MCQ
Predict the correct intermediate and product in the following reaction :
${H_3}C - C \equiv CH\,\xrightarrow[{HgS{O_4}}]{{{H_2}O,\,{H_2}S{O_4}}}\,\mathop {Intermediate}\limits_{(A)} \,$$ \to \,\mathop {\Pr oduct}\limits_{(B)} $
- A$A \,: \, \begin{array}{*{20}{c}}
{{H_3}C - C = C{H_2}} \\
| \\
{OH}
\end{array}$ $ B\,\, :\, \begin{array}{*{20}{c}}
{{H_3}C - C = C{H_2}} \\
| \\
{S{O_4}}
\end{array}$ - B$A\,\, :\,\begin{array}{*{20}{c}}
{{H_3}C - C - C{H_3}} \\
{|\,|} \\
O
\end{array}$ $B\,\,:\, H_3C-C=CH$ - ✓$A \,: \, \begin{array}{*{20}{c}}
{{H_3}C - C = C{H_2}} \\
| \\
{OH}
\end{array}$ $B\,\, :\,\begin{array}{*{20}{c}}
{{H_3}C - C - C{H_3}} \\
{|\,|} \\
O
\end{array}$ - D$ A\,\, :\, \begin{array}{*{20}{c}}
{{H_3}C - C = C{H_2}} \\
| \\
{S{O_4}}
\end{array}$ $B\,\, :\,\begin{array}{*{20}{c}}
{{H_3}C - C - C{H_3}} \\
{|\,|} \\
O
\end{array}$
