MCQ
Product $(C)$ of this reaction is


- A

- ✓

- C

- D








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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,}\\
{\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3} - C - CH = C{H_2}}\\
{\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ $\xrightarrow{{{H_2}O/{H^ \oplus }}}$ $\mathop A\limits_{{\rm{(major)}}} $ + $\mathop B\limits_{{\rm{(minor)}}} $
The major product is

$N{H_3} + {H_2}O \to NH_4^ + + O{H^ - }$ is