Claisen–Schmidt condensation template

Hydroxide forms the acetone enolate, which adds to benzaldehyde (no α-H, so it cannot enolise); protonation gives the aldol, and a second enolate expels hydroxide (E1cB) to give (E)-4-phenylbut-3-en-2-one (benzalacetone).

Mechanism step by step

A crossed aldol condensation: the acetone enolate adds to benzaldehyde, then E1cB loss of water

1. Enolate – Hydroxide removes an α-H of acetone
2. Addition – The enolate carbon attacks the aldehyde C=O
3. Protonation – Water gives the aldol (β-hydroxy ketone)
4. Second enolate – Hydroxide removes an α-H next to the new C–C bond
5. Loss of OH⁻ (E1cB) – The C=C forms, conjugated with both the ring and C=O

Key points

• Benzaldehyde has no α-H, so only acetone forms an enolate; the aldehyde is also the better electrophile, so the crossed product wins over acetone self-condensation.
• Dehydration is fast and pulls the equilibria over: the new C=C is conjugated with the ring and the C=O. The (E)-alkene forms.
• With 2 equivalents of benzaldehyde both methyl groups react: dibenzalacetone (1,5-diphenylpenta-1,4-dien-3-one), a classic teaching-lab preparation.
• Acetophenones react the same way to give chalcones, ArCH=CHCOAr.

Drawn with 137 atoms, 18 arrows, 11 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 26 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 16 Textbook