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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_3-CH=CH_2`

B

`CH_3-C equiv CH`

C

`CH_3CH_2CH(NH_2)(Cl)`

D

`CH_3CH_2C(NH_2)_2`

Text Solution

Verified by Experts

The correct Answer is:
B
`CH_3-CH_2-CHCl_2+2NaNH_2 overset(liq.NH_3)underset(196K)(rarr)CH_3C equiv CH+2NaCl+2NH_3`
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