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 is heated with alcoholic KOH?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acrylic 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.
Heating a haloalkane with alcoholic KOH brings about β-elimination: a hydrogen atom leaves the β-carbon and the halogen leaves the α-carbon, and a C=C bond forms. Here the carbon carrying Br is the α-carbon and the \( \ce{CH3} \) group is the β-carbon, which has three hydrogen atoms:
\( \displaystyle \ce{CH3-CH(Br)-COOH + KOH} \xrightarrow[\text{heat}]{\text{alc.}} \ce{CH2=CH-COOH + KBr + H2O} \)
The product, propenoic acid, is acrylic acid.
Why the other options are incorrect
Option (1) α-Hydroxypropionic acid — Incorrect. Replacement of Br by –OH is the substitution that takes place with aqueous KOH; with alcoholic KOH elimination is the main reaction.
Option (3) α-Aminopropionic acid — Incorrect. An –NH2 group needs ammonia as the reagent; KOH supplies no nitrogen.
Option (4) No reaction — Incorrect. The compound has β-hydrogen atoms on the \( \ce{CH3} \) group, so elimination does take place.
NCERT (2026-27 edition):
2. Elimination reactions – “When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β-carbon and a halogen atom from the α-carbon atom. As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.”