Reductive amination template

Propanone and methylamine at pH ≈ 6: addition, proton transfer, loss of water to the iminium ion and hydride transfer from cyanoborohydride give N-methylpropan-2-amine.

Mechanism step by step

The carbonyl and amine form an iminium ion, which NaBH₃CN reduces in the same pot

1. Amine adds – N attacks the carbonyl carbon
2. Proton transfer – H⁺ moves from N to O (through the solvent)
3. Protonation – OH becomes a leaving group
4. Iminium ion – The N lone pair pushes out water
5. Hydride transfer – B–H electrons reduce the C=N⁺ carbon

Key points

• NaBH₃CN is a weak hydride donor: at pH 6 it reduces the iminium ion much faster than the ketone.
• NaBH(OAc)₃ (Abdel-Magid, 1996) does the same job without cyanide.
• Avoids the over-alkylation of amines by alkyl halides: one new C–N bond, cleanly.
• NH₃ gives a primary amine, a primary amine a secondary, a secondary amine a tertiary.

Drawn with 82 atoms, 16 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
• Borch, Bernstein & Durst, J. Am. Chem. Soc. 1971, 93, 2897 Original paper
• Abdel-Magid, Carson, Harris, Maryanoff & Shah, J. Org. Chem. 1996, 61, 3849 Original paper