Question

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

Answer 1

$\text{Treatment of alkyl chlorides with aq. KOH}$

In an aqueous solution, $KOH$ almost completely ionize to give $O{H}^{-}$ ions. $O{H}^{-}$ ion is a strong nucleophile, which react with the alkyl chloride to undergo a nucleophilic substitution reaction to form alcohol.

$R-Cl+KOH\left(aq\right)\to R-OH+KCl$
Alkyl Alcohol
Chloride


$\text{Treatment of alkyl chlorides with alc. KOH}$

On the other hand, an alcoholic solution of $KOH$ contains alkoxide $\left(R{O}^{-}\right)$ ion, which is a stronger base than $O{H}^{-}$. Thus, it can abstract a hydrogen from the $\beta$-carbon of the alkyl chloride and form an alkene by eliminating a molecule of $HCl$.