Q6P - Organic Chemistry (Mcmurry)

Question

If methanol rather than water is added at the end of a Hell–Volhard–Zelinskii reaction, an ester rather than an acid is produced. Show how you would carry out the following transformation, and propose a mechanism for the ester forming step.

Step-by-Step Solution

Answer

Step 1: Reaction scheme

Carboxyl group converted into acid bromide and on hydrolysis α- The transformation process has various steps;

(a) Α-bromination happens.

(b) bromocarbxic acid is formed.

(c) In the last step esterification reaction takes place with methanol.

 

Step 2: Explanation of Hell-Volhard-Zelinskii (HVZ) reaction

In HVZ reaction; you can do α-bromination of carboxylic acid group with the help of and.

In the last step you need to add water for hydrolysis of the acid bromide. Lastly you will get the α-bromo carboxylic acid as product. Then on reaction with methanol esterification reaction takes place.

Step 3: Mechanism of the reaction

This reaction is a transformation of 3-methylpentanoic acid to its methyl ester through HVZ using methanol instead of water. 

When 3-methylpentanoic acid is treated with PBr₃, the carboxylic group converted into its acid bromide. Then a treatment with bromine, a bromine substitution takes place on the carboxylic acid group and yields bromosubstituted acid bromide. 

The transformation of a carboxylic acid into a bromoester 

When the bromosubstituted acid bromide is treated with methanol, a nucleophilic attack takes place on the carbonyl carbon. After this, deprotonation with a help of a base and elimination of bromide yields the desired ester.

Mechanism of ester formation