Reduce an amide to a primary amine with lithium aluminium hydride
Use LiAlH₄ to reduce the carbonyl group of an amide to a methylene group and form an amine.
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
Which reagent reduces amides to amines in the syllabus?
What you will learn
- Recall LiAlH₄ as an amide-reduction reagent
- Track the amide carbonyl carbon into the amine
- Predict the carbon-chain length of the product
Reduce C=O without losing its carbon
Lithium aluminium hydride, LiAlH₄, reduces an amide to an amine. The carbonyl carbon remains in the molecule but its C=O becomes CH₂ in the primary-amide example: RCONH₂ → RCH₂NH₂.
Ethanamide, CH₃CONH₂, gives ethylamine, CH₃CH₂NH₂. Count the carbonyl carbon as part of the product chain.
Distinguish this from nitrile reduction
Both amide and nitrile reductions can yield amines, but the starting functional group and carbon skeleton differ.
Use LiAlH₄ as the named reagent. In laboratory preparation, it is used under dry conditions followed by work-up; do not treat water as the reducing reagent.
Pause and explain
What happens to the amide carbonyl carbon in RCONH₂ → RCH₂NH₂?
Worked example
Predict the amine formed when ethanamide is reduced with LiAlH₄.
Show the worked solution
- Retain both carbons from CH₃CONH₂.
- Reduce the carbonyl carbon to CH₂.
- Keep 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 of CH₃CH₂CONH₂ reduction with LiAlH₄.
Give me a hint
Keep all three carbons and convert the carbonyl carbon to CH₂.
Compare my reasoning
- Propanamide has three carbons.
- The carbonyl carbon becomes CH₂.
- Nitrogen remains attached at the chain end.
Propan-1-amine, CH₃CH₂CH₂NH₂.
Look for these in your work
- Three carbons
- Carbonyl reduced
- Primary amine
Explain why ethanamide forms ethylamine rather than methylamine.
Give me a hint
Count the carbonyl carbon as part of the carbon chain.
Compare my reasoning
- Ethanamide has a methyl carbon and a carbonyl carbon.
- Reduction changes C=O to CH₂.
- Both carbons remain in the product.
The carbonyl carbon is retained as CH₂, so the two-carbon product is ethylamine.
Look for these in your work
- Starting atoms counted
- Reduction described
- Product chain length
A response writes CH₃CONH₂ + LiAlH₄ → CH₃NH₂. Identify the missing structural change.
Give me a hint
The carbonyl carbon must remain attached between the methyl group and nitrogen.
Compare my reasoning
- The C=O carbon is not removed.
- It becomes CH₂.
- The product is CH₃CH₂NH₂.
LiAlH₄ reduces ethanamide to ethylamine, CH₃CH₂NH₂; the carbonyl carbon becomes the second chain carbon.
Look for these in your work
- Carbon retained
- C=O reduced
- Product corrected
Common mistakes
- Deleting the carbonyl carbon
- Predicting ethanamine with three carbons
- Using NaBH₄ as the specified reagent
- Calling the product an amide still
What can you explain now?
Which reagent reduces amides to amines in the syllabus?
Compare with the explanation
LiAlH₄
Lithium aluminium hydride is the named reducing agent.
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Make it stick
How does LiAlH₄ change an amide carbonyl when it forms an amine?
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