How many of the following carboxylic acids give an α-chlorocarboxylic acid when treated with chlorine in the presence of a small amount of red phosphorus?
Methanoic acid, Ethanoic acid, Propanoic acid, 2-Methylpropanoic acid, 2,2-Dimethylpropanoic acid, Benzoic acid, Phenylethanoic acid, 3-Phenylpropanoic acid
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
Chlorine or bromine with a small amount of red phosphorus replaces a hydrogen on the α-carbon, the carbon joined to the –COOH group. This is the Hell-Volhard-Zelinsky reaction. An acid can react only if its α-carbon carries at least one hydrogen.
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{Cl2}\text{/red phosphorus}} \ce{R-CH(Cl)-COOH} \)
Has an α-hydrogen:
Ethanoic acid, \( \ce{CH3COOH} \) — gives \( \ce{ClCH2COOH} \)
Propanoic acid, \( \ce{CH3CH2COOH} \) — gives \( \ce{CH3CH(Cl)COOH} \)
2-Methylpropanoic acid, \( \ce{(CH3)2CHCOOH} \) — gives \( \ce{(CH3)2C(Cl)COOH} \)
Phenylethanoic acid, \( \ce{C6H5CH2COOH} \) — gives \( \ce{C6H5CH(Cl)COOH} \)
3-Phenylpropanoic acid, \( \ce{C6H5CH2CH2COOH} \) — gives \( \ce{C6H5CH2CH(Cl)COOH} \)
No α-hydrogen:
Methanoic acid, \( \ce{HCOOH} \) — no α-carbon at all.
2,2-Dimethylpropanoic acid, \( \ce{(CH3)3CCOOH} \) — the α-carbon holds three methyl groups.
Benzoic acid, \( \ce{C6H5COOH} \) — the carbon next to –COOH is a ring carbon with no hydrogen.
Acids that have an α-hydrogen: ethanoic, propanoic, 2-methylpropanoic, phenylethanoic and 3-phenylpropanoic acid = 5.
NCERT (2026-27 edition):
8.9.4 Substitution Reactions in the Hydrocarbon Part – “1. 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.”