Which of the following is a method of preparing an alkyl chloride from an alcohol?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Passing dry hydrogen chloride gas through a solution of the alcohol
In an alcohol, the –OH group is replaced by a halogen when it reacts with a concentrated halogen acid, a phosphorus halide or thionyl chloride. With hydrogen chloride, the reaction is done by passing dry HCl gas through a solution of the alcohol, or by heating the alcohol with concentrated aqueous HCl:
\( \displaystyle \ce{R-OH + HCl -> R-Cl + H2O} \)
Why the other options are incorrect
Option (1) Treating the alcohol with chlorine in the presence of iron(III) chloride — Incorrect. Chlorine with a Lewis acid such as FeCl3 is used to put chlorine on an aromatic ring by electrophilic substitution, giving aryl chlorides.
Option (2) Treating the alcohol with cuprous chloride — Incorrect. Cuprous chloride replaces the diazonium group of an arenediazonium salt by –Cl (Sandmeyer’s reaction); it does not replace an –OH group.
Option (3) Treating the alcohol with chlorine in bright sunlight — Incorrect. This is free radical chlorination, used on alkanes. It replaces hydrogen atoms, not the –OH group, and gives a mixture of products.
NCERT (2026-27 edition):
6.4 Methods of Preparation of Haloalkanes – “The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride.”
6.4.1 From Alcohols – “The preparation of alkyl chloride is carried out either by passing dry hydrogen chloride gas through a solution of alcohol or by heating a mixture of alcohol and concentrated aqueous halogen acid.”