A) Dehydration
B) Dehydrogenation
C) Dehydrohalogenation
D) Dehalogenation
Correct Answer: C
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-Br+\underset{\text{alk}}{\mathop{KOH}}\,\xrightarrow{\text{Dehydrohalogenation}}\]\[C{{H}_{2}}=C{{H}_{2}}+KBr+{{H}_{2}}O\] In alcoholic KOH alkoxide ions \[(R{{O}^{-}})\] are present which a strong base is. They abstract proton from b-carbon of alkyl halide and favours elimination reaction \[\underset{\text{Alcohol}}{\mathop{ROH}}\,+KOH\to \underset{\text{Potassium alkoxide}}{\mathop{ROK+{{H}_{2}}O}}\,\] \[ROK\to \underset{\text{Alkoxide ion}}{\mathop{R{{O}^{-}}}}\,+{{K}^{+}}\] \[R{{O}^{-}}+H-\overset{\beta }{\mathop{C{{H}_{2}}}}\,-\overset{\alpha }{\mathop{C{{H}_{2}}}}\,-Br\to ROH+C{{H}_{2}}=C{{H}_{2}}+Br\]
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