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Question: The products for the following reactions are (i) \(CH_{3} - \overset{Br}{\underset{H}{\underset{|}{...

The products for the following reactions are

(i) CH3−C∣HBr−CH2−CH3+alc.KOH→xCH_{3} - \overset{Br}{\underset{H}{\underset{|}{C}}} - CH_{2} - CH_{3} + alc.KOH \rightarrow x

(ii) CH3−CH∣CH3−CH=CH2→O3Y+ZCH_{3} - \underset{CH_{3}}{\underset{|}{CH}} - CH = CH_{2}\overset{\quad O_{3}\quad}{\rightarrow}Y + Z

A

x=(CH3)2C=CH2,Y=CH3CH2CHOx = (CH_{3})_{2}C = CH_{2},Y = CH_{3}CH_{2}CHO

Z = CH3CH2CHOCH_{3}CH_{2}CHO

B

x=CH2=CH2,Y=CH3CHO,Z=CH3COOHx = CH_{2} = CH_{2},Y = CH_{3}CHO,Z = CH_{3}COOH

C

x=CH3−CH=CH−CH3,Y=CH3−CH∣CH3−CHO,x = CH_{3} - CH = CH - CH_{3},Y = CH_{3} - \overset{CH_{3}}{\overset{|}{CH}} - CHO, Z = HCHO

D

x=CH3−CH=C(CH3)2,Y=HCHO,x = CH_{3} - CH = C(CH_{3})_{2},Y = HCHO,

Z = CH3CHOCH_{3}CHO

Answer

x=CH3−CH=CH−CH3,Y=CH3−CH∣CH3−CHO,x = CH_{3} - CH = CH - CH_{3},Y = CH_{3} - \overset{CH_{3}}{\overset{|}{CH}} - CHO, Z = HCHO

Explanation

Solution

: (i)

CH3−C∣H∣Br−CH2CH3→alc.KOHCH3CH=CHCH32−Butene(mainproduct′x′)CH_{3} - \overset{Br}{\overset{|}{\underset{H}{\underset{|}{C}}}} - CH_{2}CH_{3}\overset{\quad alc.KOH\quad}{\rightarrow}\underset{\begin{aligned} & 2 - Butene \\ & (mainproduct'x') \end{aligned}}{CH_{3}CH = CHCH_{3}}