####23. When $CH_3 CH_2 CHCl_2$ is treated with $NaNH_2$, the product formed is
(a) $CH_3-CH=CH_2$
[2002]
(b) $CH_3-C \equiv CH$
(c)
(d)
(b) $CH_3-CH_2-CHCl_2 \xrightarrow[\Delta]{NaNH_2}$
$CH_3-CH=CHCl \xrightarrow[\Delta]{NaNH_2} \underset{\text{ Final Product }}{CH_3-C \equiv CH}$
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