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大学-二烯烃和共轭体系习题加答案

大学-二烯烃和共轭体系习题

(一) 用系统命名法命名下列化合物：

(1)

(2)

4-甲基-1,3-戊二烯

2-甲基-2,3-戊二烯

(3)

(4)

2-甲基-1,3,5-己三烯 (3Z)-1,3-戊二烯

(二) 下列化合物有无顺反异构现象：若有，写出其顺反异构体并用Z,E-命名法命名。

解：(1) 无；

(2) 有；(3E)-1,3-戊二烯, (3Z)-1,3-戊二烯；

(3) 有；(3Z,5Z)-3,5-辛二烯， (3Z,5E) -3,5-辛二烯，(3E,5E) -3,5-辛二烯； (4) 有；(3E)-1,3,5-己三烯， (3Z)-1,3,5-己三烯； (5) 无 (三) 完成下列反应式：

解：红色括号中为各小题所要求填充的内容。

(1)

(2)

(3)

(4)

CH 2=CHCH=C(CH 3)2CH 3CH=C=C(CH 3)2CH 2=CHCH=CHC=CH 3

CH 3

C=C

CH=CH 2

CH 3

H H

(1)2-甲基-1，3-丁二烯(2)1,3-戊二烯(3)3,5-辛二烯(4)1,3,5-己三烯(5)2,3-戊二烯+ HOOCCH=CHCOOH

CH +

+COOCH 3COOCH 3H

COOCH 3H

CH 3OOC

(5)

(6)

(7)

(8)

(9)

(10)

(四) 给出下列化合物或离子的极限结构式，并指出哪个贡献最大？

(2) (3)

(4)

(5)

(6)

解： (1)

贡献最大(非电荷分离)

(2)

结构相似，二者贡献一样大

(3)

结构相似，二者贡献一样大

+ RMgX MgX

+ RH

CHO

CH 2Cl

CH 2Cl KMnO +

HOOCCH 2CHCH 2CH 2COOH

CH 2Cl

(A)(B)

CH 3

CH 33

CH 3

H CH 3CH 3

CH 3

(1)CH

N (CH 3)2C=CH C(CH 3)2CH 2=CH CH 2CH 2

C CH 3

CH 3

C CH=CH 2CH 3C

CH 3C

N (CH 3)2C=CH C(CH 3)2

(CH 3)2C CH=C(CH 3)2CH 2=CH CH 2

CH 2

CH=CH 2

(4)

各共振结构式结构相似，能量相同，对真实结构的贡献相同。

(5)

贡献最大(与电负性预计相同)

(6)

贡献最大(共价键最多)

(五) 化合物和同相比，前者C=C 双键的电子云密度降低，而后者C=C 双键的电子云密度升高。试用共振论解之。

解：的真实结构可用共振论表示如下：

其中(II)对真实结构的贡献就是使C=C 双键带部分正电荷，电子云密度降低。

的真实结构可用共振论表示如下：

其中(II)对真实结构的贡献就是使C=C 双键带部分负电荷，电子云密度升高。

(六) 解释下列反应：

(1)

(2)

解：

(1)

... ...CH 2

C CH 3O

CH 2C CH 3

O -

CH 3C CH 2CH 3

C CH CH 2O

CH 3C CH CH 2O

CH 3

C CH CH 2

CH 2=CH NO 2CH 2=CH OCH 3CH 2=CH 2CH 2=CH

NO 2CH 2=CH N

CH 2

CH=N O -O

(I)

(II)

CH 2=CH OCH 3

CH 2

CH=OCH 3(I)

(II)

Br Br

+ 2 Br 2

CH 2

CH 2Br

NBS CCl 4 ,

Br Br

Br 1,4-加成

1,6-加成

(2)

… …

(七) 某二烯烃与一分子溴反应生成2，5-二溴-3-己烯，该二烯烃若经臭氧化再还原分解则生成两分子乙醛和一分子乙二醛 (O=CH-CH=O)。试写出该二烯烃的构造式及各步反应式。解：该二烯烃的构造式为： 2,4-己二烯

各步反应式略。

(八) 3-甲基-1,3-丁二烯与一分子氯化氢加成，只生成3-甲基-3-氯-1-丁烯和3-甲基-1-氯-2-丁烯，而没有2-甲基-3-氯-1-丁烯和3-甲基-1-氯-2-丁烯。试简单解释之，并写出可能的反应机理。

解：由于C +

稳定性：

所以，只有与

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