A在浓硫酸存在下既能和甲醇反应,又能和甲酸反应,说明A中既有羧基又有羟基.A催化氧化的产物不能发生银镜反应,说明羟基不在碳链的端点上,可判断A为CH
3CH(OH)COOH,而A被氧化可得CH
3COCHO,不能发生银镜反应,A与甲醇发生酯化反应生成B,B为CH
3CH(OH)COOCH
3,A与甲酸发生酯化反应生成D,D为HCOOCH(CH
3)COOH,A在浓硫酸、加热条件下生成E,E可以溴水褪色,应发生消去反应,E为CH
2=CHCOOH,A在浓硫酸、加热条件下生成链状高分子化合物F,应是A发生缩聚反应,则F的结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4b90f603738da977e1e89fafb351f8198718e36b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,
(1)与A具有相同官能团的同分异构体的结构简式为:HOCH
2CH
2COOH,故答案为:HOCH
2CH
2COOH;
(2)由上述分析可知,B的结构简式为CH
3CH(OH)COOCH
3,F的结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4b90f603738da977e1e89fafb351f8198718e36b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,故答案为:CH
3CH(OH)COOCH
3;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4b90f603738da977e1e89fafb351f8198718e36b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(3)A→E是乳酸在浓硫酸、加热条件下发生消去反应生成CH
2=CHCOOH,反应方程式为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4e4a20a4462309f79da9b496710e0cf3d6cad66b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,
A→D是乳酸与甲酸发生酯化反应,反应方程式为:HCOOH+CH
3CH(OH)COOH
HCOOCH(CH
3)COOH+H
2O,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4e4a20a4462309f79da9b496710e0cf3d6cad66b?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
HCOOH+CH
3CH(OH)COOH
HCOOCH(CH
3)COOH+H
2O.