MCQ
When $CH_3CH_2CH_2CHCl_2$ is treated with $2\, gram$ equivalent $NaNH_2$, the product formed is
  • $C{H_3}C{H_2}C \equiv CH$
  • B
    $C{H_3}C{H_2}-CH = C{H_2}$
  • C

  • D

Answer

Correct option: A.
$C{H_3}C{H_2}C \equiv CH$
a
$CH_3-CH_2-CH_2-CHCl_2$ $\xrightarrow{{NaN{H_2}}}C{H_3} - C{H_2} - C \equiv CH$

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