A) dehydration
B) dehydrogenation
C) dehydrohalogenation
D) dehalogenation
Correct Answer: A
Solution :
\[C{{H}_{3}}-C{{H}_{2}}Br+KOH(alc.)\xrightarrow{\text{dehydrohalogenation}}\]\[C{{H}_{2}}=C{{H}_{2}}+KBr+{{H}_{2}}O.\] In alcoholic KOH, alkoxide ion \[(R\bar{O})\] is present which is a strong base favour elimination reaction.
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