Butanal with dilute NaOH gives

MCQ

Butanal with dilute $NaOH$ gives

A
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,OH} \\
\,\,\,\,\,\,\,\,|
\end{array}} \\
{C{H_3}C{H_2}C{H_2}CC{H_2}} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H}
\end{array}C{H_2}C{H_2}C{H_2}CHO$
B
$\begin{array}{*{20}{c}}
{O\,\,\,\,\,\,\,\,\,\,} \\
{||\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3}C{H_2}C{H_2}CC{H_2}C{H_2}C{H_2}CHO}
\end{array}$
C
$OHCC{H_2}C{H_2}C{H_2}C{H_2}C{H_2}C{H_2}CHO$
✓
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,CHO} \\
{|\,\,\,\,\,\,\,\,\,\,|}
\end{array}} \\
{C{H_3}C{H_2}C{H_2} - C - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{H\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$

✓

Answer

Correct option: D.

$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,\,\,\,\,OH\,\,\,\,CHO} \\
{|\,\,\,\,\,\,\,\,\,\,|}
\end{array}} \\
{C{H_3}C{H_2}C{H_2} - C - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{H\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$

d
(d) $2C{H_3} - C{H_2} - C{H_2} - CHO + {\text{dil}}{\text{. }}NaOH \to $

$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,OH\,\,\,\,CHO} \\
{|\,\,\,\,\,\,\,\,\,\,|}
\end{array}} \\
{C{H_3}C{H_2}C{H_2} - C - CH - C{H_2} - C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,} \\
{H\,\,\,\,\,\,\,\,}
\end{array}$