A) Methyl bromide
B) Ethyl bromide
C) Ethylene bromide
D) Ethylidene bromide
Correct Answer: D
Solution :
\[CH\equiv CH\xrightarrow{HBr}C{{H}_{2}}=CH-Br\xrightarrow{HBr}\]\[\underset{\text{Ethylidene dibromide}}{\mathop{C{{H}_{3}}CH<\begin{matrix} Br \\ Br \\ \end{matrix}}}\,\]
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