question_answer

\[CH\equiv CH\xrightarrow[{{H}_{2}}S{{O}_{4}}]{HgS{{O}_{4}}}A\xrightarrow[{{H}_{2}}O]{C{{H}_{3}}MgBr}B\]\[\xrightarrow{P/B{{r}_{2}}}C\]Here C is:

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 :

\[\underset{ethyne}{\mathop{CH\equiv CH}}\,\xrightarrow[{{H}_{2}}S{{O}_{4}}]{HgS{{O}_{4}}}\underset{\begin{smallmatrix}  (A) \\  ethanal \end{smallmatrix}}{\mathop{C{{H}_{3}}CHO}}\,\xrightarrow{C{{H}_{3}}MgBr}\]