Question

Write the reaction involved in: Hell-Volhard-Zelinsky reaction

Hint:

HVZ Reaction targets exclusively the $\alpha$-carbon using Red P and a halogen.

Answer 1

Step 1: Concept
$\alpha$-halogenation of aliphatic carboxylic acids.

Step 2: Meaning
Carboxylic acids possessing at least one $\alpha$-hydrogen are specifically halogenated at the $\alpha$-position.

Step 3: Analysis
A carboxylic acid is treated with chlorine or bromine in the presence of a small amount of red phosphorus, followed by hydrolysis. This selectively replaces an $\alpha$-hydrogen atom by a halogen atom to form an $\alpha$-halocarboxylic acid.

• This reaction is known as the \[ \boxed{\text{Hell--Volhard--Zelinsky (HVZ) reaction}}. \]

• It is used for the halogenation of the carbon atom adjacent to the carboxyl group (the $\alpha$-carbon).

• A carboxylic acid containing at least one $\alpha$-hydrogen is treated with chlorine or bromine in the presence of a small amount of red phosphorus.

• Red phosphorus reacts with the halogen to generate phosphorus trihalide \((PCl_3\) or \(PBr_3)\) in situ, which converts the carboxylic acid into the corresponding acyl halide.

• The acyl halide readily undergoes enolization, allowing halogenation at the $\alpha$-carbon.

• Finally, hydrolysis converts the acyl halide back into the carboxylic acid.

• Thus, an $\alpha$-hydrogen atom is replaced by a halogen atom: \[ RCH_2COOH \xrightarrow{X_2,\;P} RCHXCOOH, \] where \[ X=Cl\ \text{or}\ Br. \]

• Therefore, the HVZ reaction provides an important method for preparing $\alpha$-halocarboxylic acids.

Step 4: Conclusion
The reaction highlights the unique reactivity of the carbon adjacent to the carboxyl group.

Final Answer:
$R-CH_2-COOH + X_2 \xrightarrow{\text{(i) Red P, (ii) } H_2O} R-CH(X)-COOH + HX$ (where $X = Cl, Br$)