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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 -=CH`

C

D

Text Solution

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The correct Answer is:
B
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