Question
Write the mechanism of aldol condensation of acetaldehyde (ethanal) and identify the product. Also state the conditions under which the reaction proceeds.
Solution — Step by Step
Acetaldehyde (CH₃CHO) has three α-hydrogens on the methyl group adjacent to the carbonyl. These hydrogens are acidic enough to be removed by a dilute base — that’s what makes aldol reactions possible. Without α-hydrogens, no aldol.
Dilute NaOH abstracts one α-hydrogen, forming a nucleophilic enolate ion:
The negative charge sits on the α-carbon, which now acts as the nucleophile. This is the attacking species in the next step.
The enolate carbon attacks the electrophilic carbonyl carbon of a second acetaldehyde molecule — this is a nucleophilic addition:
The product after protonation is 3-hydroxybutanal (β-hydroxybutyraldehyde), also called the aldol product — the name comes from “aldehyde + alcohol”.
On heating, the aldol product undergoes dehydration (loss of water) to give an α,β-unsaturated aldehyde:
The final product is but-2-enal (crotonaldehyde). The double bond forms between the α and β carbons — always. This is the condensation part of aldol condensation.
Reagent: Dilute NaOH (or dilute KOH)
Condition: Cold for aldol addition; heat for dehydration to complete condensation
Final product: But-2-enal (crotonaldehyde)
Why This Works
The aldol reaction works because carbonyl compounds have two reactive sites working in opposite directions. The carbonyl carbon is electron-deficient (electrophilic), while the α-carbon — when deprotonated — becomes electron-rich (nucleophilic). One molecule plays the nucleophile, another plays the electrophile.
The dehydration step is thermodynamically driven. The β-hydroxy aldehyde loses water to form a conjugated system (C=C–C=O), which is significantly more stable. Conjugation lowers the overall energy of the molecule — this is why the reaction “wants” to go all the way to the unsaturated product on heating.
This mechanism is the template for all aldol reactions. Whether it’s acetone or benzaldehyde-acetone cross-aldol (used in perfume synthesis), the logic is identical: enolate attacks carbonyl, then dehydrate.
Alternative Method — Acid-Catalysed Mechanism
Aldol condensation can also be done with dilute acid (H⁺). Here, the carbonyl oxygen gets protonated first, which makes the α-carbon more acidic. The enol tautomer (not enolate) forms and acts as the nucleophile.
The enol attacks the protonated carbonyl of a second molecule. The outcome is the same — 3-hydroxybutanal — but the intermediate is an enol, not an enolate. For CBSE and JEE Main, the base-catalysed mechanism is more commonly asked, so prioritise that one.
In JEE Main and NEET, “aldol condensation” always means the complete reaction ending in the α,β-unsaturated carbonyl compound. If the question asks only for the “aldol addition” product, stop at 3-hydroxybutanal — don’t dehydrate.
Common Mistake
Many students write the double bond forming between the wrong carbons after dehydration — placing it between the carbonyl carbon and α-carbon (making a ketene-type structure) instead of between α and β carbons. Remember: the -OH is on the β-carbon, water is lost from β-OH and α-H, so the double bond forms between α and β. The carbonyl group (CHO) stays intact.
A second frequent error: students forget that two molecules of acetaldehyde are consumed in this reaction. In stoichiometry questions, especially in JEE Mains, this trips up the mole calculation. The 2:1 ratio is non-negotiable.