Question
An aqueous solution of formaldehyde (HCHO) is treated with concentrated NaOH. Separately, acetaldehyde (CH₃CHO) is treated with dilute NaOH. Identify the products in each case and explain why the reactions differ.
Solution — Step by Step
Before writing any mechanism, ask one question: does the aldehyde have an α-hydrogen (a hydrogen on the carbon directly adjacent to the C=O)?
Formaldehyde (HCHO) has no α-carbon at all — the carbonyl carbon is directly bonded to two H’s and that’s it. Acetaldehyde (CH₃CHO) has a CH₃ group next to the carbonyl, giving it three α-hydrogens.
Since HCHO has no α-H, aldol condensation is impossible. The reaction that happens instead is disproportionation — one molecule of HCHO gets oxidised, another gets reduced.
Products: methanol + sodium formate. One aldehyde gives up hydride (H⁻) to the other — the hydride donor gets oxidised to a carboxylate, the acceptor gets reduced to an alcohol.
Because CH₃CHO has α-hydrogens, OH⁻ can abstract one to form an enolate. That enolate attacks the carbonyl carbon of a second CH₃CHO molecule.
The initial product is 3-hydroxybutanal (aldol product). Under warming, this dehydrates to but-2-enal (crotonaldehyde) — the full condensation product.
Dilute NaOH + room temperature → stops at the β-hydroxy aldehyde (aldol addition product).
Dilute NaOH + heat → dehydration gives the α,β-unsaturated aldehyde (aldol condensation product).
JEE problems will specify conditions to force you to distinguish between these two — read carefully.
Why This Works
The aldol reaction depends entirely on enolate formation. OH⁻ acts as a base, pulling off an α-H to generate a nucleophilic carbanion. No α-H means no enolate, no aldol — simple as that.
The Cannizzaro mechanism works through a hydride transfer. NaOH attacks the carbonyl to form a tetrahedral alkoxide intermediate, which then transfers H⁻ to a second aldehyde molecule. Crucially, this only works cleanly when there’s no competing enolate pathway — which is exactly why it requires no α-H.
This is why concentrated NaOH is used for Cannizzaro: the high [OH⁻] drives the hydride-transfer mechanism. Dilute NaOH is enough for aldol because the α-H is relatively acidic (pKa ~17 for aldehydes).
Alternative Method — The Decision Tree
For MCQs, you rarely need the full mechanism. Memorise this decision tree:
Aldehyde + NaOH (or base)
↓
Has α-hydrogen?
YES → Aldol Condensation
NO → Cannizzaro ReactionCannizzaro-prone aldehydes (no α-H):
- HCHO (formaldehyde)
- C₆H₅CHO (benzaldehyde)
- (CH₃)₃CCHO (trimethylacetaldehyde / pivaldehyde)
- Cl₃CCHO (chloral)
Cross-Cannizzaro reaction (JEE Advanced favourite): HCHO + C₆H₅CHO with NaOH. Here, HCHO always acts as the reducing agent (gets oxidised to formate) because it’s the stronger hydride donor. Benzaldehyde gets reduced to benzyl alcohol. HCHO is never the alcohol product in a cross-Cannizzaro — this distinction appears directly in Advanced papers.
Common Mistake
Students see benzaldehyde reacting with NaOH and write an aldol product because “aldehydes + base = aldol.” Benzaldehyde has no α-hydrogen — the carbon next to the CHO is part of the ring, and those H’s are vinyl/aromatic H’s, not α-H’s that OH⁻ can abstract under normal conditions. The product is always benzyl alcohol + sodium benzoate (Cannizzaro). This error cost marks in JEE Advanced 2024.
For any JEE problem: write the structure, circle the α-carbon, count the α-H’s. Takes 10 seconds. If the count is zero, Cannizzaro — no exceptions under basic conditions.