Ⅰ(1)由龙胆酸甲酯的结构简式可知,分子式为C
8H
8O
4,含酚羟基、酯基,故答案为:C
8H
8O
4;酚羟基、酯基;
(2)龙胆酸甲酯与足量氢氧化钠溶液反应的化学方程式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e1fe9925bc315c60fb61975e8eb1cb134854778f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,所以1mol龙胆酸甲酯能消耗3mol氢氧化钠,故答案为:3mol;
Ⅱ(1)有龙胆酸甲酯的结构可知龙胆酸的结构,又A为(CH
3)
3COH或(CH
3)
2CHCH
2OH两种可能,X的结构可能为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/908fa0ec08fa513d4349906f3e6d55fbb3fbd9a4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
或
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/10dfa9ec8a1363273d0e3e82928fa0ec09fac7a4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;A发生消去反应生成烯烃C
4H
8,C
4H
8为纯净物且含支链,应为(CH
3)
2C=CH
2,则A为(CH
3)
3COH或(CH
3)
2CHCH
2OH,B为即为(CH
3)
2CHCH
2OH,B中有5种不同环境的H原子,B的核磁共振氢谱中应有5种吸收峰,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/10dfa9ec8a1363273d0e3e82928fa0ec09fac7a4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
或
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/908fa0ec08fa513d4349906f3e6d55fbb3fbd9a4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;消去反应;5;
(2)C
4H
8为纯净物且含支链,应为(CH
3)
2C=CH
2,其反应生成高分子化合物的反应方程式为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9d82d158ccbf6c815ec3d059bf3eb13532fa408f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;B为(CH
3)
2CHCH
2OH,C为(CH
3)
2CHCHO,B→C是发生醇的催化氧化,反应方程式为:2(CH
3)
2CHCH
2OH+O
22(CH
3)
2CHCHO+2H
2O,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9d82d158ccbf6c815ec3d059bf3eb13532fa408f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;2(CH
3)
2CHCH
2OH+O
22(CH
3)
2CHCHO+2H
2O;
(3)龙胆酸的同分异构体能发生银镜反应且是酯类,须有酚羟基和甲酸形成的酯基,能使氯化铁溶液显色,还有两个酚羟基,其结构应有
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/cdbf6c81800a19d85cf778b430fa828ba71e468f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,以及和酯基对位的酚羟基在3、5、6三个位置,共4种,还有
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/bd3eb13533fa828b42d76152fe1f4134960a5a8f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,以及将其中一个酚羟基连在酯基的对位,共2种,总共6种,故答案为:6种.