Reduce an amide carbonyl to an amine with LiAlH₄
LiAlH₄ reduces an amide's C=O group to CH₂, converting the amide into an amine.
Section 34 · Nitrogen compounds
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 34. Use the stated reagents, conditions and product scope. Written practice is self-checked This online lesson cannot demonstrate supervised laboratory skills.
Before you begin
- Recall nucleophilic substitution, reduction, acyl-chloride addition–elimination, acid–base equilibria and covalent bonding.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What reagent reduces the CO group of an amide to form an amine?
What you will learn
- Recall LiAlH₄ as the amide-reduction reagent
- Track the amide carbonyl carbon into the product
- Distinguish reduction from hydrolysis
Reduce the carbonyl group
LiAlH₄ reduces the CO group in an amide to a methylene group. A primary amide RCONH₂ becomes RCH₂NH₂, an amine.
For example, ethanamide CH₃CONH₂ becomes ethylamine CH₃CH₂NH₂. The carbonyl carbon remains in the chain; oxygen is removed during reduction.
Do not confuse reaction outcomes
Hydrolysis breaks the C(O)–N bond and forms a carboxylic acid or carboxylate. Reduction changes C=O to CH₂ and retains a C–N bond in the amine product.
Use LiAlH₄, not aqueous acid or alkali, when asked to reduce the amide carbonyl.
Pause and explain
What does the amide carbonyl carbon become during reduction?
Worked example
Predict the product of ethanamide reduction with LiAlH₄.
Show the worked solution
- Keep the carbonyl carbon.
- Change C=O to CH₂.
- Retain nitrogen attached to that carbon.
Answer Ethylamine, CH₃CH₂NH₂.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Predict the product when butanamide is reduced with LiAlH₄.
Give me a hint
Retain all four carbons and convert the carbonyl carbon to CH₂.
Compare my reasoning
- Butanamide has four carbons.
- C=O becomes CH₂.
- Nitrogen stays attached at the chain end.
Butan-1-amine, CH₃CH₂CH₂CH₂NH₂.
Look for these in your work
- Carbon count
- Carbonyl reduced
- Amine named
Contrast amide hydrolysis with amide reduction by product class.
Give me a hint
Hydrolysis cleaves C–N; reduction converts C=O.
Compare my reasoning
- Hydrolysis gives acid/carboxylate plus nitrogen product.
- LiAlH₄ reduction gives an amine.
- The carbonyl carbon is retained in reduction.
Hydrolysis breaks the amide to an acid or carboxylate; LiAlH₄ reduces C=O to CH₂ and forms an amine.
Look for these in your work
- Hydrolysis outcome
- Reduction outcome
- Carbon retained
A response says ethanamide + LiAlH₄ gives methylamine. Check atom balance and correct the amine name.
Give me a hint
Count the amide carbonyl carbon as a chain carbon.
Compare my reasoning
- Ethanamide contains two carbon atoms.
- Reduction retains both, changing C=O to CH₂.
- The product is ethylamine.
The product is ethylamine, CH₃CH₂NH₂, because the carbonyl carbon remains in the chain.
Look for these in your work
- Carbon count
- Reduction change
- Product corrected
Common mistakes
- Predicting ethanoic acid
- Losing the carbonyl carbon
- Leaving C=O in the product
- Confusing the reagent with NaOH
What can you explain now?
What reagent reduces the CO group of an amide to form an amine?
Compare with the explanation
LiAlH₄
Lithium aluminium hydride is specified.
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Make it stick
How does LiAlH₄ reduction of an amide differ from aqueous hydrolysis?
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