Esterification of carboxylic acids is typically carried out by reaction with an alcohol in the presence of an acid catalyst (usually concentrated sulfuric acid). This reaction forms an ester and water. The mechanism involves the following steps:
1. Protonation of the carboxyl group of the carboxylic acid to make it more electrophilic.
2. Nucleophilic attack of the alcohol’s hydroxyl group on the electrophilic carbonyl carbon of the carboxylic acid.
3. Formation of a tetrahedral intermediate, followed by proton transfer.
4. Loss of a proton from the tetrahedral intermediate to form the ester and water.
\[
\text{R-COOH} + \text{R'-OH} \xrightarrow{\text{H}^+} \text{R-COOR'} + \text{H}_2\text{O}
\]
(B) Hell-Volhard-Zelinsky Reaction:
The Hell-Volhard-Zelinsky (HVZ) reaction is a halogenation of α-position of carboxylic acids. In this reaction, a carboxylic acid reacts with a halogen (such as bromine) in the presence of phosphorus tribromide (PBr₃) to form an α-halogenated carboxylic acid.
\[
\text{R-COOH} + \text{Br}_2 \xrightarrow{\text{PBr}_3} \text{R-CH(Br)-COOH}
\]
The mechanism involves the formation of an acyl halide intermediate, which reacts with halogen to introduce a halogen at the α-position.
Final Answer:
The esterification of carboxylic acids involves a reaction with alcohols to form esters and water, with sulfuric acid acting as a catalyst. The Hell-Volhard-Zelinsky reaction involves the halogenation of the α-position of carboxylic acids using bromine and PBr₃.
Correct Answer: The mechanisms for esterification and Hell-Volhard-Zelinsky reaction are provided above.