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n-propyl bromide on treating with alcoho...

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 bromine")(CH_3CH_2CH_2)Br + underset(alc.)(KOH) rarr underset("propene")(CH_3CH)=CH_2 + KBr + H_(2)O`
This reaction removes a molecule of HX and therefore, 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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