# When $CH_3CH_2CHCl_2$ is treated with $NaNH_2$,the product formed is
$\begin{array}{1 1}(a)\;CH_3-CH=CH_2&(b)\;CH_3-C\equiv CH\\(c)\;CH_3CH_2CHNH_2&(d)\;CH_3CH_2CHCl\end{array}$