The shape of the lesson
By the end of the lesson, learners can
- Substitution, not addition: Look at the delocalised ring in the simulator: adding across one C=C, as ethene does with bromine, would destroy that delocalisation, so benzene substitutes instead and keeps the ring.
- The mechanism with curly arrows: Every electrophilic substitution has the same three steps.
- Where the next group goes: A group already on the ring changes both the rate and the position of a second substitution.
- Nitronium: HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−.
- Arrow 1: ring to electrophile; arrow 2: C–H into ring.
- 2,4-directing: –OH, –NH2, –CH3, –Cl.
- 3-directing: –NO2, –COOH; both deactivate.
One picture every idea returns to
In electrophilic substitution a strong electrophile is made first: for nitration, HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−. A curly arrow from the delocalised π system to the electrophile forms a new C–E bond, giving a positive intermediate (an arenium ion) in which the delocalisation is partly broken: four electrons over five carbons. A curly arrow from the C–H bond back into the ring then removes H+ and restores the aromatic ring. Benzene substitutes rather than adds because addition would destroy the stable delocalised system. Substituents already on the ring direct the next group: –OH, –NH2, –CH3 and –Cl send it to the 2- and 4-positions, while –NO2 and –COOH send it to the 3-position. Electron-donating groups activate the ring; electron-withdrawing groups (and halogens) deactivate it.
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, Break the ring's electrons, then restore them.. An electrophile pulls a pair of electrons out of the delocalised ring, a positive intermediate forms, and losing H + gives the ring back its full delocalisation. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Substitution, not addition | Look at the delocalised ring in the simulator: adding across one C=C, as ethene does with bromine, would destroy that delocalisation, so benzene substitutes instead and keeps the ring. Project tab 1 of the lab (Arenes lab, the delocalised ring) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The mechanism with curly arrows | Every electrophilic substitution has the same three steps. Project tab 2 of the lab (Arenes lab, electrophilic substitution) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Where the next group goes | A group already on the ring changes both the rate and the position of a second substitution. Project tab 5 of the lab (Arenes lab, directing effects) and let a student drive it while the class predicts. | Cooperative Concept Mapping |
| 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, 17 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 Where the next group goes, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| The first curly arrow starts from the delocalised ring (the circle), not from a C=C or a C–H bond, and ends at the electrophile's atom. | The intermediate's horseshoe must not include the sp 3 carbon, and the + goes inside it. The second arrow starts from the C–H bond, and the product is H +, not H 2 or H −. In the nitronium equation it is 2 H 2 SO 4. Halogens direct to 2 and 4 even though they deactivate the ring; –NO 2 and –COOH direct to 3. |
Electrophilic substitution: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Write equations to show how the electrophile is formed in (a) the nitration of benzene and (b) the bromination of benzene using FeBr3.
Describe, with the help of curly arrows in words, the mechanism of the nitration of benzene after the electrophile has formed.
Ethene reacts with bromine water at room temperature, but benzene does not. Benzene does react with pure bromine and FeBr3, by substitution. Explain these differences.
Predict the main organic product(s) of mononitration of (a) methylbenzene and (b) benzoic acid. State whether each reacts faster or slower than benzene.
Suggest a two-step synthesis of 1-bromo-3-nitrobenzene from benzene, giving reagents and conditions. Explain the order of the steps.
Chlorine is said to be deactivating but 2,4-directing. Explain what each part means.
Answer key and marking notes
- (a) HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−.
- (b) Br2 + FeBr3 → Br+ + FeBr4−.
- A curly arrow goes from the delocalised π electrons of the ring to the N of NO2+, forming a C–N bond.
- The intermediate is a positive ion: a horseshoe of four delocalised electrons over five carbons, with + inside; the attacked carbon holds H and NO2.
- A curly arrow goes from the C–H bond back into the ring.
- H+ is lost, the delocalised ring is restored, and H+ + HSO4− regenerates H2SO4.
- In ethene the π electrons are localised, so the electron density is high enough to polarise Br2.
- In benzene the π electrons are delocalised over six carbons, so the density is lower; FeBr3 is needed to make Br+.
- Benzene substitutes rather than adds because losing H+ restores the stable delocalised ring, whereas addition would destroy it.
- (a) 2-nitromethylbenzene and 4-nitromethylbenzene; faster, since –CH3 donates electrons and activates the ring.
- (b) 3-nitrobenzoic acid; slower, since –COOH withdraws electrons and deactivates the ring.
- The directing effect follows the group already present: alkyl is 2,4-directing, –COOH is 3-directing.
- Step 1: nitrate first with conc. HNO3 + conc. H2SO4, 25–60 °C, to give nitrobenzene.
- Step 2: Br2 with FeBr3 (halogen carrier), giving 1-bromo-3-nitrobenzene.
- –NO2 is 3-directing; brominating first would put –Br on, which is 2,4-directing, giving the wrong isomers.
- Deactivating: the electronegative Cl withdraws electron density from the ring, so chlorobenzene reacts more slowly than benzene.
- 2,4-directing: the second substituent goes mainly to the 2- and 4-positions, as it does for –OH, –NH2 and –CH3.
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.
What is the electrophile that attacks the benzene ring when benzene is warmed with a mixture of concentrated nitric and sulfuric acids?
- Athe nitronium ion, NO2+
- Bthe nitrate ion, NO3−
- Cthe nitric acid molecule, HNO3
- Dthe hydrogen sulfate ion, HSO4−
Which equation correctly shows the formation of the electrophile in the nitration of benzene?
- AHNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4−
- BHNO3 + H2SO4 → NO2+ + H3O+ + HSO4−
- CHNO3 → NO2+ + OH−
- DHNO3 + 2H2SO4 → NO3− + 2HSO4− + H3O+
The diagram shows the intermediate formed during the bromination of benzene. What happens in the next step of the mechanism?
- Athe C-H bonding pair moves into the ring and H+ is lost, restoring the delocalised ring
- BBr− adds to the positive carbon, giving an addition product
- Cthe C-Br bond breaks and Br+ is lost, re-forming benzene
- DH− is lost from the carbon holding the Br atom
Why does benzene react with bromine by substitution, whereas cyclohexene reacts by addition?
- Asubstitution keeps the delocalised π system, which makes the product much more stable than an addition product
- Bbenzene has no π electrons, so it cannot take part in addition
- Cthe C-H bonds in benzene are weaker than those in cyclohexene
- Dbenzene is a saturated hydrocarbon
Methylbenzene is nitrated with concentrated nitric and sulfuric acids at 30 °C. Which statement is correct?
- Athe main products are 2-nitromethylbenzene and 4-nitromethylbenzene, and the reaction is faster than for benzene
- Bthe main product is 3-nitromethylbenzene, and the reaction is slower than for benzene
- Cthe NO2 group replaces a hydrogen on the CH3 group
- Dthe main products are 2- and 4-nitromethylbenzene, and the reaction is slower than for benzene
Benzoic acid, shown with its ring positions numbered, is reacted with bromine and FeBr3. Which statement about the main product and the rate is correct?
- Abromine goes to position 3, and the reaction is slower than with benzene
- Bbromine goes to positions 2 and 4, and the reaction is slower than with benzene
- Cbromine goes to positions 2 and 4, and the reaction is faster than with benzene
- Dbromine goes to position 3, and the reaction is faster than with benzene
Which list puts these compounds in order of how readily they undergo electrophilic substitution, most reactive first?
- Aphenol > methylbenzene > benzene > nitrobenzene
- Bnitrobenzene > benzene > methylbenzene > phenol
- Cmethylbenzene > phenol > benzene > nitrobenzene
- Dbenzene > methylbenzene > phenol > nitrobenzene
A student wants to make 4-nitrobenzoic acid from methylbenzene in two steps. Which route works?
- Anitrate with conc. HNO3 and conc. H2SO4, separate the 4-isomer, then oxidise with hot alkaline KMnO4 and acidify
- Boxidise with hot alkaline KMnO4 and acidify, then nitrate with conc. HNO3 and conc. H2SO4
- Cnitrate with dilute HNO3 in UV light, then oxidise with acidified K2Cr2O7
- Dreact with Cl2 in UV light, then nitrate with conc. HNO3 and conc. H2SO4
Quiz key
Concentrated H2SO4 protonates HNO3, which then loses water to form NO2+, a strong electrophile that can accept a pair of π electrons from the ring. The nitrate ion is negative, so it is repelled by the electron-rich ring. HNO3 itself is not electrophilic enough to break into the delocalised system; that is why the sulfuric acid is needed. HSO4− is negative and acts as a base that removes H+ at the end, regenerating H2SO4.
Left: H 5, N 1, O 11, S 2, charge 0. Right: NO2+ + H3O+ + 2HSO4− has H 5, N 1, O 11, S 2, charge +1 +1 −2 = 0. One H2SO4 protonates HNO3; the second protonates the water that is lost. With only one H2SO4, hydrogen, oxygen, sulfur and charge do not balance. HNO3 does not split into NO2+ and OH− on its own. The last option has a total charge of −2 on the right and forms nitrate, which is not an electrophile.
A curly arrow is drawn from the C-H bond into the horseshoe; H+ leaves and the six-electron delocalised ring is restored, giving bromobenzene. The H+ reacts with FeBr4− to give HBr and regenerate the FeBr3 catalyst. Adding Br−, as in alkene addition, would leave a non-aromatic product and lose the delocalisation energy. Losing Br+ just reverses the first step and gives no product. The hydrogen leaves without its bonding pair, as H+, not H−: the electron pair stays behind to complete the ring.
Addition across one bond of the ring would destroy the delocalised system and lose about 152 kJ mol−1 of stabilisation. In substitution, loss of H+ from the intermediate restores the delocalised ring, so the product keeps that stability. Benzene does have six π electrons; they are delocalised rather than localised. The C-H bond strength is not the reason; in fact a C-H bond must break in substitution. Benzene is unsaturated (it can add 3H2), but it resists addition.
The CH3 group donates electron density into the ring, so the ring attracts NO2+ more strongly: methylbenzene is activated and nitrates faster, at a lower temperature, than benzene. Alkyl groups are 2,4-directing, so the products are 2- and 4-nitromethylbenzene. The 3-position is favoured by electron-withdrawing groups such as -NO2 and -COOH, which also slow the reaction. NO2+ is an electrophile that attacks the π system of the ring, not the side chain. Getting the positions right but calling the reaction slower confuses methyl with deactivating groups.
-COOH withdraws electron density from the ring, so the ring is deactivated and reacts more slowly than benzene; electron-withdrawing groups such as -COOH and -NO2 direct the incoming group to the 3-position, giving 3-bromobenzoic acid (position 5 is the same as 3). Positions 2 and 4 are favoured by -OH, -NH2, -CH3 and the halogens. A faster reaction would need an electron-donating, activating group.
In phenol a lone pair on O is delocalised into the ring, a strong donation: phenol reacts with bromine water at room temperature with no carrier. A methyl group donates less, by its inductive effect, so methylbenzene is activated but less so. Benzene is the reference. -NO2 strongly withdraws electron density, so nitrobenzene is the least reactive. Reversing the order treats electron-withdrawing groups as activating. Placing methylbenzene above phenol underestimates the lone-pair donation from oxygen. Putting benzene above both activated compounds ignores the effect of donating groups.
-CH3 is 2,4-directing, so nitrating methylbenzene first gives 4-nitromethylbenzene (with the 2-isomer, which is separated). Hot alkaline KMnO4 then oxidises the side chain to -COO−, and acid gives 4-nitrobenzoic acid. Oxidising first makes benzoic acid, whose -COOH is 3-directing, so nitration would give mainly 3-nitrobenzoic acid. Dilute acid in UV light does not make NO2+, and acidified dichromate does not oxidise an alkyl side chain on a benzene ring to -COOH under these conditions. Cl2 in UV light substitutes into the side chain, which does not produce a -COOH group.
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.