The reason for this is that aqueous #KOH# being alkaline in
nature gives #OH^-# ions. These #OH^-# ions then act as strong
nucleophile and attack on electrophilic carbon of alkyl halide
molecule. Thus the leaving group leaves and #OH^-# substitute the halogen atom of alkyl halide molecule, forming alcohol as a product. Consequently, aqueous #KOH# gives substitution reactions.
On the other hand, alcoholic #KOH#(mostly having ethanol) produces #C_2H_5O^-# ions which are stronger base than #OH^-# ions. So these abstract the #beta#- hydrogen of alkyl halide molecule, forming an alkene. That's the reason behind the statement; Alcoholic #KOH# give elimination reactions.
Hope it helps...
