Enamine formation template

Pyrrolidine adds to cyclohexanone; proton transfer gives the hemiaminal, acid turns the OH into a leaving group, water leaves to give the iminium ion, and loss of an α-C–H gives the enamine.

Mechanism step by step

A secondary amine and a ketone, with acid catalysis and loss of water

1. Addition – The amine nitrogen attacks the C=O carbon
2. Proton transfer – Summarised: H⁺ moves from N to O⁻ (via the solvent), giving the hemiaminal
3. Protonation – The acid catalyst (drawn as H₃O⁺) turns OH into a good leaving group
4. Loss of water – The N lone pair pushes water out: an iminium ion
5. Loss of α-H – No N–H to lose, so an α-C–H goes: the enamine

Key points

• A secondary amine gives an iminium ion with no N–H, so a proton is lost from the α-carbon instead: an enamine. A primary amine gives an imine.
• Every step is reversible: removing water (Dean–Stark trap or molecular sieves) drives the reaction; aqueous acid hydrolyses the enamine back to the ketone.
• Only a little acid is used: it is needed to make water the leaving group, but excess acid protonates the amine and stops the first step.
• The N lone pair makes the β-carbon nucleophilic: enamines are neutral enolate equivalents (Stork enamine alkylation and acylation).

Drawn with 148 atoms, 15 arrows, 11 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 11 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 16 Textbook