《有机化学》(第四版)第四章 二烯烃和共轭体系(习题答案)解析

第四章 二烯烃和共轭体系

思考题

习题4.1 下列化合物有无顺反异构体？若有，写出其构型式并命名。(P119)

(1) 1,3-戊二烯 C H 2=C H C H =C H C H 3

解：有2个顺反异构体!

C =C C H 3

H

C H 2=C H

H

C =C

H

C H 3

C H 2=C H

H

(Z)- 1,3-戊二烯 (E)- 1,3-戊二烯

(2) 2,4,6-辛三烯 C H 3C H =C H C H =C H C H =C H C H 3 解：有6个顺反异构体!

C =C

C =C

H

C =C

H

H

C H 3

H

H C H 3

H

C =C

C =C

H

C =C

H

C H 3

H

H

H C H 3

H

C =C

C =C

H

C =C

H

C H

3

H C H 3

H H

H

(Z,Z,Z)- 2,4,6-辛三烯 (Z,Z,E)- 2,4,6-辛三烯 (E,Z,E)- 2,4,6-辛三烯

C =C C =C

C =C

H

H

H

H C H 3

C H 3

H

H

C =C

C =C

C =C

H

H

C H 3H

H

H

H

C H 3

C =C

C =C

C =C

H

H

C H 3

H

H

C H 3

H

H

(E,E,E)- 2,4,6-辛三烯 (E,E,Z)- 2,4,6-辛三烯

(Z,E,Z)- 2,4,6-辛三烯

习题4.2 下列各组化合物或碳正离子或自由基哪个较稳定？为什么？(P126)

(1) 3-甲基-2,5-庚二烯 和 5-甲基-2,4-庚二烯√

C H 3C H =C C H 2C H =C H C H 3

C H 3

C H 3C H =C H C H =C C H 2C H 3

C H 3

π-π共轭

无π-π共轭

有

(2)

(C H 3)2C =C H C

H 2C H 3C H =C H C H 2C H 2=C H C H 2、和

(C H 3)2C =C H C H 2C H 3C H =C H C H 2

C H 2=C H C H 26个超共轭

σ-H 3个超共轭

σ-H 0个超共轭

σ-H 稳定性：＞＞

(3)

(C H 3)2C H C H C H =C H 2(C H 3)2C C H 2C H =C H 2(C H 3)2C C H 2C H 3、和

(C H 3)2C H C H C H =C H 2(C H 3)2C C H 2C H =C H 2

(C H 3)2C C H 2C H 3稳定性：＞＞有p -π共轭

p -π共轭无p -π共轭无给电子性：

C H 3C H =C H 2

＞

(4)

、和

C H 2C H =C H C H =C H 2C H 2C H =C H C H 2C H 3C H 3C H C H 2C H =C H 2

稳定性：＞

＞

有π-π共轭C H 2C H =C H C H =C H 2C H 2C H =C H C H 2C H 3

C H 3C H C H 2C H =C H 2

p -π有共轭

π-π无共轭π-π无共轭p -π有共轭

p -π共轭

无

(5) C H 3 , (C H 3)2C H C H 2 , C H 3C H C H 2C H 3 和 (C H 3)3C 甲基自由基 一级自由基 二级自由基 三级自由

基

(6)

(C H 2C H =C H )2C H ， C H 2=C H C H 2 和 C H 3C H =C

H

5个sp 2杂化碳共轭 3个sp 2

杂化碳共轭 成单电子不在p 轨道上，不参与共轭

共轭程度更大

共轭程度相对较小

习题4.3 解释下列事实：(P126)

(1) C H 3C H 2C H =C H C H 3

C H 3C H 2C H 2C H C H 3C l C H 3C H 2C H C H 2C H 3

C l

H C l

+

(主)

(次)

稳定性：＞C H 3C H 2C H C H 2C H 3C H 3C H 2C H 2C H C H 3

(2)

C H 3C H =C C H 3C H 3

C H 3C H 2C C H 3C H 3

C l C H 3C H C H C H 3

C H 3

C l

+

H C l

(主)

(次)