The treatment of alcohol of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

The difference in the products formed when alkyl chlorides are treated with aqueous KOH versus alcoholic KOH is due to the reaction conditions and the mechanism involved.

Aqueous KOH (aq. KOH):

When alkyl chlorides react with aqueous KOH, they undergo nucleophilic substitution via an SN1 or SN2 mechanism, depending on the nature of the alkyl halide.
In SN1 mechanism, the alkyl chloride undergoes ionization to form a carbocation intermediate, followed by attack of the hydroxide ion (OH⁻) to give an alcohol.
In SN2 mechanism, the hydroxide ion attacks the alkyl chloride directly, leading to simultaneous displacement of the chloride ion and formation of the alcohol.
Regardless of the mechanism, the ultimate product is an alcohol.
Alcoholic KOH (ethanolic KOH):

When alkyl chlorides are treated with alcoholic KOH, the reaction conditions favor elimination reactions, particularly the E2 mechanism.
In the E2 mechanism, a strong base like hydroxide ion abstracts a proton from a β-carbon adjacent to the chlorine atom, forming an alkene as the major product.
The alcoholic solvent (usually ethanol) helps in solubilizing the KOH and also stabilizes the alkoxide ions formed during the reaction.
The higher concentration of the base in alcoholic KOH compared to aqueous KOH promotes elimination reactions over substitution reactions.
So, in summary, the nature of the solvent (aqueous or alcoholic) determines whether the reaction favors substitution (aqueous KOH) or elimination (alcoholic KOH), leading to the formation of alcohols or alkenes, respectively, as the major products.