A) dehydration
B) dehydrogenation
C) dehydrohalogenation
D) dehalogenation
Correct Answer: C
Solution :
\[C{{H}_{3}}-C{{h}_{2}}-Br+\underset{alk}{\mathop{KOH}}\,\xrightarrow{dehy\,droha\log enation}\] In alcoholic KOH alkoxide ions\[(R{{O}^{-}})\]are present which is a strong base. They abstract proton from p- carbon of alkyl halide and favours elimination reaction. \[\underset{alcohol}{\mathop{ROH}}\,+KOH\xrightarrow{{}}\underset{\begin{smallmatrix} Potassium \\ alkoxide \end{smallmatrix}}{\mathop{ROK}}\,+{{H}_{2}}O\] \[ROK\xrightarrow{{}}\underset{\begin{smallmatrix} alkoxide \\ \,\,\,\,\,\,ion \end{smallmatrix}}{\mathop{R{{O}^{-}}}}\,+{{K}^{+}}\] \[R{{O}^{-}}+H-\overset{\beta }{\mathop{C}}\,{{H}_{2}}-\overset{\alpha }{\mathop{C}}\,{{H}_{2}}-Br\] \[\xrightarrow{{}}ROH+C{{H}_{2}}==C{{H}_{2}}+Br\]
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