Which products are formed when an alkyl chloride is treated with aque…

The correct answer is Option (4) → Both (1) and (2)

Aqueous KOH supplies \( \ce{OH^-} \) ions, a nucleophile. In a nucleophilic substitution, \( \ce{OH^-} \) replaces the chlorine of the alkyl chloride, and chlorine leaves as a chloride ion. The alkyl group ends up as an alcohol, and the chloride ion stays with \( \ce{K^+} \) as KCl:

    \( \displaystyle \ce{R-Cl + KOH(aq) -> R-OH + KCl} \)

So both an alcohol and KCl are formed.

Why the other options are incorrect

Option (1) Alcohol — Incorrect. The alcohol is formed, but KCl is formed along with it, so the alcohol alone is not the full answer.

Option (2) KCl — Incorrect. KCl is formed, but the alkyl group ends up as an alcohol at the same time, so KCl alone is not the full answer.

Option (3) Alkane — Incorrect. An alkane is not formed: replacing Cl by H needs a reducing agent, not KOH.

NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “… In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place and halogen atom, called leaving group departs as halide ion.”

“(a) Substitution nucleophilic bimolecular (S_N2)
The reaction between CH₃Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics, i.e., the rate depends upon the concentration of both the reactants.”