• # question_answer When$C{{H}_{3}}C{{H}_{2}}CHC{{l}_{2}}$is treated with$NaN{{H}_{2}}$, the product formed is A) $C{{H}_{3}}-CH=C{{H}_{2}}$ B) $C{{H}_{3}}-CH\equiv CH$ C) D)

$C{{H}_{3}}-C{{H}_{2}}-CHC{{l}_{2}}\xrightarrow[\Delta ]{NaN{{H}_{2}}}$$C{{H}_{3}}-CH=CHCl\xrightarrow[\Delta ]{NaN{{H}_{2}}}$$C{{H}_{3}}-C\equiv CH$