A) \[C{{H}_{2}}=CH-C{{H}_{2}}-CBr=CHBr\]
B) \[BrC{{H}_{2}}-CHBr-C{{H}_{2}}-C\equiv CH\]
C) \[C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{2}}-CB{{r}_{3}}\]
D) \[C{{H}_{3}}-CB{{r}_{2}}-C{{H}_{2}}-C\equiv CH\]
Correct Answer: A
Solution :
\[C{{H}_{2}}=CH-C{{H}_{2}}-C\equiv CH+B{{r}_{2}}\to \]\[C{{H}_{2}}=CH-C{{H}_{2}}-\underset{\underset{Br}{\mathop{|}}\,\,\,\,}{\mathop{C=}}\,\underset{\underset{Br\,}{\mathop{|\,\,\,\,}}\,}{\mathop{CH}}\,\]
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