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Answer
d
$\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_3}} \\
{|\,\,\,\,\,} \\
{\,C{H_3}}
\end{array}$ $\xrightarrow[{hv}]{{C{l_2}}}\begin{array}{*{20}{c}}
{\,\,Cl} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,C{H_3}}
\end{array}$ $+$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - C{H_2} - Cl} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_{3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}}
\end{array}$
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