The shape of the lesson
By the end of the lesson, learners can
- Making amides from acyl chlorides: Acyl chlorides are reactive because the carbonyl carbon is bonded to two electronegative atoms.
- Hydrolysis, reduction and basicity: Amides are stable at room temperature but break apart when heated under reflux with acid or alkali.
- RCOCl + 2NH3 → RCONH2 + NH4Cl.
- Acid hydrolysis: RCOOH + ammonium ion.
- Alkaline hydrolysis: RCOO− + NH3 or amine.
- LiAlH4: amide to amine; amides are neutral.
One picture every idea returns to
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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Nitrogen next to a carbonyl.. Amides are made from acyl chlorides and ammonia or amines, split apart by hot acid or alkali, and reduced to amines by LiAlH4. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Making amides from acyl chlorides | Acyl chlorides are reactive because the carbonyl carbon is bonded to two electronegative atoms. Project tab 4 of the lab (Nitrogen compounds (A Level) lab, amides) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Hydrolysis, reduction and basicity | Amides are stable at room temperature but break apart when heated under reflux with acid or alkali. Project tab 4 of the lab (Nitrogen compounds (A Level) lab, amides) and let a student drive it while the class predicts. | Quiz-Quiz-Trade |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Hydrolysis, reduction and basicity, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Quiz-Quiz-Trade
Each student holds a card with a question on the front and the answer on the back. They pair up, quiz each other, coach if needed, trade cards and find a new partner. Three rounds covers the whole set.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| 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 LiAlH 4; NaBH 4 does not reduce amides. |
Amides: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Write equations for the reactions of ethanoyl chloride with (a) ammonia and (b) ethylamine. Name the organic products.
N-ethylpropanamide, CH3CH2CONHCH2CH3, is heated under reflux with (a) HCl(aq) and (b) NaOH(aq). Give the organic products in each case.
Explain why an aqueous solution of ethylamine is alkaline but a solution of ethanamide is neutral.
7.85 g of ethanoyl chloride (Mr 78.5) reacts with excess ammonia, giving 4.72 g of ethanamide (Mr 59.0). Calculate the percentage yield.
A white solid X, C2H5NO, gives no reaction with dilute acid at room temperature. Warming X with NaOH(aq) gives a gas that turns damp red litmus blue. Identify X and the gas.
Give the reagent and the product when propanamide is reduced, and write an equation using [H].
Answer key and marking notes
- (a) CH3COCl + 2NH3 → CH3CONH2 + NH4Cl; ethanamide.
- (b) CH3COCl + 2CH3CH2NH2 → CH3CONHCH2CH3 + CH3CH2NH3+Cl−; N-ethylethanamide.
- (a) Propanoic acid, CH3CH2COOH, and ethylammonium chloride, CH3CH2NH3+Cl−.
- (b) Sodium propanoate, CH3CH2COO−Na+, and ethylamine, CH3CH2NH2.
- The C–N bond is broken in both; only the ionisation state of the products differs.
- In ethylamine the N lone pair is available (enhanced by the electron-donating ethyl group) and accepts H+ from water, releasing OH−.
- In ethanamide the N lone pair overlaps with the C=O π bond and is delocalised towards oxygen.
- So it is not available to accept a proton, and the solution stays neutral.
- n(CH3COCl) = 7.85 ÷ 78.5 = 0.100 mol, so maximum ethanamide = 0.100 × 59.0 = 5.90 g.
- Yield = 4.72 ÷ 5.90 × 100 = 80.0%.
- X is ethanamide, CH3CONH2 (neutral, so no salt forms with dilute acid).
- Alkaline hydrolysis releases ammonia, NH3, which is alkaline.
- LiAlH4 in dry ether; product propylamine, CH3CH2CH2NH2.
- CH3CH2CONH2 + 4[H] → CH3CH2CH2NH2 + H2O.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Ethanoyl chloride, CH3COCl, reacts with methylamine, CH3NH2. What is the organic product?
- AN-methylethanamide, CH3CONHCH3
- Bethanamide, CH3CONH2
- Cmethyl ethanoate, CH3COOCH3
- Dethylmethylamine, CH3CH2NHCH3
Ethanamide is heated under reflux with dilute hydrochloric acid. Which products form?
- Aethanoic acid and ammonium chloride
- Bsodium ethanoate and ammonia
- Cethylamine and water
- Dethanoyl chloride and ammonia
N-methylethanamide, CH3CONHCH3, is heated under reflux with aqueous sodium hydroxide. Which products form?
- Asodium ethanoate and methylamine
- Bethanoic acid and methylammonium ions
- Csodium methanoate and ethylamine
- Dsodium ethanoate and ammonia
Why is ethanamide a much weaker base than ethylamine?
- Athe lone pair on N is delocalised over the adjacent C=O group, so it is not available to accept a proton
- Bethanamide has no lone pair on nitrogen
- Cthe CH3 group in ethanamide withdraws electrons strongly from N
- Dethanamide is insoluble in water, so it cannot react with H+
Ethanamide is reduced with LiAlH4 in dry ether. What is the organic product?
- Aethylamine, CH3CH2NH2
- Bethanol, CH3CH2OH
- Cmethylamine, CH3NH2
- Dethanal, CH3CHO
7.85 g of ethanoyl chloride (Mr 78.5) is added to excess ammonia and 4.72 g of ethanamide (Mr 59.0) is collected. What is the percentage yield?
- A80.0%
- B60.1%
- C75.2%
- D125%
2.95 g of ethanamide (Mr 59.0) is heated with excess NaOH(aq) until hydrolysis is complete. What volume of ammonia at room conditions is released?
- A1.20 dm3
- B2.40 dm3
- C0.600 dm3
- D70.8 dm3
Compound Z, C3H7NO, is neutral. Heating Z with NaOH(aq) gives sodium ethanoate and a gas that smells fishy and turns damp red litmus blue. What is Z?
- AN-methylethanamide, CH3CONHCH3
- Bpropanamide, CH3CH2CONH2
- CN-ethylmethanamide, HCONHCH2CH3
- D1-aminopropan-2-one, CH3COCH2NH2
Quiz key
CH3COCl + 2CH3NH2 → CH3CONHCH3 + CH3NH3+Cl− (addition-elimination). The methyl group stays on N, so the product is a secondary (N-substituted) amide. Ethanamide comes from ammonia, not methylamine. Methyl ethanoate needs methanol. Ethylmethylamine has no C=O; it is what you would get by reducing the amide with LiAlH4.
CH3CONH2 + H2O + HCl → CH3COOH + NH4Cl. The C-N bond is hydrolysed; in acid the acid stays as RCOOH and the ammonia is protonated to NH4+. Sodium ethanoate and ammonia are the products of alkaline hydrolysis. Ethylamine comes from reduction, not hydrolysis. Acyl chlorides are not formed by water and HCl; the reaction runs the other way.
CH3CONHCH3 + NaOH → CH3COONa + CH3NH2. The carbonyl side becomes the carboxylate and the N side becomes the amine. Ethanoic acid with methylammonium ions is the acid hydrolysis result. Methanoate and ethylamine split the molecule on the wrong side of the C=O carbon. Ammonia would come only from an unsubstituted amide, RCONH2.
In an amide the N lone pair overlaps with the π bond of C=O and is drawn towards the electronegative oxygen. It is no longer available to bond to H+, so amides are effectively neutral, whereas the ethyl group in ethylamine pushes electron density onto N. Nitrogen in an amide still has a lone pair; it is just delocalised. Methyl groups donate electrons, they do not withdraw them. Ethanamide is soluble in water, because it hydrogen bonds to it.
CH3CONH2 + 4[H] → CH3CH2NH2 + H2O. Both carbons are kept and the nitrogen stays attached. Ethanol would need the C-N bond to break, which reduction does not do. Methylamine would mean losing a carbon. Ethanal is not formed: LiAlH4 reduces fully, and the nitrogen is not lost.
n(CH3COCl) = 7.85 ÷ 78.5 = 0.100 mol, so the maximum is 0.100 × 59.0 = 5.90 g; 4.72 ÷ 5.90 × 100 = 80.0%. 60.1% divides the mass of product by the mass of reactant. 75.2% is just the ratio of the Mr values (59.0 ÷ 78.5), not the yield. 125% inverts the ratio and is impossible.
n(ethanamide) = 2.95 ÷ 59.0 = 0.0500 mol, giving 0.0500 mol NH3 = 0.0500 × 24.0 = 1.20 dm3. 2.40 dm3 doubles the moles, as if each amide released two NH3. 0.600 dm3 halves them. 70.8 dm3 multiplies the mass by 24.0 without converting to moles.
Ethanoate means the acyl group is CH3CO-, and a fishy alkaline gas is a volatile amine, here CH3NH2. Together they make CH3CONHCH3 (C3H7NO), and a neutral solution fits an amide. Propanamide would give sodium propanoate and ammonia. N-ethylmethanamide would give sodium methanoate and ethylamine. 1-aminopropan-2-one is an amine with a free NH2, so it would be basic, and it is not hydrolysed in this way.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.