Question

Write short notes on the following:
(i) Aldol condensation
(ii) Decarboxylation
(iii) Cannizzaro reaction

Hint:

Aldol = $\alpha$-H present, Cannizzaro = $\alpha$-H absent, Decarboxylation = loss of $CO_2$.

Answer 1

(i) Aldol Condensation:
Aldol condensation is a reaction in which aldehydes or ketones having $\alpha$-hydrogen atoms undergo self-condensation in the presence of a dilute base (NaOH, KOH) to form $\beta$-hydroxy aldehydes (aldols) or $\beta$-hydroxy ketones. On heating, these dehydrate to form $\alpha,\beta$-unsaturated aldehydes or ketones.
\[ 2CH_3CHO \; \xrightarrow{NaOH} \; CH_3CH(OH)CH_2CHO \; \xrightarrow{\Delta} \; CH_3CH=CHCHO \] (ii) Decarboxylation:
Decarboxylation is the elimination of a carboxyl group ($-COOH$) from carboxylic acids as carbon dioxide. This generally takes place in the presence of soda lime (NaOH + CaO).
\[ C_6H_5COOH + NaOH \; \xrightarrow{CaO, \; \Delta} \; C_6H_6 + Na_2CO_3 \] This reaction is useful for decreasing the carbon chain length by one carbon atom. (iii) Cannizzaro Reaction:
Aldehydes which do not have $\alpha$-hydrogen undergo self-oxidation and reduction (disproportionation) in the presence of concentrated alkali to form a primary alcohol and a carboxylate salt.
\[ 2HCHO \; \xrightarrow{Conc. NaOH} \; CH_3OH + HCOONa \] Example: Formaldehyde gives methanol and sodium formate. Conclusion:
Aldol condensation forms C–C bonds, decarboxylation removes $-COOH$ groups, and Cannizzaro reaction gives alcohols and salts from aldehydes without $\alpha$-hydrogen.