question_answer

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

A)  \[C{{H}_{3}}CH(Br)C{{H}_{3}}\]                

B)  \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\]

C)   \[C{{H}_{2}}=CHBr\]    

D) \[BrCH=CHC{{H}_{3}}\]

Correct Answer: A

Solution :

                 \[\underset{Acetylene}{\mathop{CH\equiv CH}}\,\xrightarrow[{{H}_{2}}S{{O}_{4}}]{HgS{{O}_{4}}}\underset{Acetaldehyde}{\mathop{C{{H}_{3}}CHO}}\,\]      \[\xrightarrow{C{{H}_{3}}MgBr}C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix}  OMgBr \\  | \end{smallmatrix}}{\mathop{C}}}\,-H\]                           \[C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix}  Br \\  | \end{smallmatrix}}{\mathop{C}}}\,-H\xleftarrow[{}]{P/B{{r}_{2}}}\underset{Isopropyl\,alcohol}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix}  | \\  C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix}  OH \\  | \end{smallmatrix}}{\mathop{C}}}\,-H}}\,\]