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

(N/A) In an aqueous solution,$KOH$ almost completely ionizes to give $OH^{-}$ ions. The $OH^{-}$ ion acts as a strong nucleophile,which leads the alkyl chloride to undergo a nucleophilic substitution reaction ($S_N2$ or $S_N1$) to form an alcohol.
$R-Cl + KOH_{(aq)} \to R-OH + KCl$
On the other hand,an alcoholic solution of $KOH$ contains ethoxide $(C_2H_5O^-)$ ions (if ethanol is used),which are strong bases. These ions abstract a proton from the $\beta$-carbon of the alkyl chloride,leading to a dehydrohalogenation reaction (elimination) to form an alkene.
$R-CH_2(\beta)-CH_2(\alpha)-Cl + KOH_{(alc)} \to R-CH=CH_2 + KCl + H_2O$
In an aqueous medium,the $OH^{-}$ ion is highly solvated by water molecules,which reduces its basicity but allows it to act as a nucleophile. In an alcoholic medium,the base is less solvated and more reactive,favoring the abstraction of a $\beta$-hydrogen atom.