A) Ethyl alcohol
B) Dilute \[{{H}_{2}}S{{O}_{4}}\]
C) Aqueous KOH
D) Alcoholic KOH
Correct Answer: D
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-Br+\underset{\text{(alc)}}{\mathop{KOH}}\,\to \underset{\text{Ethene}}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,+KBr+{{H}_{2}}O\]
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