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When CH(3)CH(2)CHCl(2) is treated with "...

When `CH_(3)CH_(2)CHCl_(2)` is treated with `"NaNH"_(2)` the product formed is:

A

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

B

`CH_(3) - C equiv CH`

C

D

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The correct Answer is:
B
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