由A可进行两步氧化得到E,可知A为醇,E为羧酸,且A与E碳原子数相同.又A与E反应生成G(C
4H
8O
2),可推得:A为乙醇,E为乙酸,那么D为乙醛.由于G是酯,而C、G属同类有机物,可知C也为酯.由A(C
2H
6O)和B反应可生成酯C(C
10H
12O
2),结合反应前后反应的定量关系,可推得B的分子式为C
8H
8O
2,而B与A(乙醇)可发生酯化反应,可知B为羧酸,故B含一个羧基和一个甲基,而B中苯环上的一溴代物有四种,那么B苯环上羧基和甲基必处于邻位或间位,故B有两种结构,由此可推得C的结构.B中的甲基被酸性KMnO
4溶液氧化得到羧基,故F为邻苯二甲酸或间苯二甲酸,
(1)由以上分析可知D为CH
3CHO,与新制的Cu(OH)
2悬浊液反应的化学方程式是CH
3CHO+2Cu(OH)
2CH
3COOH+Cu
2O+2H
2O,
故答案为:CH
3CHO+2Cu(OH)
2CH
3COOH+Cu
2O+2H
2O;
(2)由以上分析可知C可能为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/78310a55b319ebc486378b288126cffc1e171637?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/78310a55b319ebc486378b288126cffc1e171637?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(3)F为邻苯二甲酸或间苯二甲酸,可与乙醇发生酯化反应,例如临苯二甲酸与乙醇发生酯化反应的方程式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b151f8198618367a98c6826c2d738bd4b31ce530?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b151f8198618367a98c6826c2d738bd4b31ce530?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(4)B含一个羧基和一个甲基,符合能发生水解反应和能发生银镜反应的B的同分异构体中必为甲酸某酯,故有4种同分异构体,