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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