(1)根据碳可形成4对共用电子对,可知正四面体烷的分子式为C
4H
4,四个氢原子完全等效,一氯代物只有一种,故答案为:C
4H
4;1;
(2)乙烯基乙炔为CH
2=CH-C≡CH,每个乙烯基乙炔分子中含有一个碳碳双键、一个碳碳三键,具有不饱和烃的性质.
a.每个乙烯基乙炔分子中含有一个碳碳双键、一个碳碳三键,能使酸性KMnO
4溶液褪色,故a正确;
b.1mol乙烯基乙炔分子中1mol碳碳双键、1mol碳碳三键,1mol乙烯基乙炔能与3molBr
2发生加成反应,故b正确;
c.乙烯基乙炔分子内含有碳碳双键、碳碳三键两种官能团,故c正确;
d.乙烯基乙炔为CH
2=CH-C≡CH最简式为CH,乙烯C
2H
2的最简式为CH,二者最简式相同,二者质量相同,消耗氧气相同,故d错误;
故答案为:d;
(3)环辛四烯的分子式为C
8H
8,不饱和度为5,属于芳香烃的同分异构体,含有1个苯环,侧链不饱和度为1,故含有1个侧链为-CH=CH
2,属于芳香烃的同分异构体的结构简式
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/63d9f2d3572c11df86807c92602762d0f703c22a?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/63d9f2d3572c11df86807c92602762d0f703c22a?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
(4)与苯互为同系物,侧链为烷基,一氯代物只有二种,说明该有机物中只有1种H原子,与苯互为同系物且一氯代物有二种物质,必须考虑对称结构,考虑最简单的取代基-CH
3情况,符合条件的物质的结构简式
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3b87e950352ac65c774fa5c2f8f2b21193138a3d?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
等,氯代时,可以取代苯环上的氢,也可以取代侧链上的氢,
故答案为:
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/3b87e950352ac65c774fa5c2f8f2b21193138a3d?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
.