The shape of the lesson
By the end of the lesson, learners can
- Substitution on a ring: Sort the reaction cards in the simulator, then follow an electrophile attacking a pi system in Fig. 1.
- Joining, splitting and redox: These labels describe what happens to the molecule overall, whatever the mechanism.
- Arenes: electrophilic substitution; the ring survives.
- Acyl chlorides: addition-elimination; HCl lost.
- Condensation loses a small molecule; hydrolysis uses water.
- [H] for reduction, [O] for oxidation; balance by atoms.
One picture every idea returns to
In electrophilic substitution an electrophile such as NO2+ attacks the delocalised electrons of a benzene ring and replaces an H atom, so the ring survives. In addition-elimination a nucleophile adds to the C=O carbon of an acyl chloride and then Cl− is eliminated, leaving HCl. A condensation reaction joins two molecules with the loss of a small molecule such as H2O or HCl. Hydrolysis breaks a bond by reaction with water, often catalysed by acid or alkali. Reduction adds hydrogen or removes oxygen (written [H]); oxidation adds oxygen or removes hydrogen (written [O]). Diazotisation turns phenylamine into a diazonium salt, which couples with phenol to make an azo dye.
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, Name the reaction, predict the product.. Most A Level organic reactions fit a small set of labels: electrophilic substitution, addition-elimination, condensation, hydrolysis, reduction and oxidation. 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: Substitution on a ring | Sort the reaction cards in the simulator, then follow an electrophile attacking a pi system in Fig. 1. Project tab 5 of the lab (Organic chemistry (A Level introduction) lab, reaction types) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Joining, splitting and redox | These labels describe what happens to the molecule overall, whatever the mechanism. Project tab 5 of the lab (Organic chemistry (A Level introduction) lab, reaction types) 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, 16 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 Joining, splitting and redox, 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 |
|---|---|
| Benzene undergoes electrophilic substitution, not addition: an addition would destroy the stable delocalised ring. | Addition-elimination happens at the C=O carbon of an acyl chloride, so it is not the same as nucleophilic substitution of a halogenoalkane. Hydrolysis and condensation are opposites: condensation releases a small molecule, hydrolysis uses water to split a bond. |
A Level organic reactions: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
State the type of reaction: (a) benzene with Br2 and FeBr3; (b) ethanoyl chloride with ethanol; (c) ethanamide heated with NaOH(aq); (d) nitrobenzene with Sn and concentrated HCl.
15.6 g of benzene is nitrated and 18.45 g of nitrobenzene is collected. Calculate the percentage yield.
Explain the meaning of the term addition-elimination, using the reaction of ethanoyl chloride with water.
Write a balanced equation using [O] for the oxidation of methylbenzene to benzoic acid.
Calculate the atom economy for making N-ethylethanamide (Mr 87.0) from ethanoyl chloride (78.5) and ethylamine (45.0).
Explain the difference between a condensation reaction and a hydrolysis reaction.
Answer key and marking notes
- (a) electrophilic substitution.
- (b) addition-elimination (also a condensation, as HCl is lost).
- (c) hydrolysis.
- (d) reduction.
- n(benzene) = 15.6 ÷ 78.0 = 0.200 mol.
- Maximum nitrobenzene = 0.200 × 123.0 = 24.6 g.
- Yield = 18.45 ÷ 24.6 × 100 = 75.0%.
- A lone pair on the water O attacks the δ+ carbon of C=O: addition.
- The C=O reforms and Cl− leaves, then H+ is lost: elimination.
- Overall CH3COCl + H2O → CH3COOH + HCl.
- C6H5CH3 + 3[O] → C6H5COOH + H2O.
- Reagent: hot alkaline KMnO4, then acidify.
- Total mass of reactants = 78.5 + 45.0 = 123.5.
- Atom economy = 87.0 ÷ 123.5 × 100 = 70.4%.
- Condensation: two molecules join together and a small molecule (H2O or HCl) is released.
- Hydrolysis: a bond is broken by reaction with water, so water is a reactant; it is the reverse of condensation.
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.
Benzene reacts with bromine in the presence of FeBr3 to give bromobenzene and HBr. What type of reaction is this?
- Aelectrophilic substitution
- Belectrophilic addition
- Cnucleophilic substitution
- Dfree-radical substitution
Which reagents and conditions are used to nitrate benzene to nitrobenzene?
- Aconcentrated HNO3 and concentrated H2SO4, below about 55 °C
- Bdilute HNO3 at room temperature
- CNaNO2 and HCl, below 10 °C
- Dtin and concentrated HCl, heated under reflux
Ethanoyl chloride reacts with methanol: CH3COCl + CH3OH → CH3COOCH3 + HCl. What is the mechanism type?
- Aaddition-elimination
- Belectrophilic addition
- Cfree-radical substitution
- Delectrophilic substitution
Ethanamide, CH3CONH2, is heated under reflux with aqueous sodium hydroxide. What are the products?
- Asodium ethanoate and ammonia
- Bethanoic acid and ammonium chloride
- Cethylamine and water
- Dethanenitrile and water
Which reagents reduce nitrobenzene to phenylamine?
- Atin and concentrated HCl, heated, then aqueous NaOH
- BNaBH4 in water
- Cacidified potassium dichromate(VI)
- Dconcentrated H2SO4 at 55 °C
3.90 g of benzene (Mr 78.0) is nitrated and 4.43 g of nitrobenzene (Mr 123.0) is obtained. What is the percentage yield?
- A72.0%
- B88.0%
- C114%
- D36.0%
Ethylbenzene, C6H5CH2CH3, is heated under reflux with alkaline KMnO4, then acidified. What is the organic product?
- Abenzoic acid, C6H5COOH
- Bphenylethanoic acid, C6H5CH2COOH
- Cphenol, C6H5OH
- Dethylcyclohexane, C6H11CH2CH3
Phenyl ethanoate (Mr 136.0) is made from ethanoyl chloride (Mr 78.5) and phenol (Mr 94.0): CH3COCl + C6H5OH → CH3COOC6H5 + HCl. What is the atom economy?
- A78.8%
- B100%
- C21.2%
- D65.1%
Quiz key
FeBr3 polarises Br2 to give an electrophile (Br+) that attacks the electron-rich ring; an H is replaced and the delocalised ring is kept, so it is electrophilic substitution. Electrophilic addition happens with alkenes and would destroy the ring. Nucleophilic substitution needs a nucleophile attacking a δ+ carbon, as in halogenoalkanes. Free-radical substitution needs UV light and happens with alkanes.
Concentrated H2SO4 protonates concentrated HNO3, which loses water to give NO2+; keeping the temperature below about 55 °C limits further nitration. Dilute nitric acid nitrates phenol, whose ring is much more reactive, but not benzene. NaNO2 and HCl make nitrous acid for diazotisation of phenylamine. Tin and concentrated HCl reduce nitrobenzene to phenylamine, the opposite direction.
A lone pair on the methanol O adds to the δ+ C=O carbon, then the C=O reforms and Cl− is eliminated (with H+, as HCl): addition-elimination. Overall it is also a condensation, as HCl is lost. Electrophilic addition is the reaction of an electrophile with a C=C bond. Free-radical substitution needs radicals made by UV light. Electrophilic substitution is the reaction of an arene with an electrophile.
Hydrolysis breaks the C–N bond; in alkali the acid formed is a salt, so CH3CONH2 + NaOH → CH3COONa + NH3, and ammonia gas is given off. Ethanoic acid and ammonium chloride are the products of acid hydrolysis with HCl. Ethylamine would need reduction of the amide, not hydrolysis. Ethanenitrile forms by dehydrating an amide, which removes water rather than adding it.
Tin and concentrated hydrochloric acid supply the [H]; the amine forms as its salt, C6H5NH3+, so NaOH is added to release free phenylamine. NaBH4 reduces aldehydes and ketones but not the nitro group. Acidified dichromate is an oxidising agent, the wrong direction. Concentrated H2SO4 at 55 °C is part of the nitration mixture that makes nitrobenzene.
n(benzene) = 3.90 ÷ 78.0 = 0.0500 mol, so the maximum is 0.0500 × 123.0 = 6.15 g; yield = 4.43 ÷ 6.15 × 100 = 72.0%. 88.0% divides the mass of benzene by the mass of nitrobenzene, and 114% divides them the other way; both ignore the different Mr values, and a yield above 100% is impossible. 36.0% doubles the moles of benzene, taking the maximum as 12.3 g.
Any alkyl side chain is oxidised to –COOH on the ring carbon, the extra carbon leaving as CO2: C6H5CH2CH3 + 6[O] → C6H5COOH + CO2 + 2H2O. Phenylethanoic acid would keep both side-chain carbons, but the CH2 next to the ring is the one that is oxidised. Phenol would need the whole side chain replaced by OH. Ethylcyclohexane is a reduction (hydrogenation) product, not an oxidation product.
Total reactant mass = 78.5 + 94.0 = 172.5, so atom economy = 136.0 ÷ 172.5 × 100 = 78.8%. 100% ignores the HCl that is lost in this condensation. 21.2% is the percentage wasted as HCl. 65.1% wrongly adds the HCl (36.5) to the reactants as well.
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.