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When CH(2)CH(2)CHCl(2) is treated with N...

When `CH_(2)CH_(2)CHCl_(2)` is treated with `NaNH_(2)` the product formed is

A

`CH_(3)CH=CH_(2)`

B

`CH_(3)-C-=CH_(2)`

C

D

Text Solution

Verified by Experts

The correct Answer is:
b
`CH_(3)CH_(2)CHCl_(2)overset(NaNH_(2))underset(-HCl)rarrCH_(3)CH=CHCloverset(NaNH_(2))underset(-HCl)rarrCH_(3)C-=CH`
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