Organic Chemistry · Carbonyl Alpha-Substitution Reactions
Alpha Bromination of Carboxylic Acids
On this page 9 sections
In 30 seconds
Carboxylic acids RCH₂COOH have α-hydrogens, but they do not brominate under the conditions that work for aldehydes and ketones (topic 03). A carboxylic acid cannot enolize readily enough: the hydroxyl group makes the carbonyl less electrophilic, and the α-H is far less acidic than in a ketone. The Hell–Volhard–Zelinsky (HVZ) reaction α-Bromination of a carboxylic acid using Br₂ with PBr₃ (or P + Br₂). Full entry → solves this by first converting the acid into an acyl bromide, which enolizes and brominates easily, then hydrolyzing back to the acid. The net result is clean α-bromination:
RCH2COOH + Br2 PBr3⟶ RCHBrCOOH + HBr
HVZ is the standard route to α-bromocarboxylic acids, key intermediates for α-amino acids and other α-functionalized compounds.
Why this matters
- α-Bromo acids are springboards to amino acids. Treating an α-Bromocarboxylic acid RCHBrCOOH — Br on the carbon adjacent to COOH. Full entry → with ammonia displaces Br by SN2 Bimolecular nucleophilic substitution (backside attack, one step). Full entry →, giving an α-amino acid — historically a major laboratory synthesis of amino acids, still used for non-natural analogs.
- Synthetic logic. When a functional group blocks a pathway, temporarily convert it to a more reactive derivative, run the chemistry, then convert back — a general principle.
- Real-world use. α-Halo acids are used in synthesizing pharmaceuticals, peptides, and chiral building blocks.
The college version
Core Concepts
Why carboxylic acids resist direct α-bromination
In an aldehyde or ketone, the carbonyl oxygen can be protonated and the α-H removed to form an Enol C=C–OH tautomer (of the acyl bromide here). Full entry →. In a carboxylic acid, the hydroxyl group donates electron density into the carbonyl, making it less electrophilic and disfavoring enolization. The acid's O–H is also far more acidic than its α-C–H, so reagents interact with the OH instead. Direct reaction of a carboxylic acid with Br₂ is impractically slow.
The Hell–Volhard–Zelinsky strategy
The reaction uses bromine with phosphorus tribromide (PBr₃) — or red phosphorus plus bromine, which generates PBr₃ in situ. The sequence:
- Acyl bromide formation. PBr₃ converts the acid to the acyl bromide, RCH₂COBr, releasing HBr and phosphorous acid derivatives. The acyl bromide's carbonyl is more electrophilic than the acid's, and its α-H's are more acidic — enolization is now feasible.
- α-Bromination. HBr (from step 1) catalyzes enolization of the acyl bromide to its enol; the enol attacks Br₂ at the α carbon, giving the α-bromo acyl bromide, RCHBrCOBr.
- Hydrolysis. Water hydrolyzes the α-bromo acyl bromide back to the α-bromocarboxylic acid, RCHBrCOOH.
Mechanism arrows in words: the acid's OH lone pair attacks PBr₃; P–Br bonds break stepwise; the carbonyl re-forms to give the acyl bromide. HBr protonates the acyl bromide's carbonyl; the α-H is removed to form the enol; the enol's terminal carbon attacks Br₂ (arrow from C=C to Br–Br); deprotonation re-forms the C=O; water adds and HBr is lost, giving the α-bromo acid.
Scope and limitations
- Requires an α-H; acids without one (e.g., benzoic acid, C₆H₅COOH) do not react.
- Bromine is standard; chlorine works with PCl₃; iodine is slower and less common.
- The Br lands on the α carbon adjacent to the carbonyl, at the position that forms the more stable (more substituted) enol.
- The product is still a carboxylic acid: the carboxyl group is intact, and the C–Br bond is set up for SN2 displacement.
From α-bromo acid to α-amino acid
The classic application: treat the α-bromo acid with excess ammonia. NH₃ deprotonates the carboxyl group, then the α carbon bearing Br undergoes SN2 attack by NH₃:
RCHBrCOOH + 2NH3 → RCH(NH2)COOH + NH4Br
Two equivalents of ammonia: one is the nucleophile, the second neutralizes the HBr formed.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| HVZ bromination | Ordinary ketone α-bromination | Ketones enolize directly (acid or base catalysis); acids need the acyl bromide detour via PBr₃. |
| Acyl bromide | Alkyl bromide | RCOBr has Br on the carbonyl carbon (acyl halide, electrophilic C=O); RCH₂Br is an alkyl bromide (leaving group for SN2). |
| Where Br lands | — | Br goes to the α carbon (adjacent to COOH), not the terminal carbon of the chain. |
| PBr₃ role | Br₂ role | PBr₃ converts acid → acyl bromide (activation); Br₂ provides the halogen that ends up on the α carbon. |
| α-Amino acid synthesis | Direct amination of the acid | You displace Br at the α carbon with NH₃; the carboxyl carbon is not the site of amination here. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
A carboxylic acid is like a swing set whose frame has a "Do Not Touch" sign (the OH group) — the kids (hydrogens) beside it won't leave for a bromine ball. The Hell–Volhard–Zelinsky trick swaps the sign for a "Come On Up" handle (bromine on the carbonyl), so a kid climbs off, grabs a bromine ball, and then we swap the handle back for the original sign. Net result: the kid is gone and a bromine ball sits in that spot.
Worked example
Example 1 — Synthesis of 2-bromopropanoic acid
Propanoic acid (CH₃CH₂COOH) is treated with Br₂ and PBr₃. Write the three-step sequence and the overall equation.
Step 1 — acyl bromide. CH₃CH₂COOH + PBr₃ → CH₃CH₂COBr (+ HBr, phosphorous acid).
Step 2 — α-bromination. CH₃CH₂COBr + Br₂ → CH₃CHBrCOBr + HBr.
Step 3 — hydrolysis. CH₃CHBrCOBr + H₂O → CH₃CHBrCOOH + HBr.
Overall:
CH3CH2COOH + Br2 PBr3⟶ CH3CHBrCOOH + HBr
The product has Br on C2 (the α carbon, adjacent to COOH), not on C3.
Example 2 — Stoichiometry: grams of Br₂ needed (dimensional analysis)
A student wants to brominate 5.00 g of butanoic acid (CH₃CH₂CH₂COOH, molar mass 88.11 g/mol) to 2-bromobutanoic acid via HVZ. Calculate the mass of Br₂ (molar mass 159.81 g/mol) for a 1:1 molar ratio.
Step 1 — mole relationship. The balanced reaction uses 1 mol Br₂ per mol acid:
1 mol Br21 mol acid
Step 2 — moles of acid:
5.00 g acid × 1 mol acid88.11 g acid = 5.67 × 10-2 mol acid
Step 3 — grams of Br₂:
5.67 × 10-2 mol Br2 × 159.81 g Br21 mol Br2 = 9.06 g Br2
The unit chain: g acid → mol acid → mol Br₂ → g Br₂. PBr₃ is catalytic in halogen equivalents (it converts the acid to the acyl bromide, and HBr from the reaction drives further catalysis), so the bromine requirement is set by the substrate.
Example 3 — Recognizing what cannot react
Explain why benzoic acid (C₆H₅COOH) does not undergo HVZ, while 2-methylpropanoic acid ((CH₃)₂CHCOOH) does.
Answer: benzoic acid has no α-H (the α carbon is part of the aromatic ring), so no enol can form — no reaction. 2-Methylpropanoic acid has one α-H on the tertiary α carbon; it enolizes (forming the more substituted enol), and Br lands there, giving 2-bromo-2-methylpropanoic acid, (CH₃)₂CBrCOOH.
Key takeaways
- Carboxylic acids do not enolize readily; direct α-bromination with Br₂ alone is too slow.
- HVZ reaction: RCH₂COOH + Br₂ (with PBr₃ or P + Br₂) → RCHBrCOOH + HBr; three stages: acyl bromide → α-bromo acyl bromide → hydrolysis.
- PBr₃ converts the acid to the acyl bromide (more electrophilic C=O, more acidic α-H); HBr catalyzes enolization; water hydrolyzes back to the acid.
- Requires an α-H; aromatic acids (no α-H) do not react.
- α-Bromo acid → α-amino acid via SN2 with NH₃ (2 NH₃ per α-bromo acid) is a classic synthesis.
- Stoichiometry is 1 Br₂ per α-H replaced (1:1 for mono-bromination).
- Regiochemistry: bromine goes to the α carbon that forms the more stable (more substituted) enol.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why is Br₂ alone ineffective for α-brominating a carboxylic acid?
Show answer
The acid's OH donates electron density into the carbonyl, making enolization (and α-H removal) very unfavorable; the O–H is also more acidic than the α-C–H, so reagents interact with the OH. The acyl bromide overcomes both problems.
List the three stages of the HVZ reaction and the reagent responsible for each.
Show answer
(1) Acyl bromide formation — PBr₃; (2) α-bromination — Br₂ (enol of the acyl bromide attacks Br₂); (3) hydrolysis — H₂O.
What product forms when 3-methylbutanoic acid (CH₃CH(CH₃)CH₂COOH) undergoes HVZ bromination?
Show answer
2-Bromo-3-methylbutanoic acid, CH₃CH(CH₃)CHBrCOOH — bromine goes to the α carbon (C2), the more substituted position forming the more stable enol.
How many moles of Br₂ are needed per mole of 2-methylpropanoic acid for mono-bromination?
Show answer
1 mole of Br₂ per mole of acid (1:1) for mono-bromination of the single α-H.
Show how you would synthesize alanine (2-aminopropanoic acid) from propanoic acid in two steps.
Show answer
Step 1: propanoic acid + Br₂/PBr₃ → 2-bromopropanoic acid. Step 2: 2-bromopropanoic acid + 2 NH₃ → alanine (CH₃CH(NH₂)COOH) + NH₄Br (SN2 displacement of Br by NH₃, with a second NH₃ neutralizing HBr).
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Hell–Volhard–Zelinsky (HVZ) reaction
- α-Bromination of a carboxylic acid using Br₂ with PBr₃ (or P + Br₂).
- Acyl bromide (acid bromide)
- RCOBr — the bromide analog of a carboxylic acid.
- α-Bromocarboxylic acid
- RCHBrCOOH — Br on the carbon adjacent to COOH.
- Enol
- C=C–OH tautomer (of the acyl bromide here).
- SN2
- Bimolecular nucleophilic substitution (backside attack, one step).
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.