Amides are made from acyl chlorides and ammonia or amines, split apart by hot acid or alkali, and reduced to amines by LiAlH4. Their nitrogen has a lone pair, yet they are hardly basic at all.
An amide contains the –CONH– group. Amides are made by reacting an acyl chloride with ammonia (giving RCONH2) or with a primary amine (giving an N-substituted amide, RCONHR′); the reaction is fast at room temperature and gives steamy HCl, which reacts with more base. Amides are hydrolysed by heating under reflux: with aqueous acid the products are the carboxylic acid and an ammonium salt, with aqueous alkali they are the carboxylate salt and ammonia or the amine. Amides are much weaker bases than amines, effectively neutral, because the N lone pair is delocalised over the C=O group. LiAlH4 reduces an amide to an amine.
Acyl chlorides are reactive because the carbonyl carbon is bonded to two electronegative atoms. Ammonia and amines attack it with their N lone pair in an addition-elimination reaction.
| Reactants | Amide formed | Equation |
|---|---|---|
| ethanoyl chloride + ammonia | ethanamide (primary amide) | CH3COCl + 2NH3 → CH3CONH2 + NH4Cl |
| ethanoyl chloride + methylamine | N-methylethanamide | CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3+Cl− |
| benzoyl chloride + phenylamine | N-phenylbenzamide | C6H5COCl + C6H5NH2 → C6H5CONHC6H5 + HCl |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
What forms when ethanoyl chloride reacts with ammonia?
Which products form when ethanamide is heated with NaOH(aq)?
Why is ethanamide much less basic than ethylamine?
Which reagent reduces an amide to an amine?
Amides are stable at room temperature but break apart when heated under reflux with acid or alkali. Which products you get depends on which you use.
| Reaction | Conditions | Products from ethanamide | Products from N-methylethanamide |
|---|---|---|---|
| acid hydrolysis | HCl(aq), heat under reflux | CH3COOH + NH4+ | CH3COOH + CH3NH3+ |
| alkaline hydrolysis | NaOH(aq), heat under reflux | CH3COO−Na+ + NH3 | CH3COO−Na+ + CH3NH2 |
| reduction | LiAlH4 in dry ether | CH3CH2NH2 | CH3CH2NHCH3 |
Learn the hydrolysis products by condition: acid gives the free carboxylic acid and the nitrogen as an ammonium ion; alkali gives the carboxylate salt and the free ammonia or amine. Amides are not basic like amines: their N lone pair is delocalised onto the C=O. Reduction needs LiAlH4; NaBH4 does not reduce amides.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
N-methylethanamide is heated under reflux with dilute HCl. Which pair of products forms?
7.85 g of ethanoyl chloride (Mr 78.5) gives 4.72 g of ethanamide (Mr 59.0). What is the percentage yield?
Which compound is formed when N-methylethanamide is reduced with LiAlH4?
What volume of ammonia at room conditions is released when 2.95 g of ethanamide (Mr 59.0) is fully hydrolysed by NaOH(aq)?
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