The carboxylic acid that does not undergo Hell-Volhard-Zelinsky reaction, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \((CH_3)_3C-COOH\)
In the Hell-Volhard-Zelinsky (HVZ) reaction, a carboxylic acid is treated with chlorine or bromine in the presence of a small amount of red phosphorus, and a hydrogen on the α-carbon (the carbon next to –COOH) is replaced by halogen:
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}\text{/red P}} \ce{R-CHX-COOH} \)
So only an acid with at least one α-hydrogen can react. In \( \ce{(CH3)3C-COOH} \) (2,2-dimethylpropanoic acid), the α-carbon carries three methyl groups and no hydrogen, so it does not undergo the HVZ reaction.
Why the other options are incorrect
Option (1) \(CH_3COOH\) — Incorrect. The \( \ce{CH3} \) group has three α-hydrogens; acetic acid gives chloroacetic acid, \( \ce{ClCH2COOH} \).
Option (2) \(CH_3CH_2CH_2CH_2COOH\) — Incorrect. Pentanoic acid has two α-hydrogens on the \( \ce{CH2} \) next to –COOH.
Option (3) \((CH_3)_2CHCOOH\) — Incorrect. 2-methylpropanoic acid has one α-hydrogen, enough for the reaction.
NCERT (2026-27 edition):
Halogenation – “Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids. The reaction is known as Hell-Volhard-Zelinsky reaction.”