题目内容

(15分)有机物A为芳香烃,质谱分析表明其相对分子质量为92,某课题小组以A为起始原料可以合成酯类香料H和高分子化合物I,其相关反应如下图所示:                                                       
已知以下信息:
①碳烯(:CH2)又称卡宾,它十分活跃,很容易用它的两个未成对电子插在烷烃分子的C-H键之间使碳链增长。
②通常在同一个碳原子连有两个羟基不稳定,易脱水形成羰基。
回答下列问题:

(1)A的化学名称为         
(2)由B生成C的化学方程式为                ,该反应类型为          。                                                              (3)G的结构简式为           
(4)请写出由F生成I的化学方程式                                             
(5)写出G到H的反应方程式                                       。                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                 
(6)H的所有同分异构体中,满足下列条件的共有        种;
①含有苯环        ②苯环上只有一个取代基          ③属于酯类
其中核磁共振氢谱有五种不同化学环境的氢,且峰面积比为1:1:2:2:6的是           (写结构简式)。
(1)甲苯()(2);取代反应
(3);(4)
(5)
(6)15种, 

试题分析:(1)由于A是芳香烃,结合A的相对分子质量,可确定A是甲苯;(2)根据题意可得B是乙苯;乙苯与氯气在光照下发生取代反应得到C:由B生成C的化学方程式为;(3)C与NaOH的水溶液发生反应得到D:;D与银氨溶液发生反应,然后酸化得到G:;G与乙醇发生酯化反应得到H:。该反应的方程式为;(4)D与H2发生加成反应得到E:;E在浓硫酸作用下发生消去反应得到F:;F在一定条件下发生加聚反应得到I:;反应的方程式是。(5)G与乙醇发生酯化反应得到H的方程式为。(6)H的所有符合条件的同分异构体一共有15种,它们分别是: ;;;;;;;;;;;;;;.其中核磁共振氢谱有五种不同化学环境的氢,且峰面积比为1:1:2:2:6的是
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网