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人教版选修5 第二章 第一节 第1课时 烷烃和烯烃 作业

人教版选修5 第二章  第一节  第1课时 烷烃和烯烃 作业
人教版选修5 第二章  第一节  第1课时 烷烃和烯烃 作业

第二章 第一节 第1课时 烷烃和烯烃

[A 篇 知识对点训练]

知识点1 烷烃和烯烃

1.在光照条件下,将等物质的量的甲烷和Cl 2充分反应,得到产物的物质的量最多的是( )

A.CH 3Cl B .CH 2Cl 2 C .CCl 4 D .HCl

答案 D

解析 由甲烷与氯气发生取代反应的产物中有一氯代烷和多氯代烷,每生成1 mol 一氯代烷时,可生成1 mol HCl 。每生成1 mol 多氯代烷,会同时生成多摩尔HCl ,因此得到的产物中HCl 的物质的量最多。

2.①丁烷、②2-甲基丙烷、③戊烷、④2-甲基丁烷、⑤2,2-二甲基丙烷,其沸点排列顺序正确的是( )

A.①>②>③>④>⑤

B .⑤>④>③>②>① C.③>④>⑤>①>②

D .②>①>⑤>④>③

答案 C

解析 对于烷烃而言,含有碳原子数多,相对分子质量大的烃,分子间作用力大,沸点较高,即沸点(③、④、⑤)>(①、②);对于含有相同碳原子数的烷烃,支链越多,沸点越低,即①>②;③>④>⑤,故有③>④>⑤>①>②。

3.下列关于乙烯和乙烷相比较的说法中,不正确的是( )

A.乙烯属于不饱和链烃,乙烷属于饱和链烃

B.乙烯分子中所有原子处于同一平面上,乙烷分子则为立体结构,不是所有原子都在同一平面上

C.乙烯分子中的碳碳双键和乙烷分子中的碳碳单键相比较,双键的键能大,键长长

D.乙烯能使酸性KMnO 4溶液褪色,乙烷不能

答案 C

解析 乙烯分子中碳碳双键比乙烷分子中的碳碳单键键能大,键长短,C 项错误。

4.下列不属于取代反应的是( )

A.CH 4+Cl 2――→光照CH 3Cl +HCl

B.CH 3Cl +Cl 2――→光照CH 2Cl 2+HCl

C.CH 4+4Cl 2――→光照CCl 4+4HCl

D.CH 2===CH 2+H 2――→催化剂△

CH 3CH 3 答案 D

解析 D 项反应属于加成反应。

5.下列物质不能用加成反应制得的是( )

A.CH 2Cl 2

B .CH 3CH 2Cl C.CH 3CH 2OH

D .CH 2ClCH 2Cl

答案 A

解析 在发生加成反应时,由双键变成单键,CH 2Cl 2不能由物质加成制得。

6.丙烯在一定条件下发生加聚反应的产物是( )

答案 A

解析 丙烯在加聚时生成。

7.烯烃在一定条件下发生氧化反应时碳碳双键发生断裂:RCHCHR ′――→催化剂RCHO +R ′CHO ,在该条件下,下列烯烃分别经过上述条件氧化后产物可以生成甲醛的是( )

A.CH 3CH===CH(CH 2)2CH 3

B.CH 2===CH(CH 2)2CH 3

C.CH 3CH===CH —CH===CHCH 3

D.CH 3CH 2CH===CHCH 2CH 3

答案 B

解析 CH 3CHCH(CH 2)2CH 3氧化后生成CH 3CHO 和CH 3(CH 2)2CHO ,A

错误;CH2===CH(CH2)2CH3氧化后生成HCHO和CH3(CH2)2CHO,B正确;CH3CH===CHCH===CHCH3氧化后生成CH3CHO和OHC—CHO,C错误;CH3CH2CH===CHCH2CH3氧化后生成CH3CH2CHO,D错误;答案选B。

知识点2 烯烃的顺反异构

8.下列有机物可以形成顺反异构的是( )

A.丙烯B.2-甲基-2-丁烯

C.1-氯-1-丙烯D.2,3-二甲基-2-丁烯

答案 C

解析1-氯-1-丙烯存在顺-1-氯-1-丙烯()和反-1-氯-1-

丙烯()。

9.化学式为C3H5Cl的链状化合物的同分异构体共有(包括顺反异构)( )

A.3种B.4种C.5种D.6种

答案 B

解析C3H5Cl没有碳链异构,先写出C3H5Cl中取代基(—Cl)的位置异构,再写其顺反异构。C3H5Cl的取代基(—Cl)的位置异构有三种:①

CH3CH===CHCl,

10.已知乙烯为平面结构,因此,1,2-二氯乙烯可以形成两种不同的立体异

构体:。下列各物质中,能形成类似上述两种立体异构体的是( )

A.1,1-二氯乙烯B.乙烯

C.2-丁烯D.2-丁炔

答案 C

解析1,1-二氯乙烯的结构简式为CCl2===CH2,乙烯的结构简式为

CH2===CH2,2-丁烯的结构简式为CH

CH===CHCH3,2-丁炔的结构简式为

3

。A项和B项的分子中,每个双键碳上所连的原子相同,不能

形成立体异构;C项可以形成和

两种立体异构体。D项物质只能有一种结构,不能形成立体异构体。

[B篇综合提升训练]

11.下列关于烷烃的说法正确的是( )

A.丙烷(C3H8)发生取代反应生成的C3H7Cl的结构只有一种

B.烷烃中除甲烷外,很多都能使酸性KMnO4溶液的紫色褪去

C.分子中含有六个碳原子的烷烃在常温下为液态物质

D.烷烃的分子为直链式结构,不可以带支链

答案 C

解析丙烷发生取代反应生成的C3H7Cl有2种结构:CH3CH2CH2Cl、CH3CHClCH3,A错误;烷烃都不能使酸性KMnO4溶液褪色,B错误;常温下,碳原子数多于4个的烷烃呈液态或固态(新戊烷常温下为气态),C正确;烷烃分子中可以带支链,如CH3CH(CH3)2,D错误。

12.有关烯烃的下列说法中正确的是( )

A.烯烃分子中所有的原子一定在同一平面内

B.烯烃只能发生加成反应不能发生取代反应

C.分子式是C5H10的烃分子中一定含有碳碳双键

D.烯烃既能使溴水褪色也能使酸性KMnO4溶液褪色

答案 D

解析只有乙烯中所有原子一定共面,A项错误;烯烃易发生加成反应,但不是不能发生取代反应,在特定条件下烯烃也能发生取代反应,B项错误;分子式为C5H10的烃可以是烯烃,也可能是环烷烃,C项错误。

13.甲烷中混有乙烯,欲除去乙烯得到纯净的甲烷,最好依次通过盛有下列试剂的洗气瓶( )

A.溴水,烧碱溶液,浓硫酸

B.酸性KMnO4溶液,浓硫酸

C.澄清石灰水,浓硫酸

D.浓硫酸,酸性KMnO4溶液

答案 A

解析乙烯能与溴发生加成反应而被除去,烧碱溶液能吸收挥发出来的溴蒸气,浓硫酸能吸收水。

14.某气态烃0.5 mol能与0.5 mol Cl2完全加成,加成产物又可被3 mol Cl2完全取代,则此气态烃可能是( )

A.CH2===CH2B.CH3—CH===CH2

C.CH2===CHCH2CH3D.CH2===CH—CH===CH2

答案 B

解析该气态烃与Cl2按1∶1完全加成,则该烃分子中含有一个

;0.5 mol加成产物又可被3 mol Cl2完全取代,则该加成产物分子中应含有6个氢原子,因此该气态烃分子中也应含有6个氢原子,故只有B 项正确。

15.(1)相对分子质量为70的烯烃的分子式为____________;

若该烯烃与足量的H2加成后能生成含3个甲基的烷烃,则该烯烃的可能的结构简式为_______________________________________________________。

(2)有机物A的结构简式为:

①若A是单烯烃与氢气加成后的产物,则该单烯烃可能有________种结构;

②若A是炔烃与氢气加成后的产物,则此炔烃可能有________种结构。

③若A的一种同分异构体只能由一种烯烃加氢得到,且该烯烃是一个非常对称的分子构型,有顺、反两种结构。

a.写出A的该种同分异构体的结构简式___________________________。

b.写出这种烯烃的顺、反异构体的结构简式_________________________。

答案(1)C5H10

(2)①5 ②1

③a.(CH3)3C—CH2—CH2—C(CH3)3

解析(1)由M r(C n H2n)=70,14n=70,n=5,得烯烃的分子式为C5H10;该烯烃加成后所得的产物(烷烃)中含有3个甲基,表明在烯烃分子中只含有一个支链。

当主链为4个碳原子时,支链为1个—CH3,此时烯烃的碳骨架结构为

,其双键可在①、②、③三个位置,有三种可能的结构。

(2)有机物A的碳骨架结构为,其双键可处于①、②、③、④、⑤五个位置,而三键只能处于①一个位置。

16.烯烃通过臭氧氧化并经锌和水处理得到醛或酮,例如:

(1)观察题中给出的化学方程式可知:烃分子中每有一个双键,则产物中会

有________个,A分子中有________个双键,A的分子式是

________________。

(2)已知可以简写为,其中线表示化学键,线的端点、折点或交点表示碳原子,碳原子剩余的价键用氢原子补足。写出A所有可能的结构简式:

________________________________________________________。

答案(1)2 2 C8H12

(2)、

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化学选修五练习烷烃和烯烃

第二章第一节第1课时 一、选择题 1.下列有关简单的烷烃的叙述中正确的是() ①都是易燃物②特征反应是取代反应③相邻两个烷烃在分子组成上相差一个甲基 A.①和③B.②和③ C.只有①D.①和② 解析:烷烃含有碳和氢两种元素,都能燃烧生成二氧化碳和水,①正确;烷烃都易发生取代反应,②正确;相邻两个烷烃在分子组成上相差一个CH2,③不正确。 答案:D 2.(2014·大连检测)(双选)关于有机物的下列说法中,正确的是() A.甲烷和乙烷可用酸性KMnO4溶液鉴别 B.烃R催化加氢可得有机物,则烃R只可能是烯烃 C.乙烯和乙炔都能使溴的四氯化碳溶液褪色 D.乙基与氢氧根结合可得有机物甲醇 解析:甲烷和乙烷都不与酸性KMnO4溶液反应,故无法鉴别;乙基是电中性的游离基,而氢氧根是带负电荷的离子,两者无法结合生成电中性分子甲醇。 答案:BC 3.有机物的名称和一氯代物的种数是() A.2,4,5-三甲基-4-乙基己烷,7种 B.2,4-二甲基-4-异丙基己烷,8种 C.2,4-二甲基-3-异丙基己烷,7种 D.2,3,5-三甲基-3-乙基己烷,8种 解析:有机物的命名应选取含支链最多的最长碳链为主链,同时应使支链编号之和为最小,据此该有机物应命名为2,3,5-三甲基-3-乙基己烷,其分子中有8种不同化学环境的氢原子,故其一氯代物的结构有8种。 答案:D 4.下列关于碳氢化合物的叙述正确的是() A.碳氢化合物的通式为C n H2n+2 B.燃烧产物为二氧化碳和水的化合物一定是碳氢化合物 C.碳原子间以单键相连的烃是烷烃

D.碳氢化合物分子的相对分子质量一定是偶数 解析:C n H2n+2是烷烃的通式,A项错误;燃烧产物为二氧化碳和水的化合物不一定是碳氢化合物,化学式为C x H y O z的有机物的燃烧产物也是二氧化碳和水,B项错误;碳原子间全部以单键相连的链烃才是烷烃,C项错误;因为碳原子的相对原子质量(12)为偶数,烃分子中的氢原子个数也一定为偶数,所以碳氢化合物分子的相对分子质量一定是偶数是正确的。 答案:D 5.(2014·江西重点中学联考)下列反应中属于加成反应的是() A.甲烷与氯气的混合气体在光照条件下的反应 B.丙烯和氯气在一定条件下生成ClCH2CH==CH2的反应 C.乙烯使酸性高锰酸钾溶液褪色的反应 D.乙烯与HCl气体生成一氯乙烷的反应 解析:甲烷与氯气的混合气体在光照条件下的反应属于取代反应,选项A错误;丙烯和氯气在一定条件下生成ClCH2CH===CH2的反应属于取代反应,选项B错误;乙烯使酸性高锰酸钾溶液褪色,乙烯发生了氧化反应,选项C错误;乙烯与HCl气体生成一氯乙烷的反应属于加成反应,选项D正确。 答案:D 6.既可以用来鉴别乙烷和乙烯,又可以用来除去乙烷中混有的乙烯,得到纯净乙烷的方法是() A.与足量溴反应B.通入足量溴水中 C.在一定条件下通入氢气D.分别进行燃烧 解析:乙烷和乙烯均是无色气体,但前者易发生取代反应,后者易发生加成反应,若与足量溴作用,乙烷可以发生取代反应而生成溴乙烷,乙烯发生加成反应生成二溴乙烷,不仅不易鉴别,还会损失大量的乙烷且混入大量的溴蒸气杂质,显然不合理;若在一定条件下通入氢气,虽可将乙烯转变为乙烷,但通入氢气的量不易控制,很难得到纯净的乙烷,用氢气无法鉴别它们;若分别进行燃烧,显然无法再得到乙烷;因此只能选B,因为乙烷不和溴水反应,而乙烯能和溴水发生加成反应而使溴水褪色,且生成的CH2BrCH2Br为液态,便于分离。 答案:B 7.使1 mol 乙烯与氯气发生完全加成反应,然后使该加成反应的产物与氯气在光照条件下发生取代反应,则两个过程中消耗的氯气的总的物质的量是() A.3 mol B.4 mol C.5 mol D.6 mol 解析:认真分析乙烯和氯气的加成反应和甲烷与氯气发生的取代反应的结构变化,不难发现加成所消耗的氯气与双键数之比为1∶1,而取代反应所消耗的氯气与分子中的氢原子数之比为1∶1,1 mol乙烯中含有碳碳双键1 mol,氢原子4 mol,所以该两个过程中消耗的氯气

英语选修8unit5重点单词

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