A) \[C{{H}_{3}}CH(Br)C{{H}_{3}}\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\]
C) \[C{{H}_{2}}=CH-Br\]
D) \[BrCH=CH-C{{H}_{3}}\]
Correct Answer: A
Solution :
\[CH\equiv CH\underset{{{H}_{2}}S{{O}_{4}}}{\mathop{\xrightarrow{HgS{{O}_{4}}}}}\,C{{H}_{3}}CHO\xrightarrow{C{{H}_{3}}MgBr}\] \[\begin{matrix} \,\,\,\,\,\,\,\,\,\,OMgBr \\ | \\ C{{H}_{3}}CC{{H}_{3}} \\ | \\ H \\ \end{matrix}\xrightarrow{{{H}_{2}}O}\begin{matrix} \,\,\,\,OH \\ | \\ C{{H}_{3}}CH\,\,\,\,\,\, \\ | \\ \,\,\,\,\,\,\,C{{H}_{3}} \\ \end{matrix}\xrightarrow{P/B{{r}_{2}}}\]\[\begin{matrix} \,\,\,Br \\ | \\ C{{H}_{3}}CH\,\,\,\,\, \\ | \\ \,\,\,\,\,\,C{{H}_{3}} \\ \end{matrix}\]
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