The reagent used to convert propanol to 1-bromopropane is:

The correct answer is Option (1) → \(PBr_3\)

To convert an alcohol into an alkyl bromide, the –OH group must be replaced by –Br. This is done with phosphorus tribromide (often made in the flask from red phosphorus and bromine) or with HBr. Three molecules of alcohol react with one of \( \ce{PBr3} \):

    \( \displaystyle \ce{3CH3CH2CH2OH + PBr3 -> 3CH3CH2CH2Br + H3PO3} \)

So propan-1-ol gives 1-bromopropane with \( \ce{PBr3} \).

Why the other options are incorrect

• Option (2) \(Br_2\) — Incorrect. Bromine adds to C=C bonds or substitutes hydrogen; it does not replace an –OH group.

• Option (3) \(CH_3Br\) — Incorrect. Methyl bromide is itself an alkyl halide; with an alcohol (as its alkoxide) it would give an ether (Williamson synthesis), not 1-bromopropane.

• Option (4) \(CBr_3\) — Incorrect. There is no such compound; carbon forms \( \ce{CBr4} \) or \( \ce{CHBr3} \).

NCERT (2026-27 edition):
From Alcohols – “The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride. … Phosphorus tribromide and triiodide are usually generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine respectively.”