What will be the product of the given reaction?

The correct answer is Option (1) → 1-(Chloromethyl)-3-methoxybenzene

The starting compound is 3-methoxybenzyl alcohol. Thionyl chloride replaces the –OH group of an alcohol by –Cl; the side products \( \ce{SO2} \) and HCl are gases, and pyridine, a base, takes up the HCl. The methoxy group on the ring is not touched:

    \( \displaystyle \ce{CH3O-C6H4-CH2OH + SOCl2} \xrightarrow{\text{pyridine}} \ce{CH3O-C6H4-CH2Cl + SO2 + HCl} \)

The –CH₂OH group becomes –CH₂Cl, still meta to –OCH₃: the product is 1-(chloromethyl)-3-methoxybenzene.

Why the other options are incorrect

Option (2) 3-Methoxybenzaldehyde — Incorrect. An aldehyde would need an oxidising agent such as PCC; thionyl chloride substitutes the –OH group and does not oxidise the alcohol.

Option (3) 3-Methoxytoluene — Incorrect. A –CH₃ group would mean the –OH was replaced by hydrogen (a reduction); here it is replaced by chlorine.

Option (4) 3-(Chloromethyl)phenol — Incorrect. A phenol would need the ether C–O bond to be cleaved, which takes concentrated HI or HBr at high temperature; thionyl chloride does not cleave ethers.

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. Thionyl chloride is preferred because in this reaction alkyl halide is formed along with gases SO₂ and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides.”