MCQ
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
    $C{H_3} - C \equiv CH$
  • C
    $C{H_3}C{H_2}CH(N{H_2})(Cl)$
  • $C{H_3}C{H_2}C{(N{H_2})_2}$

Answer

Correct option: D.
$C{H_3}C{H_2}C{(N{H_2})_2}$
d
It’s obvious.

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