rhe treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols hut in presence of alocholic KOH, alkenes are major products. Explain.

Explanation

In aqueous KOH, KOH is almost completely ionized to give OH- ions. These being strong nucleophiles result into substitution reaction on alkyl chlorides to from alcohols.
Moreover in aqueous solution the `OH^(-)` ions are highly hydrated (solvated). The hydration reduces the basic character of `OH^(-)` ions which therefore fails to abstract a hydrogen from the `beta` carbon of the alkyl halide to form an alkene.
e.g. `underset("ethyl Bromide")(CH_3 CH_2 Br + KOH (aq)) to underset("ethanol")(CH_3 CH_2 OH) + KBr`
On the other hand an alcoholic solution of KOH contains alkoxide `(RO^(-))` ions which being stronger base than `OH^(-)` ions preferably abstract a hydrogenfrom `beta` carbon of the alkyl halide eliminates a molecules of `H_2O` from an alkyl halide to form an alkene
`underset("Ethyl Bromide")(CH_3CH_2Br) + KOH (alc.) to underset("ethene")(CH_2 = CH_2 ) + KBr + H_2O`