Reductive amination: practice and mechanism
The amine forms an imine or iminium ion with the carbonyl group. NaBH₃CN selectively reduces this C=N bond (and hardly the C=O): an amine forms.
Example
The amine and the carbonyl compound form an imine or iminium ion. Sodium cyanoborohydride selectively reduces this C=N bond, hardly the carbonyl group itself. An amine forms in one step.
Mechanism step by step
- 1
Methylamine adds to the carbonyl group; after loss of water an imine forms, present as the iminium ion in weak acid.
- 2
NaBH₃CN transfers a hydride to the iminium carbon.
- 3
A secondary amine (N-methylamine) forms at the former carbonyl carbon.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
Often confused with: Imine formation · Reduction with sodium borohydride
Frequently asked questions
Why NaBH₃CN instead of NaBH₄?
The CN group makes the hydride weaker. It reduces the positively charged iminium ion but hardly the neutral C=O, so there is no premature reduction to the alcohol.
Why is this better than SN2 with an amine?
No overalkylation: each step adds exactly one group. This builds primary, secondary or tertiary amines selectively.