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第十六章重氮化合物和偶氮化合物一、命名下列化合物：1.N2HSO42.CH3CONHN2Cl3.CH3N=NOH4.CH3N=NN(CH3)25.NHNHCH3H3C6..CCl2.二氯碳烯二、当苯基重氮盐的邻位或对位连有硝基时，其偶合反应的活性事增强还是降低？为什么？解：偶合反应活性增强。邻位或对位上有硝基使氮原子上正电荷增多，有利于偶合反应。NN+N=N+N=N++OHN=NOH三、试解释下面偶合反应为什么在不同PH值得到不同产物？H2NOH+C6H5N2+PH=5PH=9H2NOHN=NC6H5H2NOHN=NC6H5解：PH等于5时，氨基供电子能力强于羟基，偶合反应在氨基的邻位发生。PH等于9时，为弱碱性，羟基以萘氧负离子形式存在，氧负离子供电子能力大于氨基，偶合反应发生在羟基的邻位。四、完成下列反应式：重氮苯硫酸盐对乙酰氨基重氮苯盐酸盐4－甲基－4‘－羟基偶氮苯4－甲基－4‘－二甲氨基偶氮苯2，2‘－二甲基氢化偶氮苯FCH3NaBF4CH3N=NCH3NH2CH3CNCH3CuCNKCNBrCH3CuBrHBrCH3C2H5OHCH3H2O,H+NaNO2,HClFe,HClN2ClNH2NO2CH3CH3CH31.NHOH2.OHOCH3+CH2N23.CH3CHCl3(CH3)3COKCH3ClCl4.CH3CH2CH2C=CCH2CH2CH3HH+CH2..CH2CH2CH3CH2CH2CH3HH五、指出下列偶氮染料的重氮组分和偶连组分。解：重氮组分偶连组分1.N=NHSO3N(CH3)2HSO3N2+N(CH3)22.N=NN=NOHN=NN2+OH3.CH3CONHN=NCH3OHCH3CONHN2+HOCH34.NaSO3N=NHONaSO3N2+HO5.SO3HNH2N=NSO3HN2N=NNH2SO3HN2++NH2六、完成下列合成：1.NO2ClClClNO2ClClNH2NH2N2ClFe/HCl3Cl2ClClClNaNO2,HClClClClH3PO2Cl2.CH3CH3BrBrCH3CH3CH3CH3BrCH3HNO3H2SO4NO2NH2NH2BrBr[H]2Br21,HNO22,H3PO2BrCNCNNO2NO2NH2NH2N2ClN2ClCNCN[O]NaNO2,HClKCN,CuCN3.NHCOCH3OHBrNHCOCH3Br2NHCOCH3BrBrBrNH2N2HSO4H2O,H+NaNO2H2SO450%H2SO4OHBr4.5.CH3CH3NO2CH3CH3CH3CH3HNO3H2SO4NO2NH2NHCOCH3[H]CH3COClHNO3H2SO4CH3NO2CH3NHCOCH3NO2CH3H2O,H+NH2N2ClNaNO2,HClNO2CH3NO2H3PO26.NH2NO2BrBrNH2CH3COClNHCOCH3HNO3H2SO4NHCOCH3NO2NHCOCH3NO2Br2BrBrNO2BrBrH2O,H+NH2NO2BrBrNaNO2HClN2ClNO2BrBrH3PO27.NH2BrCOOHBrNH2NH2NH2H2SO4SO3HBr22BrBrNH2SO3HH2O,H+BrBrBrBrNaNO2,HClN2ClKCN/CuCNBrBrBrBrCNCOOHH2O,H+七、以苯，甲苯，萘和小于或等于两个碳原子的有机化合物为原料合成下列化合物：1.(CH3)2NN=NHNO3H2SO4NO2[H]NH2N(CH3)22CH3INH2HNO2N2ClN(CH3)2N2Cl+(CH3)2NN=N2.CH3N=NH2NNH2CH3HNO3H2SO4NO2CH3CH3[H]NH2CH3NaNO2HClN2ClHNO3H2SO42NO2O2NH2NNH2[H]CH3N2ClCH3N=NH2NNH23.OHCH3N=NCH3CH3SO3HCH3CH3H2SO4NaOHONaCH3H3O+OHCH3HNO3H2SO4CH3CH3NO2NO2+CH3NO2[H]CH3CH3NH2N2ClNaNO2HClCH3OH+OHCH3N=NCH3N=N4.HSO3OHNH2H2SO4NH2SO3HHNO2SO3HN2Cl+H2SO4650CSO3H1.NaOH2.H3O+OHOHSO3HN2Cl+N=NHSO3OH5.(CH3)2NN=NNO2NO2NH2NHCOCH3NO2NHCOCH3NO2NO2NH2N2ClN(CH3)2(CH3)2NN=NNO2NH22CH3IN(CH3)2N=NCH3CH3CH3NO22ZnNaOH/C2H5OHN=NCH3CH36.7.COONaN=NNaSO3OHOH+CO2NaCO3OHCOONaNaSO3N2ClCOONaN=NNaSO3OH8.N=NCH3HOCH3OHCH3+ClN2CH3N=NHOCH3CH3八、推测下列化合物的结构：1，某芳烃分子式为C6H3NO2ClBrFe,HClNaNO2,HClC2H5OHClBrNaOH,H2OC6H3OHClNO2ClBrNO22、某芳香性族化合物分子式为,,试根据下列反应确定其结构：C6H4NO2CH3Fe,HClNaNO2,HClCuCNH2O,H+KMnO4COCOONO2CH3NH2CH3CH3CH3CH3COCOOFe,HClNaNO2,HClN2ClCuCNCNH2O,H+COOHKMnO4COOHCOOH九、下列结构的偶氮染料，以氯化亚锡－盐酸溶液还原分解后，生成那些化合物？1.N=N(CH3)2NSO3H(CH3)2NNH2H2NSO3HSnCl2,HClN=NSnCl2,HClHSO3NH2SO3H2.HSO3NH2NH2SO3HH2N+C6H3NO2ClBr，试根据下列反应确定其结构：解：结构式为：C6H4NO2CH3解：结构及反应：N=NSnCl2,HClNH2H2N+3.CH3OHCH3OH十、某化合物以氯化亚锡盐酸还原可得对甲基苯胺和N,N－二甲基对苯二胺，试推测原化合物的结构，并以苯，甲苯及甲醇为原料合成之。N=NCH3N(CH3)2CH3CH3CH3CH3HNO3H2SO4NO2NH2N2Cl[H]HNO2HNO3H2SO4NO2NH2N(CH3)2[H]十一、某化合物以氯化亚锡盐酸还原得到间甲基苯胺和4－甲基－1，2－苯二胺，试推测原化合物的结构，并以甲苯为原料合成之。H2SO4HNO3CH3COCl[H]NHCOCH3NH2NO2H2SO4HNO3CH3CH3CH3CH3N=NCH3NH2CH3原化合物为：解：原化合物为：NHCOCH3CH3NO2H2O,H+NaNO2,HClH3PO2CH3NO2Fe,HClCH3NH2NaNO2,HClCH3N2ClCH3NH2N=NCH3NH2CH3