MCQ
Which reaction is incorrectly written
- A$C{H_3} - CH = C{H_2}$ $\xrightarrow[{{H_2}{O_2}|\mathop O\limits^\Theta H}]{{{{{\text{(B}}{{\text{H}}_3}{\text{)}}}_2}}}$ $\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - C{H_2}}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|}\\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,OH}
\end{array}$ - B$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,O}\\
{\,\,\,\,\,\,\,\,||}\\
{C{H_3} - C - H}
\end{array}$ $\xrightarrow[{{H_2}O}]{{{\text{C}}{{\text{H}}_3}{\text{MgBr}}}}$ $\begin{array}{*{20}{c}}
{OH}\\
|\\
{C{H_3} - CH - C{H_3}}
\end{array}$ - C$\begin{array}{*{20}{c}}
{C{H_3} - C = C{H_2}}\\
|\\
{\,\,\,\,\,C{H_3}}
\end{array}$ $\xrightarrow{{{\text{dil}}{\text{. }}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}$ $\begin{array}{*{20}{c}}
{\,\,\,\,OH}\\
|\\
{C{H_3} - C - C{H_3}}\\
|\\
{\,\,\,\,\,\,\,C{H_3}}
\end{array}$ - ✓$C{H_3} - C \equiv N$ $\xrightarrow[\Delta ]{{{H_2}O/{H^ \oplus }}}$ $\begin{array}{*{20}{c}}
O \\
{||} \\
{C{H_3} - C - N{H_2}}
\end{array}$
