第二章 烷烃
1、用系统命名法命名下列化合物:
(1) (2)
(3) (4)
(5) (6)
答案:
(1)2,5-二甲基-3-乙基己烷 (2)2-甲基-3,5,6-三乙级辛烷
(3)3,4,4,6-四甲基辛烷 (4)2,2,4-三甲基戊烷
(5)3,3,6,7-四甲基癸烷(6)4-甲基-3,3-二乙基-5-异丙基辛烷
2.写出下列化合物的构造式和键线式,并用系统命名法命名之。
(1) C5H12仅含有伯氢,没有仲氢和叔氢的
(2) C5H12仅含有一个叔氢的
(3) C5H12仅含有伯氢和仲氢
答案:
键线式 构造式 系统命名
(1)
(2)
(3)
3.写出下列化合物的构造简式:
(1) 2,2,3,3-四甲基戊烷
(2) 由一个丁基和一个异丙基组成的烷烃:
(3) 含一个侧链和分子量为86的烷烃:
(4) 分子量为100,同时含有伯,叔,季碳原子的烷烃
(5) 3-ethyl-2-methylpentane
(6) 2,2,5-trimethyl-4-propylheptane
(7) 2,2,4,4-tetramethylhexane
(8) 4-tert-butyl-5-methylnonane
答案:
(1) 2,2,3,3-四甲基戊烷
简式:CH3CH2(CH3)2(CH3)3
(2) 由一个丁基和一个异丙基组成的烷烃:
(3) 含一个侧链和分子量为86的烷烃:
因为CnH2n+2=86 所以 n=6该烷烃为 C6H14,含一个支链甲烷的异构体为:
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(4) 分子量为100,同时含有伯,叔,季碳原子的烷烃
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(5) 3-ethyl-2-methylpentane
(6) 2,2,5-trimethyl-4-propylheptane
(7) 2,2,4,4-tetramethylhexane
(8) 4-tert-butyl-5-methylnonane
4.试指出下列各组化合物是否相同?为什么?
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答案:
(1)两者相同,从四面体概念出发,只有一种构型,是一种构型两种不同的投影式
(2)两者相同,均为己烷的锯架式,若把其中一个翻转过来,使可重叠.
5.用轨道杂化理论阐述丙烷分子中C-C和C-H键的形式.
答案:
解:丙烷分子中C-C键是两个C以SP3杂化轨道沿键轴方向接近
到最大重叠所形成的δ化学键.
6.(1) 把下列三个透视式,写成楔形透视式和纽曼投影式,它们是不是不同的构象呢?
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(2)把下列两个楔形式,写成锯架透视式和纽曼投影式,它们是不是同一构象?
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(3)把下列两个纽曼投影式,写成锯架透视式和楔形透视式,它们是不是同一构象?
答案:
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为同一构象。
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(3)
7、写出2,3-二甲基丁烷的主要构象式(用纽曼投影式表示)。
答案:
8.试估计下列烷烃按其沸点的高低排列成序(把沸点高的排在前面)
(1)2-甲基戊烷 (2)正己烷 (3)正庚烷 (4)十二烷
答案:
十二烷>正庚烷>正己烷>2-甲基戊烷 因为烷烃的沸点随C原子数的增加而升高;同数C原子的烷烃随着支链的增加而下降。
9.写出在室温时,将下列化合物进行一氯代反应预计到的全部产物的构造式:
⑴.正己烷:⑵.异己烷:⑶.2,2-2甲基丁烷:
答案:
⑴.正己烷: 一氯代产物有3种 CH3CH2CH2CH2CH2CH2Cl
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⑵.异己烷:一氯代产物有5种
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⑶.2,2-2甲基丁烷: 一氯代产物有3种
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10、根据以下溴代反应事实,推测相对分子质量为72的烷烃异构体的构造简式。
(1)只生成一种溴代产物;(2)生成三种溴代产物;(3)生成四种溴代产物。
答案:
11、写出乙烷氯代(日光下)反应生成氯乙烷的历程。
答案:
解 :Cl2 →2Cl CH3CH3