# When $CH_3CH_2CHCl_2$ is treated with $NaNH_2$,the product formed is

$\begin{array}{1 1}(a)\;CH_3-CH=CH_2&(b)\;CH_3-C\equiv CH\\(c)\;CH_3CH_2CHNH_2&(d)\;CH_3CH_2CHCl\end{array}$

$CH_3CH_2CHCl_2+2NaNH_2\quad\large\xrightarrow{liq.NH_3/196K}\quad$$CH_3C\equiv CH+2NaCl+2NH_3$
Hence (b) is the correct answer.
edited Apr 3, 2014