A) halogenation
B) hydrogenation
C) dehydrogenation
D) dehydrohalogenation
Correct Answer: D
Solution :
\[C{{H}_{3}}\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}C{{H}_{3}}+KOH(alc.)\xrightarrow[{}]{{}}\] \[{{H}_{2}}C=\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{C}}\,-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-H\] \[{{H}_{2}}O+KBr+C{{H}_{3}}-CH=CH+C{{H}_{3}}\] \[\because \]In this reaction hydrogen as well as halogen is removed. \[\therefore \]It is dehydrohalogenation reaction.
You need to login to perform this action.
You will be redirected in
3 sec
