A) \[aq.KOH\]
B) \[alc.KOH\]
C) \[conc.{{H}_{2}}S{{O}_{4}}\]
D) All of these
Correct Answer: B
Solution :
\[\underset{\begin{smallmatrix} ethylidene \\ dibromide \end{smallmatrix}}{\mathop{C{{H}_{3}}CHBr}}\,\xrightarrow[-2KBr,\,\,-2{{H}_{2}}O]{Alc.KOH}\underset{acetylene}{\mathop{CH\equiv CH}}\,\] Hence, \[A=Alc.KOH.\] (a dehydrohalogenating reagent).
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