有机化学 第四版 课后练习答案 - 图文

C2H5H3CCOH解：CH3(CH2)2(R)-3-甲基-3-己醇浓HXC2H5H3CCCH3(CH2)2+XXC2H5CCH3(CH2)2CH3

(R)-3-甲基-3-卤代己烷(S)-3-甲基-3-卤代己烷外消旋体(七) 将3-戊醇转变为3-溴戊烷(用两种方法，同时无或有很少有2-溴戊烷)。

解：方法一：

CH3CH2CHCH2CH3OHPBr3CH3CH2CHCH2CH3Br

方法二：

CH3CH2CHCH2CH3OHTsCl吡啶CH3CH2CHCH2CH3OTsKBrCH3CH2CHCH2CH3Br

(八) 用高碘酸分别氧化四个邻二醇，所得氧化产物为下面四种，分别写出四个邻二醇的构造式。

OHOH(1) 只得一个化合物CH3COCH2CH3； CH3CH2CCCH2CH3

CH3CH3(2) 得两个醛CH3CHO和CH3CH2CHO；

OHCH3CHOHCHCH2CH3

OH(3) 得一个醛HCHO和一个酮CH3COCH3； HOCH2CCH3 CH3OHCH3 or

(4) 只得一个含两个羰基的化合物(任举一例)

OHOHOH CH3(九) 用反应机理解释以下反应

Cl(1) CH3CH2CHCH2CH2OHHClZnCl2CH3CH2CCH3CH2CH3+CH3CH2C=CHCH3

CH3CH3解：CH3CH2CHCH2CH2OHH+CH3CH2CHCH2CH2OH2CH3-H2OHCH3CH2CCH2CH2

CH3CH3Cl-氢迁移ClCH3CH2CCH2CH3

CH3CH2CCH2CH3CH3-H+CH3CH3CH3CH2C=CHCH3(2)

(CH3)2CIC(CH3)2OHAg+CH3CH3CCCH3 CH3O解：(CH3)2CC(CH3)2OHAg+CH3(CH3)2CCH3COHCH3甲基迁移 ICH3 (CH3)2CCCH3OH-H+CH3CCCH3

CH3O溶解在H2SO4中(3)

CH2=CHCHCH=CHCH3OH

CH2CH=CHCH=CHCH3+CH2=CHCH=CHCHCH3OH解

H+- H2OOH

：

CH2=CHCHCH=CHCH3OHCH2=CHCHCH=CHCH3OH2CH2=CHCHCH=CHCH3

CH2=CHCHCH=CHCH3CH2CH=CHCH=CHCH3 H2O- H+CH2=CHCH=CHCHCH3 H2O- H+

CH2CH=CHCH=CHCH3OHCH2=CHCH=CHCHCH3OH