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Reductive amination: the mechanism, step by step

The amine and the carbonyl condense to a C=N, and the reducing agent then delivers a hydride to that carbon. The order matters: nothing reduces the carbonyl itself.

rule-derived4 stepsNot a record of what was observed

The pathway runs in 4 steps. What sits between them is a real structure: it is what the reaction passes through, and on a bad day it is what you isolate instead of the product.

Below is the sequence as it is taught. On the platform it is drawn on whatever structures you supply, with the curly arrows resolved to the actual atoms involved rather than to a general case with R groups on it.

The steps

  1. The amine adds to the carbonyl

    The nitrogen shares its pair of electrons with the carbon of the C=O group, and the double bond's electrons move up onto the oxygen.

    Addition of the amine lone pair into the carbonyl π*, giving a zwitterion.

  2. The proton moves from nitrogen to oxygen

    The spare hydrogen on the nitrogen hops across to the oxygen, leaving a hydroxyl and a neutral nitrogen.

    Proton transfer giving the neutral hemiaminal, in practice through the solvent.

  3. Water leaves and the C=N forms

    The nitrogen pushes its electrons into the bond to carbon, making a double bond, and water is forced off.

    Loss of water to give the imine. This is the species the reducing agent acts on, and it is far more easily reduced than the carbonyl it came from.

  4. The reducing agent delivers a hydride

    The reducing agent hands a hydrogen, with both its electrons, to the carbon of the C=N. The nitrogen takes the electrons of the double bond, and an amine is left.

    Hydride addition to the C=N. The reagent is chosen to be mild enough to leave an unactivated carbonyl alone, which is why the condensation has to happen first.

What this drawing assumes

Read these before you quote the mechanism
  • The reducing agent is chosen so that it reduces the C=N and leaves the C=O alone. Drawing the hydride arriving after the condensation is not a simplification, it is the whole point of the method.
  • The proton transfers between nitrogen and oxygen are collected into one step.

Other pathways this class runs by

  • With a secondary amine the species reduced is an iminium rather than a neutral imine, and it is reduced faster still.
  • Run stepwise, the imine can be isolated first and reduced separately; done in one pot, it never accumulates.

A class is not one pathway. Which of these runs depends on the substrate, the solvent and what else is in the flask, and the drawing above does not decide that for you.

Run it on your own structures

Enter the reactants and the product you expect. If the transformation is one the engine can perform and this pathway reaches that product from those structures, it is drawn on them — with the arrows on the right atoms and the intermediates you would actually pass through. If it does not reach the product, you are told that instead of being shown a drawing that does not apply.