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Most halogenation reactions in organic chemistry need a double bond or an aromatic ring. The Hell-Volhard-Zelinsky (HVZ) reaction is different: it puts a halogen on the carbon next to a COOH group. It turns up in conversion questions and in the NCERT chapter's reaction list. This post explains what the reaction does, why red phosphorus is needed, how to predict products and how it ties to acid strength.
Is HVZ on the syllabus?
The 2026-27 reprint of the NCERT Class 12 chapter Aldehydes, Ketones and Carboxylic Acids keeps it in Section 8.9.4: "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 exam syllabi differ. The Syllabus for JEE (Main) 2026 opens its oxygen-compounds unit with "general methods of preparation, properties, reactions and uses", but for carboxylic acids it names only "acidic strength and factors affecting it". The NEET (UG) 2026 syllabus has the same wording. Neither names HVZ. The JEE (Advanced) 2026 brochure lists "reduction, halogenation, formation of esters, acid chlorides and amides" among the reactions of carboxylic acids. So HVZ is part of the NCERT chapter and falls under JEE Advanced's halogenation of acids. Check the current year's documents.
What does the reaction do?
RCH₂COOH + X₂ (red P) → RCHXCOOH + HX, where X is Cl or Br.
Only a hydrogen on the α-carbon, the one next to COOH, is replaced. The COOH group itself stays.
The reaction carries the names of three chemists. Hell's 1881 paper in the Berichte was titled Ueber eine neue Bromirungsmethode organischer Säuren ("on a new bromination method for organic acids"), and Volhard wrote on the preparation of α-brominated acids in 1887.
Why is red phosphorus needed?
A carboxylic acid barely enolises, so Br₂ alone does little at the α-carbon. The OpenStax Organic Chemistry section on alpha bromination of acids/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.04%3A_Alpha_Bromination_of_Carboxylic_Acids) explains the way round it:
PBr₃ converts the acid into an acid bromide, RCH₂COBr, plus HBr.
HBr catalyses enolisation of the acid bromide.
The enol reacts with Br₂ at the α-carbon, giving the α-bromo acid bromide.
Water hydrolyses the acid bromide back to a carboxylic acid: the α-bromo acid.
Where does PBr₃ come from? NCERT's haloalkanes chapter notes that phosphorus tribromide is usually generated in situ by the reaction of red phosphorus with bromine. That is why only a small amount of red phosphorus is needed in HVZ.
Worked example 1: propanoic acid
CH₃CH₂COOH with Br₂ and red P.
Find the α-carbon: the CH₂ next to COOH. It has two hydrogens.
Replace one: CH₃CHBrCOOH, 2-bromopropanoic acid.
The CH₃ group is the β-carbon. It is not touched.
Worked example 2: which acids react, and at which carbon?
Methanoic acid, HCOOH. No α-carbon at all, so no HVZ product.
2-Methylpropanoic acid, (CH₃)₂CHCOOH. One α-hydrogen, giving (CH₃)₂CBrCOOH, 2-bromo-2-methylpropanoic acid.
Phenylethanoic acid, C₆H₅CH₂COOH. The CH₂ is the α-carbon, giving C₆H₅CHBrCOOH. The ring is not brominated by this route.
Benzoic acid, C₆H₅COOH. The carbon next to COOH is a ring carbon with no hydrogen, so no HVZ product. Ring substitution of aromatic acids is a separate reaction, also in NCERT's Section 8.9.4.
Worked example 3: HVZ and acid strength
Ethanoic acid with Cl₂ and red P gives chloroethanoic acid, ClCH₂COOH. NCERT's acidity order, in the same chapter, ranks
CCl₃COOH > CHCl₂COOH > ClCH₂COOH > CH₃COOH
Each chlorine on the α-carbon is electron-withdrawing and stabilises the carboxylate ion, so the acid gets stronger. NCERT gives the pKa of ethanoic acid as 4.76; a smaller pKa means a stronger acid. This is the part of the topic that JEE Main and NEET-UG name directly, so it is worth linking the two.
What can the product be used for?
The α-halo acid has a reactive C–X bond. OpenStax notes that if methanol is added instead of water at the end, the α-bromo acid bromide gives an α-bromo ester. The C–Br bond can then undergo substitution like a haloalkane, covered in our post on SN1 and SN2 in haloalkanes.
Common mistakes and how to avoid them
Halogenating the β-carbon. Only the carbon next to COOH reacts.
Applying HVZ to acids with no α-hydrogen. Check methanoic and benzoic acids first.
Confusing HVZ with the haloform reaction. Haloform needs a CH₃CO group in an aldehyde or ketone and splits off CHX₃. HVZ keeps the carbon chain and the COOH.
Forgetting the phosphorus. Without it, there is no acid bromide and little reaction.
For the rest of the chapter, see our guides to carboxylic acids and their derivatives and named reactions in organic chemistry. If you keep mixing up which reagent does what, Fermi's AI tutor can work out what you already know and build a plan for the session.
A practice routine
Write the HVZ product for five acids, marking the α-carbon first.
Write the four mechanism steps from memory and check against the list above.
Rank three chloro-substituted ethanoic acids by acid strength and explain the order.
Write two conversions that start with HVZ and end with a substitution at the α-carbon.
Try it now: name the product when butanoic acid reacts with Br₂ and red phosphorus, then say which carbon was not touched.
