Question
Identify the products , , in the following sequence and explain the mechanism of each step:
Solution — Step by Step
Two molecules of acetaldehyde combine in dilute NaOH. The base abstracts an -H from one acetaldehyde to form an enolate, which attacks the carbonyl carbon of another acetaldehyde.
The -hydroxy aldehyde loses water on heating, producing an -unsaturated aldehyde:
H₂/Ni reduces both C=C and C=O simultaneously:
Aldol step: dil. NaOH removes acidic -H () → enolate ion → nucleophilic addition to second acetaldehyde → protonation by water gives aldol product .
Dehydration step: heat eliminates water (E1cb-like through enolate) — driven by formation of conjugated -system.
Reduction step: H₂ adsorbs on Ni surface, syn-adds across both unsaturations, giving fully saturated alcohol.
Final answers: 3-hydroxybutanal, crotonaldehyde, 1-butanol.
Why This Works
The aldol condensation converts simple carbonyls into longer-chain -unsaturated carbonyls — a fundamental C-C bond-forming reaction. Aldehydes with -H undergo it; aldehydes without -H (like benzaldehyde, formaldehyde) do not — they instead undergo Cannizzaro.
Catalytic hydrogenation reduces both C=C and C=O. If only the C=C needed reduction, we’d use NaBH₄ or special catalysts.
Alternative Method
Use selective reduction. NaBH₄ reduces only the C=O of to give (crotyl alcohol). Then H₂/Pd reduces the C=C separately. Two-step gives the same .
Common Mistake
Writing the aldol product as (only carbons) — this would be the -condensation, which doesn’t happen. The mechanism dictates that the nucleophilic carbon (-carbon of one aldehyde) attacks the carbonyl carbon of the other aldehyde, giving a -carbon chain.
Stopping at and reporting butanal. H₂/Ni reduces both unsaturations — the C=C and the C=O — to give -butanol, not butanal.
JEE Main and NEET both ask multi-step organic conversions where you must track each intermediate. Practise drawing arrows for each step; don’t memorise products without understanding mechanism.