The shape of the lesson
By the end of the lesson, learners can
- Three routes to an amine: Making an amine from a halogenoalkane is covered in the AS nitrogen compounds lab.
- Why amines are bases: Compare ethylamine, ammonia and phenylamine in the simulator and watch how available the lone pair is.
- RX + excess NH3, ethanol, sealed tube.
- Nitrile or amide + LiAlH4 (nitrile also H2/Ni).
- RNH2 + RX gives the secondary amine R2NH.
- Basicity: ethylamine > NH3 > phenylamine.
One picture every idea returns to
An amine is ammonia with one or more H atoms replaced by alkyl or aryl groups: a primary amine is RNH2, a secondary amine is R2NH. Primary alkyl amines are made by heating a halogenoalkane with excess ammonia in ethanol in a sealed tube, by reducing an amide with LiAlH4 or by reducing a nitrile with LiAlH4 or H2/Ni. Because the amine product is itself a nucleophile, it can react with more halogenoalkane to give a secondary amine. The lone pair on nitrogen accepts H+, so amines are bases: ethylamine is stronger than ammonia because the electron-donating alkyl group makes the lone pair more available, while phenylamine is much weaker because its lone pair is delocalised into the benzene ring.
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, Ammonia with an alkyl group.. Swap a hydrogen of ammonia for an alkyl group and you have an amine. 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: Three routes to an amine | Making an amine from a halogenoalkane is covered in the AS nitrogen compounds lab. Project tab 2 of the lab (Nitrogen compounds (A Level) lab, making phenylamine) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Why amines are bases | Compare ethylamine, ammonia and phenylamine in the simulator and watch how available the lone pair is. Project tab 1 of the lab (Nitrogen compounds (A Level) lab, base strength) and let a student drive it while the class predicts. | Jigsaw |
| 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 Why amines are bases, 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.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
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 |
|---|---|
| Basicity is about how available the lone pair on N is to accept H +, not about how many H atoms there are. | Alkyl groups push electron density onto N (stronger base); a benzene ring pulls the lone pair into its delocalised system (weaker base). In the halogenoalkane route, forgetting the excess ammonia means a mixture of primary, secondary and tertiary amines and quaternary salts. |
Primary and secondary amines: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
Give the reagents and conditions for making propylamine, CH3CH2CH2NH2, from (a) 1-bromopropane, (b) propanenitrile, (c) propanamide.
Explain why the order of base strength is ethylamine > ammonia > phenylamine.
10.89 g of bromoethane (Mr 108.9) is heated with excess ammonia and 3.15 g of ethylamine (Mr 45.0) is obtained. Calculate the percentage yield.
Write an equation for the formation of diethylamine from ethylamine and name the type of reaction.
Calculate the volume of hydrogen at room conditions needed to reduce 0.0500 mol of ethanenitrile to ethylamine.
Ethylamine is a gas that dissolves well in water, but ethylammonium chloride is a white solid. Explain what happens when HCl(aq) and then excess NaOH(aq) are added to ethylamine.
Answer key and marking notes
- (a) Excess NH3 in ethanol, heated in a sealed tube.
- (b) LiAlH4 in dry ether, or H2 with a nickel catalyst.
- (c) LiAlH4 in dry ether.
- Base strength depends on how available the N lone pair is to accept a proton.
- The ethyl group is electron-donating, so it increases the electron density on N in ethylamine: stronger than NH3.
- In phenylamine the lone pair overlaps with the ring π system and is partly delocalised, so it is less available: weaker than NH3.
- n(bromoethane) = 10.89 ÷ 108.9 = 0.100 mol, so maximum ethylamine = 0.100 × 45.0 = 4.50 g.
- Yield = 3.15 ÷ 4.50 × 100 = 70.0%.
- CH3CH2NH2 + CH3CH2Br → (CH3CH2)2NH + HBr.
- Nucleophilic substitution: the N lone pair of ethylamine attacks the δ+ carbon.
- CH3CN + 2H2 → CH3CH2NH2, so n(H2) = 0.100 mol.
- Volume = 0.100 × 24.0 = 2.40 dm3.
- HCl: the N lone pair accepts H+, giving the ionic salt CH3CH2NH3+Cl−.
- NaOH: OH− removes H+ from CH3CH2NH3+, releasing ethylamine again (smell of fish).
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.
The chart shows pKb values for ethylamine, ammonia and phenylamine. Which statement explains the order?
- Athe ethyl group donates electron density to N, making its lone pair more available, while in phenylamine the lone pair is partly delocalised into the benzene ring
- Bphenylamine has the largest Mr, so it attracts protons most strongly
- Cethylamine is the strongest base because it has the most hydrogen atoms on nitrogen
- Dthe benzene ring donates electrons to N, so phenylamine is the strongest base
Which reagent and conditions convert bromoethane into ethylamine?
- Aexcess ammonia dissolved in ethanol, heated in a sealed tube
- Baqueous sodium hydroxide, heated under reflux
- Cpotassium cyanide in ethanol, heated under reflux
- Dconcentrated sulfuric acid, heated to 170 °C
Propanenitrile, CH3CH2CN, is reduced with LiAlH4 in dry ether. What is the organic product?
- Apropylamine, CH3CH2CH2NH2
- Bethylamine, CH3CH2NH2
- Cpropanamide, CH3CH2CONH2
- Dpropanoic acid, CH3CH2COOH
Ethylamine is heated with bromoethane. Which product forms and what type of amine is it?
- A(CH3CH2)2NH, a secondary amine
- BCH3CH2NH2, a primary amine
- C(CH3CH2)2NH, a primary amine
- DCH3CH2CH2CH2NH2, a primary amine
Ethylamine is added to dilute hydrochloric acid. What forms?
- Aethylammonium chloride, CH3CH2NH3+Cl−
- Bchloroethane and ammonia
- Cethanamide and hydrogen
- Dno reaction, because amines are neutral
What volume of hydrogen at room conditions is needed to reduce 0.0500 mol of ethanenitrile, CH3CN, completely to ethylamine using a nickel catalyst?
- A2.40 dm3
- B1.20 dm3
- C4.80 dm3
- D0.600 dm3
10.89 g of bromoethane (Mr 108.9) is heated with excess ammonia and 3.15 g of ethylamine (Mr 45.0) is collected. What is the percentage yield?
- A70.0%
- B28.9%
- C30.0%
- D143%
Which two-step route converts bromomethane, CH3Br, into ethylamine, CH3CH2NH2?
- AKCN in ethanol, heat under reflux; then LiAlH4 in dry ether
- Bexcess NH3 in ethanol in a sealed tube; then LiAlH4
- CNaOH(aq), heat; then NH3 at room temperature
- DKCN in ethanol; then dilute HCl, heat under reflux
Quiz key
Base strength depends on how available the N lone pair is to accept H+. Ethylamine (pKb 3.27) is stronger than ammonia (4.74) because the alkyl group pushes electron density onto N. Phenylamine (9.37) is much weaker because its lone pair overlaps with the ring π system and is partly delocalised. The largest pKb means the WEAKEST base, so phenylamine does not attract protons most strongly, and Mr is not the cause. Ethylamine has two H atoms on N, fewer than ammonia's three, so counting hydrogens gets it backwards. The ring withdraws the lone pair rather than donating to N, so phenylamine is the weakest, not the strongest.
CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br. The ammonia lone pair attacks the δ+ carbon (nucleophilic substitution); excess ammonia reduces further substitution, and a sealed tube stops the gaseous ammonia escaping. NaOH(aq) gives ethanol instead. KCN in ethanol gives propanenitrile, which adds a carbon. Concentrated sulfuric acid at 170 °C dehydrates alcohols; it does not make amines.
CH3CH2CN + 4[H] → CH3CH2CH2NH2. All three carbons are kept, so the product is propylamine; H2 with a nickel catalyst does the same job. Ethylamine would mean losing a carbon, which reduction does not do. Propanamide would need partial hydrolysis, not reduction. Propanoic acid comes from hydrolysing the nitrile with acid, not from reducing it.
CH3CH2NH2 + CH3CH2Br → (CH3CH2)2NH + HBr. Diethylamine has two alkyl groups on N, so it is a secondary amine. The second option is the starting material. Calling diethylamine primary confuses the classification, which counts carbon groups on N, not on a carbon. Butylamine would need a new C-C bond, but substitution forms a C-N bond.
The N lone pair accepts H+: CH3CH2NH2 + HCl → CH3CH2NH3+Cl−, an ionic salt, just as NH3 gives NH4Cl. The C-N bond is not broken, so no chloroethane forms. Amides need an acyl chloride, and no hydrogen is released. Amines are bases, not neutral; it is amides that are neutral.
CH3CN + 2H2 → CH3CH2NH2. Four H atoms are added, which is 2 mol H2 per mol nitrile, so 0.100 mol H2 × 24.0 = 2.40 dm3. 1.20 dm3 uses a 1 : 1 ratio (only reducing the triple bond to a double bond). 4.80 dm3 uses 4 mol H2, counting H atoms as molecules. 0.600 dm3 halves the moles wrongly.
n(bromoethane) = 10.89 ÷ 108.9 = 0.100 mol, so the maximum is 0.100 × 45.0 = 4.50 g; yield = 3.15 ÷ 4.50 × 100 = 70.0%. 28.9% divides the product mass by the reactant mass, ignoring the different Mr values. 30.0% is the percentage lost, not the percentage obtained. 143% inverts the ratio and is impossible.
CH3Br + CN− → CH3CN + Br− adds the extra carbon (nucleophilic substitution); reducing the nitrile then gives CH3CH2NH2. Ammonia gives methylamine, CH3NH2, with no new carbon, and LiAlH4 cannot add one. NaOH(aq) gives methanol, which does not react with ammonia to make an amine. Hydrolysing the nitrile with acid gives ethanoic acid, not an amine.
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.