Overall reaction
R–CH₂–COOH + Br₂ (red P or PBr₃) → R–CHBr–COOH + HBr
α-Bromination of a carboxylic acid through its acyl bromide enol.
Name reaction · mechanism
Hell–Volhard–Zelinsky Reaction
The Hell–Volhard–Zelinsky (HVZ) reaction brominates a carboxylic acid at the α-carbon. Phosphorus tribromide (or red phosphorus + Br₂) turns the acid into the acyl bromide, which enolises far more easily than the acid; the enol attacks bromine, and water finally converts the α-bromo acyl bromide back to the acid.
Mechanism step by step
- Protonation. PBr₃ has made the acyl bromide; HBr protonates its C=O
- Enol. Loss of an α-H
- α-Bromination. The enol attacks Br₂
- Deprotonation. The α-bromo acyl bromide
Each step is checked electron by electron: the curly arrows give exactly the structures drawn.
Key points
- Carboxylic acids barely enolise; the acyl bromide enolises readily.
- The α-bromo acyl bromide swaps Br for OH with more acid, so PBr₃ is catalytic.
- Water at the end hydrolyses the acyl bromide to the acid (addition–elimination).
- The α-bromo acid is a route to α-amino acids (NH₃) and α-hydroxy acids.
For JEE and NEET
- Needs at least one α-hydrogen.
- Only the α-position is halogenated (Cl₂ works similarly).
- Products are routes to α-amino acids (NH₃) and α-hydroxy acids (aqueous base).
- Red phosphorus + Br₂ makes PBr₃ in situ.
FAQ
Why is PBr₃ needed?
Carboxylic acids barely enolise. PBr₃ converts the acid to the acyl bromide, whose enol forms readily and reacts with Br₂.
Does HVZ work on acids without α-hydrogen?
No — for example, (CH₃)₃CCOOH or benzoic acid do not give HVZ products.
What is the product from propanoic acid?
2-Bromopropanoic acid, CH₃CHBrCOOH.
Draw it in MolDraw
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