(1)
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/9358d109b3de9c82da18550e6f81800a19d84335?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
含有酯基,属于酯类;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d31b0ef41bd5ad6ee78bc86b82cb39dbb7fd3cc4?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
含有Cl,属于卤代烃;
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/caef76094b36acaff4a13a7d7fd98d1000e99cea?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
含有醛基,属于醛;
故答案为:酯; 卤代烃;醛;
(2)①
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0b7b02087bf40ad1ea966b62542c11dfa8eccec5?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
含有苯环、乙基,名称为乙苯;
②
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/500fd9f9d72a605991624d9b2b34349b023bbac5?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
最长的碳链为5个,2号C上有1个甲基,3号C上一个乙基,名称为2-甲基-3-乙基戊烷;
③
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0bd162d9f2d3572c9aa2305d8913632763d0c3eb?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
含有碳碳三键的最长碳链有5个碳原子,编号从右侧开始,3号C上一个甲基,名称为4-甲基-2-戊炔;
故答案为:①乙苯②2-甲基-3-乙基戊烷③4-甲基-2-戊炔;
(3)①新戊烷的系统命名法为2,2-二甲基丙烷,因此其结构简式为C(CH
3)
4 ,
故答案为:C(CH
3)
4 ;
②间三甲苯中苯环连接3个甲基且均处于间位位置,其结构简式为
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/4e4a20a4462309f710ba21a3710e0cf3d6cad6ae?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/4e4a20a4462309f710ba21a3710e0cf3d6cad6ae?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
;
③含有C=C双键的碳链有5个碳原子,双键处于1与2、3与4号C之间的位置,从C=C双键一端编号,甲基处于2号碳原子上,结构简式为CH
2=CH(CH
3)CH=CHCH
3,
故答案为:CH
2=CH(CH
3)CH=CHCH
3.