有机物A~F有如下转化关系:已知一个碳原子上连有两个羟基时不稳定,转化如下:请回答:(1)A的结构简式

2025-03-14 05:34:25
推荐回答(1个)
回答1:

A发生水解反应C、D,D酸化得到E,C氧化得到E,则C的结构简式为CH3CHO,D为CH3COONa,E为CH3COOH,A的结构简式为.B的相对分子质量为162,其完全燃烧的产物中n(CO2):n (H2O)=2:1,则B中碳氢原子个数之比为1:1,所以n=m,结合其相对分子质量知:n=m=
162?32
13
=10,所以B的分子式为C10H10O2,B水解生成E(乙酸)和F,F的分子式为C10H10O2+H2O-C2H4O2=C8H8O,不饱和度为5,F具有如下特点:能跟FeCl3溶液发生显色反应,能发生加聚反应;苯环上的一氯代物只有两种,则F含有酚羟基、含有1个-CH=CH2,且2个取代基处于对位,则F为,故B为
(1)由上述分析可知,A的结构简式为,反应④是酯的水解反应,也属于取代反应,
故答案为:;取代反应(或水解反应);
(2)C与新制Cu(OH)2的化学方程式为:CH3CHO+2Cu(OH)2+NaOH
CH3COONa+Cu2O↓+3H2O,
故答案为:CH3CHO+2Cu(OH)2+NaOH
CH3COONa+Cu2O↓+3H2O;
(3)由上述分析可知,B的分子式为C10H10O2,故答案为:C10H10O2
(4)F为,F与浓溴水反应的化学方程式为:
B与NaOH反应的化学方程式为:
故答案为:
(5)F()的一种同分异构体G含苯环且能发生银镜反应,含有醛基,满足条件的G的结构简式为(邻、间、对),
故答案为:

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