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Hell-Volhard-Zelinsky (HVZ) Reaction

By Chemca Editorial Team • Last Updated: January 2026 • 8 min read

The Hell-Volhard-Zelinsky (HVZ) Reaction is a chemical reaction used for the $\alpha$-halogenation of carboxylic acids. It involves treating carboxylic acids containing at least one $\alpha$-hydrogen with chlorine or bromine in the presence of a small amount of Red Phosphorus.

1. General Reaction

The hydrogens attached to the $\alpha$-carbon (carbon adjacent to $-COOH$) are replaced by halogen atoms.

$$ R-CH_2-COOH \xrightarrow{1. \ X_2, \ Red \ P} \xrightarrow{2. \ H_2O} R-CH(X)-COOH + HX $$

Conditions:

• Reactant: Carboxylic Acid with $\alpha$-Hydrogen.
• Reagent: Halogen ($Cl_2$ or $Br_2$) + Red Phosphorus.
• Work-up: Hydrolysis with water.
• Product: $\alpha$-Halo Carboxylic Acid.

2. Detailed Mechanism

The reaction proceeds via the formation of an acyl halide, which undergoes enolization.

Step 1: Formation of Acyl Halide

Red Phosphorus reacts with the halogen ($X_2$) to form Phosphorus Trihalide ($PX_3$), which converts the carboxylic acid into an acid halide.

$$ 2P + 3X_2 \rightarrow 2PX_3 $$ $$ 3R-CH_2-COOH + PX_3 \rightarrow 3R-CH_2-COX + H_3PO_3 $$

Step 2: Keto-Enol Tautomerism

The acid halide undergoes tautomerization to form the enol form. This is easier for acid halides than for carboxylic acids due to the electron-withdrawing nature of the halogen.

$$ R-CH_2-C(=O)X \rightleftharpoons R-CH=C(OH)X \text{ (Enol)} $$

Step 3: Halogenation

The nucleophilic double bond of the enol attacks a halogen molecule ($X_2$) to form an $\alpha$-halo acyl halide.

$$ R-CH=C(OH)X + X_2 \rightarrow R-CH(X)-C(=O)X + HX $$

Step 4: Hydrolysis

Upon addition of water, the acyl halide group is hydrolyzed back to the carboxylic acid group.

$$ R-CH(X)-COX + H_2O \rightarrow R-CH(X)-COOH + HX $$

3. Examples

A. Synthesis of 2-Chloroacetic Acid

$$ CH_3COOH + Cl_2 \xrightarrow{Red \ P} Cl-CH_2-COOH $$

Note: With excess halogen, di- and tri-chloroacetic acids can form.

B. Synthesis of 2-Bromopropanoic Acid

$$ CH_3CH_2COOH + Br_2 \xrightarrow{Red \ P} CH_3CH(Br)COOH $$

4. Limitations

• Requirement of $\alpha$-H: Acids like Formic Acid ($HCOOH$) and Benzoic Acid ($C_6H_5COOH$) do not undergo HVZ reaction because they lack $\alpha$-hydrogens.
• Polyhalogenation: If excess halogen is used, all available $\alpha$-hydrogens can be replaced.

5. Applications

The resulting $\alpha$-halo acids are valuable synthetic intermediates. For example, they can be treated with excess ammonia to prepare $\alpha$-Amino Acids.

$$ R-CH(X)-COOH + 2NH_3 \rightarrow R-CH(NH_2)-COOH + NH_4X $$

HVZ Reaction Quiz

Test your concepts on Alpha-Halogenation. 10 MCQs with explanations.

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