第八章卤代烃作业答案

1、第八章 卤代烧2、完成下列反应式NaCN (1mol)CH=CHBr CH2CNCH3CH3HC2H534、按与AgNO3-酒精(Sn 1)反应活性顺序排列:C6H5BrCH3cH2cH2cH2Br(2)>(1)>(4)>(3)>(5)>(6)5、比较下列每对反应的速度：b>a(1) a. CH3cH2cHCH2Br + CN- I CH3b. CH3cH2cH2cH2Br +CN-(2) a. CH3CH2Br + SH-CH3OHb>aOb. CH3cH2Br + SH-HCN(CH 3)2H2Oa. CH3I + NaOHb. CH3I + Na

2、SH HaOb>a(4) a. CH3Br + (CH3”Nb>ab. CH3Br + (CH3)3P (5) a CH3CH2I + CH3S- (1.0mol/L) b>ab. CH3CH2I + CH3s- (2.0mol/L) (6) a. CH3Br + CH3OH b>ab. CH3Br + CH3SH "6、完成下列转化：a (CH3)2CHCH 2cH 2Br(1) (CH3)2CHCH 2cH 2cl(CH 3)2CHCH 2cH 2IKOHa (CH 3)2CHCH 2cH 2cl(CH 3)2CHCH=CH 2C2 H50HHBr过氧化物

3、餐(CH3)2CHCH 2cH 2Br(CH3)2CHCH 2cH2clKI丙酮一(CH3)2cHCH 2cH2I。CH3CHCH 2ClI(2) CH3cH2cH2cl .OHb 、CH2CHCH 2I I ICl OHClKOHa CH3cH 2cH 2cl CH3cH=CH 2C2H5OHCl2+H2OOH0HCH3CHCH 2ClKOHCl2CH3cH2cH2cl-CH3cH=CH 2202H50Hhv1Cl2+H2OCH 2cH=CH 2 OHCl OH ClCH2CHCH2CH3CH3(5)CH3cH2cH2Bra C CH3CH2CH2D b b-CH3cH2cH2cH2cH2c

4、H3-c、小叫叫产3OMg _DoOa cH3cH2cH2Br “ CH3cH2cH2MgBr 3 砥3cH2cH2Db 2cH3cH2cH2Br + 2Na cH3cH2cH2cH2cH2cH3 + 2NaBr7、分别写出下列反应产物生成的可能过程:aOH+ cH2=cHcH 2cH20H2-ch=ch 2H.cH2Br -Br cH215、解释下列立体化学结果:H2O产CH3C6H5cHBrCC6H5cHOH(1)2 %的构型转化，98 %外消旋化fHtC6H5cH0cH 327 %的构型转化，73 %外消旋化(1)作为亲核试剂CH3OH>H2O ,亲核性强，利于Sn2.作为溶剂，H20极性,CH3OH溶剂极性强，利于Sn 1。19、有一化合物分子式为C8Hio,在铁的存在下与1 mol澳作用，只生成一种化 合物A, A在光照下与1 mol氯作用，生成两种产物B和C,试推断A,B,C 的结构：CH3C ICH2C1c.MgCH3CH2CH2Br 无水乙7.