E1cB elimination template

In base, 3-hydroxy-1,3-diphenylpropan-1-one loses its α-H to give the enolate (the conjugate base), which expels hydroxide to give (E)-chalcone.

Mechanism step by step

Base removes the acidic α-H first; the enolate then expels the β-leaving group

1. Conjugate base – Fast, reversible: hydroxide removes the α-H; the charge moves onto O (enolate)
2. Loss of OH⁻ – Slow step: the enolate pushes out hydroxide from the β-carbon, forming the conjugated C=C

Key points

• E1cB = Elimination, Unimolecular, via the conjugate Base: the C–H breaks first, giving a carbanion (here an enolate) that loses the leaving group in a second step.
• It needs an acidic C–H (next to C=O, NO₂ or CN), and it lets a poor leaving group such as OH⁻ or RO⁻ go, which E2 rarely does.
• In D₂O/NaOD the α-H exchanges for D faster than elimination occurs: evidence for the reversible carbanion step. The conjugated (E)-alkene forms.
• Compare E2 (one concerted step) and E1 (carbocation first). This is the dehydration step of every base-catalysed aldol condensation; tertiary aldols also undergo retro-aldol in base.

Drawn with 105 atoms, 9 arrows, 8 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 17 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 17 Textbook