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Hydride reduction — nitrile to amine

LiAlH₄ reduction of a nitrile to a primary amine

How the LiAlH₄ reduction of a nitrile to a primary amine works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: LiAlH₄ (excess), then H₂O workup. Conditions: Dry Et₂O or THF, 0 °C → reflux; then H₃O⁺ or Fieser workup. Bonds formed: C–H ×2, N–H ×2; bonds broken: C≡N (to C–N single), Al–N. Stereochemistry: None (achiral CH₂NH₂). Rate and selectivity: Second hydride addition fast relative to the first.

At a glance

ReagentsLiAlH₄ (excess), then H₂O workup
Solvent / conditionsDry Et₂O or THF, 0 °C → reflux; then H₃O⁺ or Fieser workup
Bonds formedC–H ×2, N–H ×2
Bonds brokenC≡N (to C–N single), Al–N
StereochemistryNone (achiral CH₂NH₂)
Rate / selectivitySecond hydride addition fast relative to the first

Step 1: First hydride adds to C≡N

+HAlHHHRNδ⁺δ⁻RNAlHHH1 · FIRST HYDRIDE ADDS TO C≡Nlithium aluminium hydrideAlH₄⁻REDUCTANTnitrileELECTROPHILEiminoaluminate (metalated imine)INTERMEDIATEAnhydrous: the N⁻ is captured by aluminium, not protonated.

Hydride adds to the nitrile carbon, and the resulting nitrogen anion is stabilized as an aluminum ate complex — a metalated imine intermediate, not a free proton source, since the reaction runs under anhydrous conditions with no water present until workup.

Step 2: Second hydride adds to the C=N

+RNAlHHHHAlHHHRNAlHHH2 · SECOND HYDRIDE ADDS TO THE C=NiminoaluminateELECTROPHILEAlH₄⁻AlH₄⁻REDUCTANTamide–aluminate (N⁻)INTERMEDIATEThe metalated imine is still electrophilic, so reduction does not stop at the imine.

A second equivalent of hydride adds to this metalated imine's electrophilic carbon exactly as it would to any ordinary imine, since the reaction doesn't stop after the first addition.

Step 3: Workup protonates N⁻

++RNAlHHHHOH₂RNH₂OHAlOHOH3 · WORKUP PROTONATES N⁻amide anionNUCLEOPHILEhydronium (workup)H₃O⁺ACIDprimary aminePRODUCTaluminium hydroxidesBYPRODUCTNet: R–C≡N + 2 H₂ equivalents → R–CH₂–NH₂.

Aqueous workup protonates the resulting amide anion, giving a primary amine — this two-hydride reduction converts a nitrile directly into a CH₂-NH₂ group, extending the carbon chain by one nitrogen-bearing carbon relative to the starting material it was typically made from.

Abbreviations and terms

R
R group. Any carbon substituent (alkyl, aryl…) whose identity does not matter for the mechanism.
THF
Tetrahydrofuran. Cyclic ether solvent; its oxygen lone pairs solvate Li⁺/Mg²⁺ so organometallics stay dissolved and reactive.
Et2O
Diethyl ether. Low-boiling ether solvent, classic for Grignard reagents.
LiAlH4
Lithium aluminium hydride. Powerful hydride (H⁻) donor; reduces esters, acids, amides to alcohols/amines.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.

How it compares with related reactions

LiAlH₄ reduction of a nitrile to a primary amineLiAlH₄ reduction of an ester to a primary alcoholNaBH₄ reduction of a ketone
ReagentsLiAlH₄ (excess), then H₂O workupLiAlH₄ (≥ 0.5 equiv., 2 H⁻ per ester), then H₃O⁺ or Fieser workupNaBH₄ (0.25–1 equiv)
ConditionsDry Et₂O or THF, 0 °C → reflux; then H₃O⁺ or Fieser workupDry Et₂O or THF, 0 °C to refluxEtOH or MeOH, 0–25 °C; aqueous workup
StereochemistryNone (achiral CH₂NH₂)No stereocentre (primary alcohol)Hydride attacks the less hindered face of ring ketones
Rate / selectivitySecond hydride addition fast relative to the firstFast; aldehyde reduces faster than the esterAldehydes > ketones; esters and amides untouched

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the LiAlH₄ reduction of a nitrile to a primary amine?

LiAlH₄ (excess), then H₂O workup. Conditions: Dry Et₂O or THF, 0 °C → reflux; then H₃O⁺ or Fieser workup.

Which bonds are formed and which are broken?

Formed: C–H ×2, N–H ×2. Broken: C≡N (to C–N single), Al–N.

What is the stereochemical outcome?

None (achiral CH₂NH₂)

What controls the rate or the selectivity?

Second hydride addition fast relative to the first

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