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 α-bromopropionic acid undergoes substitution with ammonia?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → α-Aminopropionic acid
In the passage, the Hell-Volhard-Zelinsky reaction puts a halogen on the α-carbon of a carboxylic acid. α-Bromopropionic acid is \( \ce{CH3-CH(Br)-COOH} \): the bromine sits on an sp3 carbon, so it reacts like a haloalkane.
Ammonia is a nucleophile. As with an alkyl halide (ammonolysis), the nitrogen of \( \ce{NH3} \) replaces the bromine and an –NH2 group takes its place:
\( \displaystyle \ce{CH3-CH(Br)-COOH + 2NH3 -> CH3-CH(NH2)-COOH + NH4Br} \)
The second ammonia molecule takes up the HBr formed. The product, 2-aminopropanoic acid, is the amino acid alanine.
Why the other options are incorrect
Option (1) Acrylic acid — Incorrect. \( \ce{CH2=CH-COOH} \) is formed by elimination of HBr with alcoholic KOH; in substitution the bromine is replaced and no double bond forms.
Option (3) α-Hydroxypropionic acid — Incorrect. An –OH group comes in when the nucleophile is the hydroxide ion (aqueous alkali), not ammonia.
Option (4) α,α-Dibromopropionic acid — Incorrect. Ammonia replaces the bromine; it does not add another bromine atom.
NCERT (2026-27 edition):
Table 6.4: Nucleophilic Substitution of Alkyl Halides (R–X)
