(1)甲含有的官能团有碳碳双键和羟基,故答案为:羟基;
(2)由转化关系可知Y为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b17eca8065380cd7f142915fa244ad3459828127?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,发生消去反应生成
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/203fb80e7bec54e741b029acba389b504fc26a27?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,反应的方程式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4a36acaf2edda3cc13b88b9202e93901213f9227?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/4a36acaf2edda3cc13b88b9202e93901213f9227?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(3)根据乙的结构简式和题干中所给信息,可知A是乙二醛,B是苯甲醛,乙在加热条件下与新制备氢氧化铜反应生成酸,应为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/b58f8c5494eef01fb67f2b77e3fe9925bc317d1c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,则C应为醇,因B为苯甲醛,则C为苯甲醇,E是相对分子质量为76,结合题给信息可知应为OH-CH
2-COOH,本身能发生缩聚反应生成
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3801213fb80e7bec228630532c2eb9389b506b27?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,
①C为苯甲醇,含有羟基,可发生取代反应、氧化反应、可与钠反应生成氢气,不具有酸性,不能与碳酸钠反应,故答案为:c;
②由以上分析可知F为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3801213fb80e7bec228630532c2eb9389b506b27?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/3801213fb80e7bec228630532c2eb9389b506b27?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
③乙为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/203fb80e7bec54e741b029acba389b504fc26a27?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,含有醛基,可在加热条件下与新制备氢氧化铜浊液反应,
方程式为C
6H
5CH=CHCHO+2Cu(OH)
2+NaOH
C
6H
5CH=CHCOONa+Cu
2O↓+3H
2O,
故答案为:C
6H
5CH=CHCHO+2Cu(OH)
2+NaOH
C
6H
5CH=CHCOONa+Cu
2O↓+3H
2O;
④该物质能发生银镜反应,一定含有醛基,可以和Br
2/CCl
4发生加成反应,含有碳碳碳双键,遇FeCl
3溶液显示紫色,说明含有酚羟基,所以该物质的结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9825bc315c6034a8cd898b33c81349540923761c?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/9825bc315c6034a8cd898b33c81349540923761c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
⑤a.G难溶于水,分子中含有碳碳双键,能使Br
2/CCl
4溶液褪色,故a正确;
b.D含有羧基,能与NaHCO
3溶液反应,故b错误;
c.甲和C机构不同,不是同系物,故c错误;
d.G含有碳碳双键,可发生加聚反应生成H,故d正确.
故答案为:a、d;