The shape of the lesson
By the end of the lesson, learners can
- Making and testing phenylamine: Phenylamine cannot be made from a halogenoarene and ammonia, because the C–Cl bond in chlorobenzene is too strong.
- Diazotisation and coupling: Two steps make an azo dye.
- Nitrobenzene: Sn + conc. HCl, reflux, then NaOH.
- Br2(aq): white 2,4,6-tribromophenylamine, no catalyst.
- HNO2: NaNO2 + HCl, below 10 °C.
- Couple with phenol in NaOH(aq): –N=N– dye.
One picture every idea returns to
Phenylamine, C6H5NH2, is made by reducing nitrobenzene with tin and concentrated HCl, heated under reflux, then adding NaOH to release the free amine from its salt. Its N lone pair is partly delocalised into the ring, which makes it a weak base but activates the ring: with aqueous bromine at room temperature it gives a white precipitate of 2,4,6-tribromophenylamine. With nitrous acid (NaNO2 + HCl) below 10 °C it forms benzenediazonium chloride, which couples with phenol in alkaline solution to give an azo dye containing –N=N–. The azo link joins two benzene rings into one extended delocalised system, which absorbs visible light, so azo compounds are used as dyes and indicators.
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, From nitrobenzene to colour.. Reduce nitrobenzene to phenylamine, watch bromine water give an instant white precipitate, then turn phenylamine into a diazonium salt in an ice bath and couple it with phenol to make a bright orange azo dye. 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 and testing phenylamine | Phenylamine cannot be made from a halogenoarene and ammonia, because the C–Cl bond in chlorobenzene is too strong. 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: Diazotisation and coupling | Two steps make an azo dye. Project tab 3 of the lab (Nitrogen compounds (A Level) lab, an azo dye) and let a student drive it while the class predicts. | Corners |
| 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, 15 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 Diazotisation and coupling, 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.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
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 reducing agent for nitrobenzene is tin and concentrated HCl, followed by NaOH; do not write NaBH 4 or LiAlH 4. | Diazotisation must be kept below 10 °C (but not frozen), otherwise phenol and N 2 form. Coupling with phenol needs alkaline conditions. Phenylamine with bromine water needs no catalyst and gives three Br atoms on the ring. |
Phenylamine and azo dyes: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Describe how phenylamine is made from nitrobenzene. Include reagents, conditions and an equation using [H].
Phenylamine reacts with bromine water at room temperature, but benzene does not. Explain, and give the equation.
Give the reagents and conditions, and an equation, for converting phenylamine into benzenediazonium chloride.
A student warms the benzenediazonium chloride solution to 30 °C. State what is observed and write an equation.
0.930 g of phenylamine (Mr 93.0) is converted into the azo dye C6H5N=NC6H4OH (Mr 198.0). 1.58 g of dye is collected. Calculate the percentage yield.
Explain why the azo dye is coloured and give one use of azo compounds.
Answer key and marking notes
- Heat nitrobenzene under reflux with tin and concentrated hydrochloric acid.
- C6H5NO2 + 6[H] → C6H5NH2 + 2H2O.
- Add excess NaOH(aq) to release phenylamine from its salt, C6H5NH3+Cl−.
- The N lone pair is partly delocalised into the ring, increasing its electron density (especially at positions 2, 4 and 6).
- The ring polarises Br2 and is attacked without a halogen carrier; benzene's ring is not electron-rich enough.
- C6H5NH2 + 3Br2 → C6H2Br3NH2 + 3HBr (white precipitate).
- NaNO2 and HCl(aq), which make nitrous acid in situ.
- Keep the temperature below 10 °C (ice bath).
- C6H5NH2 + HNO2 + HCl → C6H5N2+Cl− + 2H2O.
- Bubbles of nitrogen gas are seen as the diazonium ion decomposes.
- C6H5N2+ + H2O → C6H5OH + N2 + H+.
- n(phenylamine) = 0.930 ÷ 93.0 = 0.0100 mol, so maximum dye = 0.0100 × 198.0 = 1.98 g.
- Yield = 1.58 ÷ 1.98 × 100 = 79.8%.
- The –N=N– group joins two benzene rings into one extended delocalised system, so the gap between energy levels is small enough to absorb visible light.
- Uses: dyes for textiles or food colourings, and acid-base indicators such as methyl orange.
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 scheme shows the conversion of nitrobenzene into an orange dye Y. Which reagents are used in step 1?
- Atin and concentrated hydrochloric acid, heated under reflux, followed by NaOH(aq)
- Bconcentrated nitric and sulfuric acids at 50 °C
- CNaBH4 in water at room temperature
- Dexcess ammonia in ethanol in a sealed tube
Aqueous bromine is added to phenylamine at room temperature. What is observed and what forms?
- Athe bromine is decolourised and a white precipitate of 2,4,6-tribromophenylamine forms
- Bno change unless an FeBr3 catalyst is added
- Cthe bromine is decolourised and 4-bromophenylamine forms as a colourless solution, with no precipitate
- Dan orange precipitate of an azo dye forms
Phenylamine is converted into compound X, which then couples with phenol to give an orange dye. What are X and the conditions for making it?
- Abenzenediazonium chloride; NaNO2 and HCl(aq) below 10 °C
- Bbenzenediazonium chloride; NaNO2 and HCl(aq) heated under reflux
- Cnitrobenzene; concentrated HNO3 at 50 °C
- Dphenol; NaOH(aq) at room temperature
Benzenediazonium chloride is added to phenol dissolved in NaOH(aq) to make an azo dye. Why is the phenol dissolved in alkali?
- Ait forms the phenoxide ion, whose ring is more electron-rich and so reacts with the weakly electrophilic diazonium ion
- Balkali keeps the temperature below 10 °C
- Calkali reduces the diazonium ion to phenylamine
- Dthe azo dye is only coloured when it is ionised in acid
Why is the azo dye C6H5N=NC6H4OH coloured?
- Athe –N=N– group links two benzene rings into one extended delocalised system, which absorbs light in the visible region
- Bit contains a transition metal ion with partly filled d orbitals
- Cthe N=N bond is very weak and gives out light as it breaks
- Dthe benzene rings absorb only ultraviolet light, which we see as orange
12.3 g of nitrobenzene (Mr 123.0) is reduced and 6.51 g of phenylamine (Mr 93.0) is obtained. What is the percentage yield?
- A70.0%
- B52.9%
- C75.6%
- D143%
0.930 g of phenylamine (Mr 93.0) reacts with excess bromine water. What mass of 2,4,6-tribromophenylamine (Mr 329.7) forms?
- A3.30 g
- B1.10 g
- C9.89 g
- D0.930 g
A solution containing 0.0100 mol of benzenediazonium chloride is warmed to 40 °C and decomposes completely. What volume of gas at room conditions is released?
- A0.240 dm3 of nitrogen
- B0.480 dm3 of nitrogen
- C0.240 dm3 of nitrogen dioxide
- D0.120 dm3 of hydrogen chloride
Quiz key
Step 1 is a reduction: C6H5NO2 + 6[H] → C6H5NH2 + 2H2O. Tin and concentrated HCl supply the [H]; the product is present as C6H5NH3+, so NaOH(aq) is added to release the free amine. Concentrated HNO3/H2SO4 makes nitrobenzene from benzene; it is the step before. NaBH4 reduces C=O groups, not nitro groups. Ammonia substitutes halogenoalkanes, but there is no halogen here, and halogenoarenes do not react that way.
C6H5NH2 + 3Br2 → C6H2Br3NH2 + 3HBr. The N lone pair is partly delocalised into the ring, which makes the ring electron-rich at positions 2, 4 and 6, so substitution happens at all three without a catalyst and the insoluble product precipitates. A halogen carrier is needed for benzene, not phenylamine. Mono-substitution does not stop at one Br with excess bromine water. Azo dyes need a diazonium salt and a coupling agent, not bromine.
C6H5NH2 + HNO2 + HCl → C6H5N2+Cl− + 2H2O. This diazotisation must be kept below 10 °C because the diazonium ion decomposes when warm. Heating under reflux would destroy X, giving phenol and nitrogen. Re-forming nitrobenzene would undo step 1. NaOH(aq) does not convert phenylamine to phenol.
C6H5OH + OH− → C6H5O− + H2O. The O− donates electron density into the ring even more strongly than OH, so the diazonium ion (a weak electrophile) substitutes at the 4-position, forming C6H5N=NC6H4OH. Temperature is controlled by an ice bath, not by alkali. NaOH is not a reducing agent. The coupling is done in alkali, not acid, and the colour comes from delocalisation, not ionisation in acid.
Joining two rings through the azo group extends the delocalised π system across the whole molecule. This lowers the energy gap between electron energy levels so that visible light is absorbed, and the transmitted light gives the orange colour. That is why azo compounds are used as dyes and indicators such as methyl orange. There is no metal ion. The N=N bond does not break, and no light is emitted. Benzene alone absorbs only ultraviolet and is colourless; UV absorption cannot give a visible colour.
n(nitrobenzene) = 12.3 ÷ 123.0 = 0.100 mol, so the maximum is 0.100 × 93.0 = 9.30 g; 6.51 ÷ 9.30 × 100 = 70.0%. 52.9% compares the product mass with the reactant mass. 75.6% is 9.30 ÷ 12.3, the theoretical mass ratio, not the yield. 143% inverts the ratio and is impossible.
n(phenylamine) = 0.930 ÷ 93.0 = 0.0100 mol, giving 0.0100 mol product = 0.0100 × 329.7 = 3.30 g. 1.10 g divides by 3 because of the three Br atoms, but those come from Br2, not from extra phenylamine. 9.89 g multiplies by 3 instead. 0.930 g assumes the mass stays the same, ignoring the much larger Mr of the product.
C6H5N2+ + H2O → C6H5OH + N2 + H+. One mole of N2 per mole of diazonium ion: 0.0100 × 24.0 = 0.240 dm3. 0.480 dm3 counts the two N atoms as two moles of gas. No NO2 forms: the nitrogen leaves as N2, which is why the ion must be kept cold. HCl is not released; the chloride stays in solution.
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.