The treatment of ethyl bromide with aqueous KOH results ethyl alcohol but in presence of alcoholic KOH, ethylene is the major product. Explain

Explanation

Difference lies in the mode and extent of dissociation of KOH in the presence of different solvents. `OH^(-)` is a good nucleophile while `C_(2)H_(5)O^(-)` is a strong base.
In the presence of water, KOH dissociates completely into `OH^(-)` ions which being a strong nucleophile brings about substitution on alkyl halides and produces alcohols from alkyl halides. Further, in the aqueous solution,
`OH^(-)` ions are highly solvated (hydrated). This solvation reduces the basic character of `OH^(-)` ions which therefore, fails to abstract a H-atom from the `beta`-carbon of the alkyl chloride to form an alkene. In alcoholic medium, solution also contains `C_(2)H_(5)O^(-)`, ethoxide ions in addition to `OH^(-)` ions. Being a stronger base than `OH^(-)` they abstract `H^(+)` ion from `beta`-C-atom giving rise to an alkene as a major product (dehydrohalogenation).