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When CH3CH2CHCl2 is treated with NaNH2, ...

When `CH_3CH_2CHCl_2` is treated with `NaNH_2`, the product formed is

A

`CH_3CH=CH_2`

B

`CH_3C-=CH`

C

D

Text Solution

Verified by Experts

The correct Answer is:
B
It is a double dehydrohalogentation in the presence of a strong base `NaNH_2`.
`CH_3CH_2CHCl_2underset(-2HCl)overset(2NaNH_2)toCH_3C-=Chunderset(-NH_3)overset(NaNH_3)toCH_3C-=CNaoverset(H_2O)toCH_3C-=CH`
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