The product of Hell-Volhard-Zelinsky reaction of propanoic acid is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → α-halocarboxylic acid
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 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) β-halocarboxylic acid — Incorrect. The halogen goes only to the α-carbon, the one activated by the neighbouring C=O group; the β-carbon is not attacked.
Option (2) α-haloaldehyde — Incorrect. The –COOH group is kept in the product; nothing reduces it to –CHO.
Option (4) β-haloaldehyde — Incorrect. Wrong on both counts: wrong carbon and wrong functional group.
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:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 19-05 S2 | 13-05 S1 | – | – | – |