When $CH_3CH_2CH_2CHCl_2$ is treated with $2 \ g$ equivalent of $NaNH_2$,the product formed is:

  • A
    $CH_3CH_2C \equiv CH$
  • B
    $CH_3CH_2CH=CH_2$
  • C
    $CH_3CH_2CH_2CH(NH_2)_2$
  • D
    $CH_3CH_2CH_2CH(Cl)(NH_2)$

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