What is the product of the following reaction?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(α\)-Halocarboxylic acid
In the Hell-Volhard-Zelinsky reaction the halogen goes to the α-carbon, the carbon next to –COOH. Phosphorus and the halogen form \( \ce{PX3} \), which turns some of the acid into the acyl halide; the acyl halide enolises easily and takes up the halogen at its α-carbon, and water finally hydrolyses it back to the acid:
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}\text{/red P}} \ce{R-CHX-COOH + HX} \)
So the product is an α-halocarboxylic acid.
Why the other options are incorrect
Option (2) \(β\)-Halocarboxylic acid — Incorrect. Only the α-carbon is activated by the neighbouring carbonyl group; the β-carbon is not attacked.
Option (3) \(α,β\)-Dihalocarboxylic acid — Incorrect. Halogens on two neighbouring carbons come from addition of a halogen to a C=C bond, which this saturated acid does not have.
Option (4) 1,1-Dihalocarboxylic acid — Incorrect. C1 of a carboxylic acid is the –COOH carbon itself, which carries no hydrogen to be replaced.
NCERT (2026-27 edition):
Halogenation
