n- propyl bromide on treating with alcoholic KOH produces – InterviewSolution

1.	n- propyl bromide on treating with alcoholic KOH produces
Answer» propyne propanol propane propene Solution :`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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