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复旦大学研究生入学考试1994有机化学试题一,写出下列反应的主要产物,注意产物的立体构型(2×20=40)1.ClCH=CHCH2ClNaI丙酮?2.OCH3HHCH31.LiAlH42.H2O?3.(CH3)3CCHCl2(CH3)3COK?4.β-D-吡喃葡萄糖Br2H2O?5.OCH3CHO+100℃?6.OCH3Br2NaOAc/HOAc?7.(CH3)2CHP(C6H5)3[]I1.n-BuLi2.OOEt?8.CH3CO(CH2)4COCH31.Mg-Hg2.H3O?9.CH3COOH?10.NHOOBr2OH,H2O?11.PhCHO+CH3(CH2)3NH2H2Ni?12.光气+过量苯AlCl3?13.CH2=C=O+CH3NH2?14.EtOOEtOOEtONaEtOH?15.NNHCH3I?16.OOAlCl3?17.CH3ClKNH2NH3(l)?18.(CH3)3CCHO+HCHO浓碱?19.COOHBr2HgO?20.CH3CH3?二,写出下列反应机理(5×4=20)1.HOCH2CH2CH2CHOCH3OHHCl(g)OHOCH32.(R)-CH3CHBrCOONaH2O,OH(R)-CH3CHOHCOONa3.CH=CHCH2ClONaOHCHCH=CH24.CH2=CHCH2Br*HOBrCH2CHCH2OHBrBr*70%CH2CHCH2OHBrBr*30%(Br为同位素标志)*三,写出下列合成反应式,可用必要的有机及无机试剂(5×5=25)1.从≤C5的有机物开始合成AcOHHOAc2.的有机物开始合成C5从≤OOOCHOHH3.从≤C3的有机物开始合成OO4.从单取代苯开始合成CH3BrBrCH2CH35.从甲苯开始合成四,写出下列化合物的结构式(5×3=15)1.化合物A(C10H12O2),IR在3100-2900cm-1(s),1670(s)1500,1600(m),830(s)HNMR1.2(t,3H)ppm,2.9(q,2H)3.9(s,3H),7.0(b,2H),8.0(b,2H)2.化合物B,(C6H8O),HNMR中显示有一个甲基的单峰,常压Pd/C催化氢化得到C,(C6H10O,IR在1745cm-1有强吸收),C用NaOD/D2O处理得到D(C6H7D3O)。C用过氧酸处理得到E的HNMR中显示有甲基的双峰(J=8Hz)。3.化合物F(C11H16O2N2)和冷的稀碱不反应;和HNO2反应后滴加β-萘酚,生成红色产物;若用NaOH水溶液加热,则溶解,冷却后用醚萃取,从醚层中得到G(C14H11ON),可溶于水,显碱性。G用HNO2处理,无N2及油状物生成。G和硝酸铈铵有正反应,和乙酐反应则得到H(C6H13O2N);后者不溶于稀盐酸,但可溶于浓盐酸。从上述醚萃反应后的水层中,经酸化得到I(C7H7O2N),mp185-9℃,可溶于稀盐酸。复旦大学研究生入学考试1994有机化学试题答案一,写出下列反应的主要产物,注意产物的立体构型(2×20=40)1.ClCH=CHCH2I2.CH3CH3HOHHD3.C(CH3)3Cl4.HOCOOHOHCH2OHOHOH5.OCH3CHO6.OOAcCH3BrHOOAcBrCH3H7.C(CH3)2OEt8.OO9.CH310.COONH211.PhCH2NH(CH2)3CH312.PhCPhO13.CH3CONHCH314.COOEtO15.NCH3NHI16.OOH17.CH3NH2CH3NH218.(CH3)3CCH2OHHCOOH19.Br20.CH3CH3二,写出下列反应机理(5×4=20)1.HOCH2CH2CH2CHOHHOCH2CH2CH2C=OH..OOHHHOOH2HOHCH3OHOOCH3HH-HOOCH3H2.CBrHCOONaCH3(R)C=OOCHCH3-Br(邻基参与)CBrHC=OOCH3OHCOHHCOONaCH3(R)3.CH=CHCH2Cl-ClCH=CHCH2OOOHOCHOHCH=CH24.CH2=CHCH2BrBrOHCH2CHBrCH2BrHOCH2OHCHCH2BrBrCH2CHBrCH2BrCH2CHBrCH2BrCH2OHCHCH2BrBrOH******(70%)(30%)三,写出下列合成反应式,可用必要的有机及无机试剂(5×5=25)1.CH3CH2CH2CCH1.NaNH22.CH3CH2CH2BrCH3CH2CH2CCCH2CH2CH3Na/NH3(l)1.RCO3H2.H3OHOAcOAcHAcCl2.OOO+OOOHH1.O32.Zn/H2OCHOCHOOOOHHOHOOOHHCHO3.2CH3COCH3OHOCH(CO2Et)2CH2(CO2Et)2EtOO1.OH/H2O2.EtOH/HOCO2EtEtOOO4.CH3CH3COClAlCl3CH3COCH3Br2FeCH3COCH3BrBrZn/HgHClCH3CH2CH3BrBr5.CH3MnO2H2SO4CHO(CH3CO)2OCH3COONaCH=CHCOOHPPAONH2NH2/OHLi四,写出下列化合物的结构式(5×3=15)1.A.COCH2CH3OCH3OCH32.BC.OCH3D.OCH3DDDE.OOCH33.FNH2COCH2CH2N(CH3)2O(C11H16O2N2)G.HOCH2CH2N(CH3)2(C4H11ON)H.CH3COOCH2CH2N(CH3)2(C6H13O2N)I.NH2COOH
本文标题:复旦大学有机化学真题1994
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