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n- propyl bromide on treating with alcoh...

n- propyl bromide on treating with alcoholic KOH produces

A

propyne

B

propanol

C

propane

D

propene

Text Solution

Verified by Experts

The correct Answer is:
D
`underset("Propyl bromide")(CH_(3)CH_(2)CH_(2)Br)+underset(("alc"))(KHO)tounderset("propene"" ")(CH_(3)CH=CH_(2)+KBr+H_(2)O)`
This reaction removes a molecule of HX and therefor, the reaction is called dehydrohalogenation. The hydrogen atom is eliminated from `beta` carbon atom (carbon atom next to the carbon to which halogen is attached). Therefore the reaction is also called `beta` - elimination reaction.
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