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.
What is formed when phenylacetic acid reacts with chlorine in the presence of a small amount of red phosphorus?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → α-Chlorophenylacetic acid
The passage states that carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of a small amount of red phosphorus (Hell-Volhard-Zelinsky reaction).
In phenylacetic acid, \( \ce{C6H5-CH2-COOH} \), the α-carbon is the \( \ce{CH2} \) group next to –COOH, and it carries two α-hydrogen atoms. One of them is replaced by chlorine:
\( \displaystyle \ce{C6H5-CH2-COOH} \xrightarrow[\text{(ii) } \ce{H2O}]{\text{(i) } \ce{Cl2}\text{/red P}} \ce{C6H5-CH(Cl)-COOH} \)
The product, 2-chloro-2-phenylethanoic acid, is α-chlorophenylacetic acid.
Why the other options are incorrect
Option (2) β-Chlorophenylacetic acid — Incorrect. The carbon after the α-carbon is a ring carbon, and the reaction takes place only at the α-position.
Option (3) m-Chlorophenylacetic acid — Incorrect. Chlorine with red phosphorus substitutes the α-hydrogen of the side chain, not a hydrogen of the benzene ring.
Option (4) p-Chlorophenylacetic acid — Incorrect. This is also a ring-substituted product; the Hell-Volhard-Zelinsky reaction does not touch the ring.