Q.Write only the chemical equation of Hunsdiecker reaction.

Carboxylic Acid Halogenation: The Hell–Volhard–Zelinsky Reaction

Imagine you have a carboxylic acid — say, propanoic acid ($\mathrm{CH_3CH_2COOH}$). You want to replace one of the hydrogen atoms on the carbon chain with a halogen (like bromine or chlorine). But here's the catch: the carboxylic acid group ($-\mathrm{COOH}$) is already quite reactive. If you just add bromine directly, nothing useful happens — the $\alpha$-carbon (the carbon right next to the $-\mathrm{COOH}$ group) is not reactive enough to attack bromine on its own.

The trick is to activate the $\alpha$-carbon first. This is exactly what the Hell–Volhard–Zelinsky (HVZ) reaction does.

The Intuition

The $-\mathrm{COOH}$ group is electron-withdrawing. That makes the $\alpha$-carbon slightly positive (electrophilic), but not enough to react with a halogen directly. To make it work, we convert the acid into an acyl halide (like $\mathrm{RCOBr}$) using $\mathrm{PBr_3}$ or $\mathrm{PCl_3}$. The acyl halide is even more electron-withdrawing, which makes the $\alpha$-hydrogen more acidic — it can be removed by a base (like a catalytic amount of $\mathrm{PBr_3}$ or $\mathrm{Br_2}$ itself) to form an enol or enolate intermediate. This enol then attacks a halogen molecule, giving an $\alpha$-haloacyl halide. Finally, water hydrolyses it back to the $\alpha$-halo carboxylic acid.

In short: activate → enolize → halogenate → hydrolyse.

The Precise Statement

The reaction is regioselective: halogenation occurs exclusively at the $\alpha$-carbon (the carbon adjacent to the $-\mathrm{COOH}$ group). No other position on the chain is halogenated.

Step-by-Step Mechanism

1. Formation of acyl bromide The carboxylic acid reacts with $\mathrm{PBr_3}$ (or $\mathrm{PCl_3}$) to form an acyl bromide:

2. Enolization A catalytic amount of $\mathrm{PBr_3}$ or $\mathrm{Br_2}$ acts as a Lewis acid, making the $\alpha$-hydrogen more acidic. A base (often $\mathrm{Br}^-$ from the reaction) abstracts this hydrogen, forming an enol:

3. Halogenation The enol attacks a $\mathrm{Br_2}$ molecule, giving the $\alpha$-bromoacyl bromide:

4. Hydrolysis Water hydrolyses the acyl bromide back to the carboxylic acid:

Why This Matters

The $\alpha$-halo carboxylic acid is a versatile intermediate. You can:

• Substitute the halogen with $\mathrm{OH}$ to get $\alpha$-hydroxy acids (like lactic acid).
• Substitute with $\mathrm{NH_3}$ to get $\alpha$-amino acids (the building blocks of proteins). …