Question
Acetaldehyde (CH₃CHO) undergoes self-condensation in the presence of dilute NaOH. Write the mechanism, identify the product, and predict what happens when the product is heated. Also, what product forms in a cross aldol condensation between formaldehyde and acetaldehyde?
Solution — Step by Step
In acetaldehyde (CH₃CHO), the three hydrogens on the methyl group are α-hydrogens — they sit next to the carbonyl. Dilute NaOH abstracts one of these to form the enolate ion (CH₂⁻CHO).
Why does NaOH attack here specifically? The carbonyl pulls electron density away, making the α-carbon slightly acidic (pKₐ ≈ 17 for aldehydes). The enolate is resonance-stabilised — negative charge delocalised onto oxygen.
The enolate (nucleophile) attacks the carbonyl carbon of a second CH₃CHO molecule. This is a standard nucleophilic addition — the π bond breaks, electrons go to oxygen.
The intermediate formed is an alkoxide ion: CH₃CH(O⁻)CH₂CHO.
The alkoxide picks up a proton from water (the NaOH solution contains water), giving the β-hydroxy aldehyde: 3-hydroxybutanal (acetaldol).
This compound has both an aldehyde and an alcol group — that’s literally where the name “aldol” comes from.
On heating, the β-hydroxy aldehyde loses water (dehydration). The OH on the β-carbon and an α-hydrogen are eliminated, forming a conjugated α,β-unsaturated aldehyde.
Product: but-2-enal (crotonaldehyde) — the condensation product. The word “condensation” comes from this elimination of water.
In cross aldol, we mix two different carbonyl compounds. The key rule: the compound WITHOUT α-hydrogens acts as the electrophile (it can’t form an enolate, so it can only be attacked).
Formaldehyde (HCHO) has no α-hydrogens. Acetaldehyde forms the enolate. So: enolate of CH₃CHO attacks HCHO.
Product: 3-hydroxypropan-1-al (β-hydroxypropionaldehyde).
Why This Works
The entire mechanism rests on one idea: the carbonyl group is electron-withdrawing, which makes the adjacent α-carbon nucleophilic after deprotonation, while making the carbonyl carbon electrophilic. The same functional group creates both the nucleophile (enolate) and the electrophile (another molecule’s carbonyl). That’s why self-condensation is possible at all.
The conjugated product (crotonaldehyde) is more stable than the aldol — the C=C and C=O are in conjugation, lowering overall energy. This thermodynamic drive is what makes the dehydration step favourable on heating.
Cross aldol is more useful in synthesis because it’s directed — we can choose which compound is the nucleophile and which is the electrophile. In practice, the compound without α-hydrogens (formaldehyde, benzaldehyde) is used in excess to push the reaction toward the desired cross product.
Alternative Method — Recognising Product by Carbon Count
There’s a faster way to write aldol products in exams without drawing the full mechanism. The two reactant molecules join α-carbon to carbonyl carbon:
- Self-condensation of CH₃CHO (2 carbons + 2 carbons) → 4-carbon β-hydroxy aldehyde
- Cross aldol of HCHO (1 carbon) + CH₃CHO (2 carbons) → 3-carbon β-hydroxy aldehyde
Count the carbons, place OH on the β-position, CHO at the end. You can write the product in under 10 seconds for MCQs.
For JEE Main, aldehydes with no α-hydrogens (formaldehyde, benzaldehyde) always act as electrophiles in cross aldol. This has appeared as a direct MCQ — HCHO + CH₃COCH₃ in JEE Main 2024 asked for the major product.
Common Mistake
Mixing up aldol addition vs. aldol condensation. Many students write crotonaldehyde as the direct product of NaOH treatment at room temperature. Wrong — at room temperature, the product is the β-hydroxy aldehyde (aldol addition product). Crotonaldehyde forms only on heating. In exams, watch for whether the question says “dilute NaOH” (aldol) or “dilute NaOH followed by heating” (condensation). This exact distinction was tested in CBSE 2023 and NEET 2022.