Q.(क) Explain the mechanism of esterification of carboxylic acids.

Esterification: acid protonates C=O $\to$ alcohol adds $\to$ water eliminated $\to$ ester. Cannizzaro: $2HCHO \to CH_3OH + HCOONa$. HVZ: $CH_3COOH + Cl_2/P \to ClCH_2COOH$. (OR — identify A, B, C below.)

(k) Mechanism of esterification (Fischer esterification). A carboxylic acid reacts with an alcohol in the presence of a mineral-acid catalyst (conc. H$_2$SO$_4$) to give an ester and water. The steps are:

1. Protonation of the carbonyl oxygen of the acid by H$^+$, making the carbonyl carbon more electrophilic.
2. Nucleophilic addition of the alcohol oxygen to the carbonyl carbon, giving a tetrahedral intermediate.
3. Proton transfer so that one of the two $-$OH groups becomes $-$OH$_2^+$ (a good leaving group).
4. Elimination of water and loss of the extra proton to regenerate the catalyst, giving the ester. Overall: $RCOOH + R'OH \underset{}{\overset{H^+}{\rightleftharpoons}} RCOOR' + H_2O$ The reaction is reversible; excess alcohol or removal of water drives it forward.

(kh) Name reactions.

• (i) Cannizzaro reaction — an aldehyde with no $\alpha$-hydrogen undergoes self oxidation$-$reduction (disproportionation) with concentrated alkali: $2HCHO \xrightarrow{\text{conc. } NaOH} CH_3OH + HCOO^-Na^+$ (one molecule is reduced to alcohol, the other oxidised to the acid salt).
• (ii) Hell$-$Volhard$-$Zelinsky (HVZ) reaction — a carboxylic acid having an $\alpha$-H is halogenated at the $\alpha$-carbon using Cl$_2$/Br$_2$ in the presence of a little red phosphorus: $CH_3COOH \xrightarrow[\text{red } P]{Cl_2} ClCH_2COOH + HCl$ …