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Q. Alcoholic KOH is used for

Jharkhand CECEJharkhand CECE 2012

Solution:

$ C{{H}_{3}}-C{{H}_{2}}Br+KOH(alc.)\xrightarrow{\text{dehydrohalogenation}} $
$ C{{H}_{2}}=C{{H}_{2}}+KBr+{{H}_{2}}O. $
In alcoholic KOH, alkoxide ion
$ (R\bar{O}) $ is present which is a strong base favour elimination reaction.