCOOC2H5_C_OC2H5C2H5OH_C2H5O_CHOCH3_CHOHCH32525_OO_OOOC2H5___C2H5OCCH3COCH3O1.(S)-3-溴-3-甲基己烷在H2O-丙酮中得外消旋的3-甲基-3-己醇。解答：(S)-3-溴-3-甲基己烷为一叔卤代烷，它在H2O-丙酮中的反应为SN1反应，反应历程为：H3CC3H7CH3..C3H7慢C+HOHCBr[]C2H5C2H5H3CCH3C3H7C3H7COH+HOC2H5C2H5所以，以上反应的结果为外消旋

7、的3-甲基-3-己醇。72.试解释化合物A的乙酸解反应的结果为构型保持产物B。OOO-C-CH3O-C-CH3CH3COONaCH3COOHOTsO-C-CH3OAB解答：其原因是在反式异构体中有乙酰氧基的邻基参与，导致反应速率加快，顺式异构体中则无此效应。因为在反式异构体中乙酰氧基位于离去基团-OTs的反位，可以从背面进攻乙酰氧基鎓离子，可得构型保持的产物（相当于二次构型转化）CH3H3CCOC+O：O__OTsOOTsOCOCH3_OAcOAc(+_)反式OCOCH3OAc3.试说明为什么内消旋（赤式）-1，2-二溴-1，2-二苯乙烷和碱作用主要生成顺式-1