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