Explain Hell-volhard Zelinsky reaction with example

HVZ is a reaction used to prepare alpha-halo acids. The essential condition for this reaction is the presence of alpha hydrogen. Thus Formic acid does not undergo this reaction. The reagent used is Red phosphorous/X₂   where X is a halogen. In the first step, -OH in acid group is replaced by X. On further reaction, alpha hydrogen is replaced by another X atom. On hydrolysis, the terminal X atom will be replaced by the -OH group. Thus the product obtained is alpha-halo acids.

RCH₂COOH + Br₂/P ----> RCH₂COBr  + Br₂/P ----> RCHBrCOBr + H₂O ----> RCHBrCOOH ( alpha-halo acid)

Unlike other halogenation reactions, this reaction takes place in the absence of a halogen carrier. The reaction is initiated by addition of a catalytic amount of PBr₃, after which one molar equivalent of Br₂ is added.

PBr₃ replaces the carboxylic OH with a bromide, resulting in a carboxylic acid bromide. The acyl bromide can then tautomerize to an enol, which will readily react with the Br₂ to brominate a second time at the αposition.

In neutral to slightly acidic aqueous solution, hydrolysis of the α-bromo acyl bromide occurs spontaneously, yielding the α-bromo carboxylic acid in an example of a nucleophilic acyl substitution. If an aqueous solution is desirable, a full molar equivalent of PBr₃ must be used as the catalytic chain is disrupted.

If little nucleophilic solvent is present, reaction of the α-bromo acyl bromide with the carboxylic acid yields the α-bromo carboxylic acid product and regenerates the acyl bromide intermediate. In practice a molar equivalent of PBr₃ is often used anyway to overcome the slow reaction kinetics.

The mechanism for the exchange between an alkanoyl bromide and a carboxylic acid is below. The α-bromoalkanoyl bromide has a strongly electrophilic carbonyl carbon because of the electron-withdrawing effects of the two bromides.