Amide Reduction Lialh4

Amide Reactions: Reduction of Amides to Amines using LiAlH4

LiAlH₄ (lithium aluminum hydride) reduces amides (R-C(=O)-NR'R'') to amines (R-CH₂-NR'R'') after aqueous workup. The carbonyl carbon is converted to a methylene (CH₂) group, and all substituents on nitrogen are retained. The key mechanistic feature is formation of a tetrahedral intermediate that collapses to an iminium ion, which is then reduced to the amine.


Quick Summary

FeatureDetails
TransformsR-C(=O)-NR'R'' → R-CH₂-NR'R''
Reagents1) LiAlH₄, ether 2) H₂O
ProductAmine (class depends on starting amide)
Key featureIminium intermediate (not aldehyde)

Product Type Rule

Starting AmideProduct Amine
Primary amide (RCONH₂)Primary amine (RCH₂NH₂)
Secondary amide (RCONHR')Secondary amine (RCH₂NHR')
Tertiary amide (RCONR'R'')Tertiary amine (RCH₂NR'R'')

Mechanism (5 Steps)

The mechanism involves hydride addition to form a tetrahedral intermediate, collapse to an iminium ion, a second hydride reduction, and aqueous workup to release the free amine.

Step 1 - Hydride Addition

Hydride from LiAlH₄ attacks the electrophilic carbonyl carbon. The π electrons shift to oxygen, forming a tetrahedral alkoxide intermediate.

Hydride attacks amide carbonyl, pi electrons shift to oxygen
Step 1: H⁻ attacks carbonyl C; π electrons shift to O forming tetrahedral alkoxide.

Step 2 - Tetrahedral Intermediate Collapses

Li⁺ and AlH₃ coordinate to oxygen. The nitrogen lone pair attacks the carbonyl carbon (N→C) while the C-O bond breaks (C-O→O), setting up iminium formation.

Tetrahedral intermediate with O coordinated to Li and Al; N attacks C while C-O breaks
Step 2: N lone pair attacks C (N→C); C-O bond breaks (C-O→O); Al/Li coordinate to O.

Step 3 - Iminium Ion Formation

The tetrahedral intermediate collapses to form an iminium ion (C=N⁺). Oxygen departs as an aluminum-bound leaving group. This is the key difference from ester reduction (which gives aldehyde).

Iminium ion formed after loss of oxygen leaving group
Step 3: Iminium ion (C=N⁺) formed; O departs as Al-bound species.

Step 4 - Second Hydride Reduces Iminium

A second hydride attacks the electrophilic iminium carbon. The π electrons shift to nitrogen, reducing the C=N double bond to a C-N single bond.

Second hydride attacks iminium carbon
Step 4: Second H⁻ attacks iminium C; C=N becomes C-N.

Step 5 - Aqueous Workup

Water protonates nitrogen and cleaves any Al-N interactions, releasing the free amine product.

Aqueous workup releases free amine
Step 5: H₂O protonates N; free amine released.

Worked Examples

Example A - Benzamide → Benzylamine. Primary amide reduces to primary amine.
Benzamide
Reactant
Reagent: LiAlH4 then H2O
Reagent
Benzylamine
Product
Example B - N-methylacetamide → Ethylmethylamine. Secondary amide reduces to secondary amine. The N-methyl group is retained.
N-methylacetamide
Reactant
Reagent: LiAlH4 then H2O
Reagent
Ethylmethylamine
Product
Example C - N-methyl-γ-butyrolactam → N-methylpyrrolidine. Lactams (cyclic amides) reduce to cyclic amines.
N-methyl-gamma-butyrolactam
Reactant
Reagent: LiAlH4 then H2O
Reagent
N-methylpyrrolidine
Product

Amide vs Ester vs Nitrile: Why Different Products?

Functional GroupLiAlH₄ ProductKey Intermediate
Amide (R-CONR'R'')Amine (R-CH₂-NR'R'')Iminium ion
Ester (R-COOR')Alcohol (R-CH₂OH)Aldehyde
Nitrile (R-C≡N)Primary amine (R-CH₂-NH₂)Imine

Why amides form iminium (not aldehyde): In amides, nitrogen is directly bonded to the carbonyl carbon. When the tetrahedral intermediate collapses, nitrogen's lone pair can push electrons into the C-N bond, forming a C=N double bond (iminium). In esters, oxygen cannot do this effectively, so the aldehyde is formed instead.


Common Exam Traps

  1. NaBH₄ does not reduce amides - Under standard conditions, NaBH₄ is too weak. LiAlH₄ is required.

  2. Chemoselectivity - LiAlH₄ reduces many functional groups. If other reducible groups are present, expect additional reductions:

    • Reduced: aldehydes, ketones, esters, acids, nitriles, epoxides, acid chlorides
    • Not reduced: alkenes, alkynes, aromatic rings, ethers

  3. Product class follows starting amide - Primary amide → primary amine; secondary amide → secondary amine; tertiary amide → tertiary amine.

  4. Alternative reagent: BH₃ - Borane also reduces amides to amines, sometimes with different chemoselectivity.


Product Prediction Checklist

  1. Identify the amide: R-C(=O)-NR'R''
  2. Replace C=O with CH₂: product is R-CH₂-NR'R''
  3. Determine amine class from N substitution
  4. Check for other LiAlH₄-reducible groups

Tips

  • "Carbonyl carbon becomes CH₂" - quick connectivity prediction
  • Count substituents on N to predict amine class
  • Amides go through iminium; esters go through aldehyde
  • LiAlH₄ is not selective - it reduces almost everything

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