Question

Write short notes on the following:
(i) Aldol condensation
(ii) Cannizzaro's reaction

Hint:

Aldol needs an alpha-hydrogen with dilute base; Cannizzaro is disproportionation of aldehydes without alpha-hydrogen using concentrated alkali.

Answer 1

(i) Aldol condensation
Step 1: Requirement. Aldehydes and ketones that contain at least one alpha-hydrogen (a hydrogen on the carbon next to the carbonyl group) undergo this reaction.
Step 2: Reagent and first product. In the presence of a dilute base (such as dilute NaOH or KOH), two such molecules combine. One acts as a carbanion (nucleophile) and adds to the carbonyl carbon of the other, giving a beta-hydroxy aldehyde or ketone called an aldol.
\( 2\,CH_3CHO \xrightarrow{dil.\ NaOH} CH_3CH(OH)CH_2CHO \) (3-hydroxybutanal)
Step 3: Condensation step. On warming, the aldol loses a water molecule to form an alpha,beta-unsaturated carbonyl compound.
\( CH_3CH(OH)CH_2CHO \xrightarrow{\Delta} CH_3CH=CHCHO + H_2O \) (but-2-enal)

(ii) Cannizzaro's reaction
Step 1: Requirement. Aldehydes that do NOT have any alpha-hydrogen undergo this reaction.
Step 2: Reagent. On treatment with concentrated alkali, such aldehydes undergo self oxidation and reduction (disproportionation).
Step 3: Products. One molecule is oxidised to the carboxylic acid salt and the other is reduced to the alcohol.
\( 2\,HCHO + NaOH \rightarrow CH_3OH + HCOONa \)
With benzaldehyde: \( 2\,C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \)