Amides: formation, hydrolysis and reduction
Key idea: Amide nitrogen's lone pair is delocalised into the carbonyl group, reducing basicity. Hydrolysis breaks the C–N bond; reduction changes C=O to CH₂ while keeping C–N intact.
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Guided lesson
Guided steps
- Find the amide group.
- Track which reactant supplies each part of the amide.
- Read how the carbonyl changes the nitrogen lone pair.
- Compare the two nitrogen lone pairs.
- Explain how delocalisation changes basicity.
- Follow the products in acid and alkali.
- Write both hydrolysis equations.
- Track the carbonyl carbon during reduction.
- Form the amide, then reduce it.
- Give the products from each set of conditions.
- Rebuild the amide from its hydrolysis products.
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ContinueIn an amide, the nitrogen is bonded directly to a carbonyl carbon: RCONH₂, RCONHR' or RCONR'R''. That one change turns a basic amine into a neutral compound. An amide link also joins the units in nylon and proteins. This lesson covers how amides are made, why they are neutral, and their hydrolysis and reduction.
Check your understanding 1 Which nitrogen compound?
Compare CH₃CH₂NH₂ and CH₃CONHCH₃. Which contains an amide group?
Show answer
CH₃CONHCH₃ is the amide: trace the C=O carbon directly to nitrogen. CH₃CH₂NH₂ is an amine.
Making amides
An acyl chloride reacts with ammonia or a primary amine at room temperature, in a condensation reaction that eliminates hydrogen chloride:
With a primary amine, RNH₂, the product is an N-substituted amide: the acyl group comes from the acyl chloride and the group on nitrogen from the amine. The HCl reacts with more ammonia or amine, so in practice an excess is used:
Each scheme shows the main organic species. The surrounding equations include the other reactants and products.
Form an N-substituted amide
R'COCl → R'CONHRConditions: Excess primary amine; room temperature.
The acyl fragment supplies C=O; the amine supplies the group on nitrogen.
Reduce and retain the C–N bond
R'CONHR → R'CH₂NHRConditions: LiAlH₄ in dry ether, followed by water.
The carbonyl carbon becomes CH₂. Neither carbon group is removed.
Why amides are neutral
In ethanamide, CH₃CONH₂, the lone pair on nitrogen overlaps with the π bond of the neighbouring C=O group and is delocalised over the N–C–O system. The electronegative oxygen draws this electron density towards itself. The nitrogen lone pair is much less available to accept a proton. Amides are treated as neutral in these acid–base comparisons, unlike amines.
Spot the mistake 1
Why ethanamide is not basic
Learner claim
Compare the two lone pairs
Show solution step by step
Look at the neighbour
Method
The amide nitrogen is bonded to a C=O group.Reason
Its lone pair overlaps with the C=O π bond and is delocalised towards the electronegative oxygen.Working
Answer
The lone pair is not confined to nitrogen.Link to basicity
Method
The nitrogen lone pair is much less available to accept a proton.Reason
In ethylamine, the lone pair stays on nitrogen and the ethyl group pushes electrons towards it.Working
Answer
Ethanamide is neutral; ethylamine is a base.
Common mistake
Check your understanding 2 Explain the difference
Explain why ethanamide and ethylamine differ in basicity. Link the position of the nitrogen lone pair to proton acceptance.
Show answer
In ethanamide, the nitrogen lone pair overlaps with the neighbouring C=O π system and is delocalised over N–C–O. It is much less available to accept a proton. In ethylamine the pair is localised on nitrogen; the ethyl group donates electron density, making proton acceptance easier. Include the neighbouring carbonyl, delocalisation and proton acceptance in your explanation.
Hydrolysis
Heating an amide under reflux with aqueous acid or aqueous alkali breaks the C–N bond. The conditions decide the form of each product.
With aqueous acid, the products are the carboxylic acid and an ammonium ion:
With aqueous alkali, they are the carboxylate ion and ammonia, which turns damp red litmus blue. Test the gas with the paper; do not smell it directly:
For an N-substituted amide, the nitrogen product is the amine, or its protonated ion in acid, not ammonia.
Check your understanding 3 Hydrolysis in acid and in alkali
Write ionic equations for the hydrolysis of N-methylethanamide, CH₃CONHCH₃, by heating with (a) aqueous acid and (b) aqueous alkali.
Show answer
(a) CH₃CONHCH₃ + H₂O + H⁺ → CH₃COOH + CH₃NH₃⁺
(b) CH₃CONHCH₃ + OH⁻ → CH₃COO⁻ + CH₃NH₂
In acid, the amine is protonated and the acid stays as CH₃COOH. In alkali, the acid is deprotonated and the amine is released.
Reduction
Lithium aluminium hydride, LiAlH₄, in dry ether, followed by an aqueous work-up, reduces an amide to an amine. The C=O group becomes CH₂; the C–N bond stays:
So hydrolysis breaks the amide into two pieces, while reduction keeps every carbon in one molecule.
Try it yourself 2
Make an amide, then reduce it
Problem
Try this before viewing the solution
Hints
Hint 1
first step
Find the acyl carbon and the atom of methylamine that bonds to it.Hint 2
second step
Reduction changes C=O to CH₂ and keeps the C–N bond.Show solution step by step
Make the amide
Method
The product is N-methylethanamide.Reason
One methylamine molecule forms the amide; a second takes up the HCl.Working
Answer
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺ + Cl⁻Reduce it
Method
Use LiAlH₄ in dry ether, then an aqueous work-up.Reason
The C=O becomes CH₂; the C–N bond and both carbon groups stay.Working
Answer
CH₃CONHCH₃ + 4[H] → CH₃CH₂NHCH₃ + H₂OName the product
Method
The product is N-methylethylamine, a secondary amine.Reason
Its nitrogen carries an ethyl group, a methyl group and one hydrogen.Working
Answer
CH₃CH₂NHCH₃
Quick check
Check your understanding 4
For CH₃CH₂CONHCH₃, give the products of (a) heating with aqueous sodium hydroxide and (b) reduction with LiAlH₄ in dry ether.
Show answer
(a) Propanoate ions, CH₃CH₂COO⁻, and methylamine, CH₃NH₂: hydrolysis breaks the C–N bond.
(b) CH₃CH₂CH₂NHCH₃: reduction keeps the C–N bond and turns C=O into CH₂, so all four carbons stay in one molecule.
Check your understanding 5 Rebuild the starting amide
An amide X gives ethanoate ions and ethylamine when heated with aqueous sodium hydroxide. Deduce X, then give the organic product when X is reduced with LiAlH₄ in dry ether followed by water. Explain why hydrolysis and reduction give different carbon skeletons.
Show answer
Ethanoate supplies the CH₃CO– part and ethylamine supplies the -NHC₂H₅ part, so X is CH₃CONHCH₂CH₃, N-ethylethanamide. Reduction gives CH₃CH₂NHCH₂CH₃, diethylamine. Hydrolysis breaks C–N to give two two-carbon fragments; reduction replaces C=O by CH₂ and keeps all four carbons in one molecule. Check both fragments, the retained carbonyl carbon and the nitrogen substituents in your structures.
Common mistakes
- Calling an amide a base because it contains nitrogen with a lone pair.
- Writing ammonia as the product of hydrolysing every amide. An N-substituted amide gives an amine.
- Giving the carboxylic acid as the product of alkaline hydrolysis instead of the carboxylate ion.
- Losing the carbonyl carbon when an amide is reduced.
- Mixing up hydrolysis, which breaks the C–N bond, with reduction, which keeps it.
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Syllabus and review details
- GCE A-Level H2 Chemistry 2027 · 9476-2027
9476 (2027), complete syllabus
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