(1)同系物:结构相似,分子组成上相差一个或若干个CH
2原子团的化合物,则环己烷(C
6H
12)与环辛烷(C
8H
16)均为环烷烃,分子组成上相差2个CH
2,互为同系物;
同分异构体:分子式相同,结构不同的化合物,则苯与棱晶烷(分子式均为C
6H
6),环辛四烯与立方烷(分子式均为C
8H
8)互为同分异构体.
故答案为:环己烷;环辛烷;苯;棱晶烷;环辛四烯;立方烷;
(2)由正四面体烷的结构可知分子中有4个碳原子和4个氢原子,分子式为C
4H
4;正四面体烷完全对称,只有一种氢原子,一氯代物只有1种,一氯代物中氢原子也只有1种,所以故二氯代物只有1种;
金刚烷分子中有两种氢原子,一种是次甲基氢,一种亚甲基氢,所以其一氯代物有2种;
立方烷分子中只有一种氢原子,其一氯代物有1种,而一氯代物有3种氢原子,分为在
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/5882b2b7d0a20cf4d5c1416e75094b36acaf992a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,则立方烷的二氯代物有3种;
故答案为:C
4H
4;1;2;3;
(3)苯的同系物含有1个苯环,侧链为烷基,一氯代物只有两种的物质,说明只有2种氢原子,则与苯互为同系物且一氯代物只有两种的物质的结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/80cb39dbb6fd52669f7083a5a818972bd4073623?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,名称为1,4-二甲苯;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e61190ef76c6a7ef31771797fefaaf51f3de662a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
,名称为1,3,5-三甲苯;
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/80cb39dbb6fd52669f7083a5a818972bd4073623?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
1,4-二甲苯;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/e61190ef76c6a7ef31771797fefaaf51f3de662a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
1,3,5-三甲苯;
(4)立方烷的一种属于芳香烃的一种同分异构体为苯乙烯,苯乙烯能发生加聚反应,方程式为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/838ba61ea8d3fd1f67591d8b334e251f95ca5f2a?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/838ba61ea8d3fd1f67591d8b334e251f95ca5f2a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(5)环辛四烯中含有碳碳双键能被酸性KMnO
4氧化,-CH=双键断裂,被氧化成-COOH,最终得到产物为HOOC-COOH,故答案为:HOOC-COOH.