Hell-Volhard Zelinsky reaction is used for the formation of

The correct answer is Option (4) → \( α \)-halocarboxylic acids

In the Hell-Volhard-Zelinsky (HVZ) reaction, a carboxylic acid that has an α-hydrogen is treated with chlorine or bromine in the presence of a small amount of red phosphorus, then with water. The halogen replaces an α-hydrogen, i.e. a hydrogen on the carbon next to the –COOH group:

    \( \displaystyle \ce{R-CH2-COOH + X2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) red P}} \ce{R-CH(X)-COOH + HX} \)

For example, with propanoic acid the α-carbon is C-2:

    \( \displaystyle \ce{CH3CH2COOH + Br2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) red P}} \ce{CH3CH(Br)COOH + HBr} \)

The product, 2-bromopropanoic acid, is an α-halocarboxylic acid.

Why the other options are incorrect

• Option (1) alcohols — Incorrect. The –COOH group is kept in the product. A carboxylic acid gives an alcohol only on reduction, for example with \( \ce{LiAlH4} \), not by halogenation.

• Option (2) aldehydes — Incorrect. Nothing in the reaction reduces –COOH to –CHO; only an α-hydrogen is replaced by a halogen.

• Option (3) ketones — Incorrect. The reaction forms no new carbon–carbon bond and keeps the acid group, so no ketone can form.

NCERT (2026-27 edition):
Substitution Reactions in the Hydrocarbon Part

Note: The same/similar question has been asked in different years by NTA as per the details given below: