Question

（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）。

Answer 1

（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。