2, ,2- Dimethylpropanoic acid on heating with sodalime gives - Vidyadip

MCQ

$2,\,2-$ Dimethylpropanoic acid on heating with sodalime gives

A
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,\,C{H_3}\,}
\end{array}$
B
$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,C{H_3}\,} \\
{|\,\,} \\
{C{H_3} - C = C{H_2}}
\end{array}$
✓
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}\,\,} \\
{|\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_3}}
\end{array}$
D
$\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,\,OH\,}
\end{array}$

✓

Answer

Correct option: C.

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

c
$\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{C{H_3}} \\
{|\,\,\,\,\,\,\,}
\end{array}} \\
{C{H_3} - C - COOH} \\
{|\,\,\,\,\,\,\,} \\
{C{H_3}}
\end{array}$ $\xrightarrow[\Delta ]{{NaOH + CaO}}$ $\begin{array}{*{20}{c}}
{\,\,C{H_3}} \\
{|\,\,\,\,} \\
{C{H_3} - CH - C{H_3}}
\end{array}$

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