The shape of the lesson
By the end of the lesson, learners can
- Making phenol, and phenol as an acid: Compare the acids in the simulator, then use the table to explain where phenol sits and why it fizzes with sodium but not with carbonate.
- An activated ring: Compare each reaction of phenol with the matching reaction of benzene.
- Making it: NaNO2/HCl below 10 °C, then warm with water.
- Acid: reacts with Na and NaOH, not with carbonates.
- Br2(aq): white 2,4,6-tribromophenol, no catalyst.
- Esters: acyl chloride only; azo dyes with diazonium salts.
One picture every idea returns to
Phenol, C6H5OH, is made by diazotising phenylamine with NaNO2 and HCl below 10 °C, then warming the diazonium salt with water, which releases N2. A lone pair on the O overlaps with the delocalised ring. This spreads the charge of the phenoxide ion into the ring, so phenol (pKa 10.0) is a weak acid: it reacts with Na and NaOH but is too weak to release CO2 from carbonates. The same overlap raises the electron density of the ring, activating it at the 2-, 4- and 6-positions: phenol decolourises bromine water at room temperature to give a white precipitate of 2,4,6-tribromophenol, is nitrated by dilute HNO3, and couples with diazonium salts to give azo dyes. Phenol forms esters with acyl chlorides, not with carboxylic acids.
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, An –OH that activates the ring.. Phenol is made from phenylamine through a diazonium salt. 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 phenol, and phenol as an acid | Compare the acids in the simulator, then use the table to explain where phenol sits and why it fizzes with sodium but not with carbonate. Project tab 1 of the lab (Alcohols and phenol lab, ranking the O-H acids) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: An activated ring | Compare each reaction of phenol with the matching reaction of benzene. Project tab 5 of the lab (Alcohols and phenol lab, phenol's activated ring) and let a student drive it while the class predicts. | Mix-Pair-Share |
| 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 An activated ring, 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.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
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 |
|---|---|
| Phenol does not react with carbonates: it is too weak an acid to release CO 2, which is how it is told apart from a carboxylic acid. | Bromine water gives 2,4,6-tri bromophenol (three Br, 3HBr) with no halogen carrier. Nitration of phenol uses dilute HNO 3 and gives a mixture of 2- and 4-nitrophenol. Esters of phenol are made with an acyl chloride, never by heating with a carboxylic acid. |
Phenol: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Describe how phenol is made from phenylamine, giving reagents, conditions and equations.
Explain why phenol is more acidic than ethanol.
Calculate the pH of a 0.100 mol dm−3 solution of phenol (pKa 10.0).
Compare the bromination of phenol with that of benzene: reagents, conditions, product and reason for the difference.
4.70 g of phenol reacts with excess ethanoyl chloride and 5.44 g of phenyl ethanoate (Mr 136.0) is collected. Calculate the percentage yield.
Benzenediazonium chloride is added to an alkaline solution of phenol below 10 °C. Name the type of reaction, the type of product and give its structural formula.
Answer key and marking notes
- React phenylamine with NaNO2 and HCl(aq) below 10 °C: C6H5NH2 + HNO2 + HCl → C6H5N2+Cl− + 2H2O.
- Warm the benzenediazonium chloride solution with water.
- C6H5N2+ + H2O → C6H5OH + N2 + H+.
- A lone pair on the O of the phenoxide ion overlaps with the delocalised π electrons of the ring.
- So the negative charge is spread over the ring and the phenoxide ion is stabilised.
- In ethoxide the charge stays on O and is intensified by the electron-donating ethyl group, so ethanol ionises far less.
- Ka = 1.00 × 10−10 mol dm−3; [H+] = √(1.00 × 10−10 × 0.100) = 3.16 × 10−6 mol dm−3.
- pH = −log(3.16 × 10−6) = 5.50.
- Phenol: bromine water at room temperature, no catalyst; white precipitate of 2,4,6-tribromophenol (+ 3HBr).
- Benzene: liquid Br2 with a FeBr3 (or AlBr3) halogen carrier; bromobenzene only.
- The O lone pair in phenol increases the electron density of the ring (especially at 2, 4 and 6), so it can polarise Br2 itself.
- n(phenol) = 4.70 ÷ 94.0 = 0.0500 mol, so the maximum is 0.0500 × 136.0 = 6.80 g.
- Yield = 5.44 ÷ 6.80 × 100 = 80.0%.
- A coupling reaction (electrophilic substitution at the 4-position of phenol), forming an azo dye.
- 4-hydroxyazobenzene, C6H5N=NC6H4OH, a yellow-orange solid.
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.
Which reagents and conditions convert phenylamine into phenol?
- ANaNO2 and HCl(aq) below 10 °C, then warm the solution with water
- BNaOH(aq), heat under reflux
- Cconcentrated HNO3 and concentrated H2SO4 at 55 °C
- DNaNO2 and HCl(aq) below 10 °C, then add alkaline phenol
Which reagent gives a visible reaction with phenol but NOT with ethanol?
- Abromine water
- Bsodium metal
- Cethanoyl chloride
- Dsodium carbonate solution
What is the organic product when phenol reacts with excess bromine water?
- A2,4,6-tribromophenol
- B4-bromophenol only
- Cbromobenzene
- D1,2-dibromocyclohexanol
In the diagram of phenol, the ring carbons are numbered from the C bearing –OH. At which positions does electrophilic substitution mainly occur?
- A2, 4 and 6
- B3 and 5
- C1 only
- D4 only
Which test distinguishes phenol from ethanoic acid?
- Aadd sodium carbonate solution: only ethanoic acid gives a gas
- Badd sodium metal: only phenol gives a gas
- Cadd sodium hydroxide solution: only ethanoic acid reacts
- Dadd universal indicator: only phenol turns it red
1.88 g of phenol (Mr 94.0) reacts completely with bromine water. What mass of 2,4,6-tribromophenol (Mr 330.7) forms, and how many moles of Br2 react?
- A6.61 g; 0.0600 mol
- B6.61 g; 0.0200 mol
- C2.20 g; 0.0200 mol
- D19.8 g; 0.0600 mol
4.65 g of phenylamine (Mr 93.0) is diazotised and the solution warmed until no more gas is released. What is the maximum volume of nitrogen at room conditions (24.0 dm3 mol−1)?
- A1.20 dm3
- B2.40 dm3
- C0.600 dm3
- D1.12 dm3
Phenol reacts with dilute nitric acid at room temperature, but benzene needs concentrated nitric and sulfuric acids at 55 °C. Which explanation is correct?
- Aa lone pair on the O overlaps with the ring, raising its electron density so it attracts electrophiles more strongly
- Bthe O–H bond of phenol breaks easily, releasing H+ which catalyses nitration
- Cphenol is a stronger acid than nitric acid
- Dthe –OH group withdraws electrons from the ring, making it more reactive
Quiz key
Nitrous acid made in situ turns phenylamine into C6H5N2+Cl− below 10 °C; warming it with water gives phenol and N2. Hot NaOH does not replace an –NH2 on a ring. Nitrating mixture nitrates the ring and does not make phenol. Adding the diazonium salt to alkaline phenol makes an azo dye, not phenol.
Bromine water is decolourised by phenol at room temperature, with a white precipitate of 2,4,6-tribromophenol, because the ring is activated; ethanol has no ring and does not react. Both phenol and ethanol fizz with sodium, and both form esters with ethanoyl chloride (steamy HCl fumes). Neither is acidic enough to release CO2 from sodium carbonate.
The ring is so activated at the 2-, 4- and 6-positions that all three are substituted: C6H5OH + 3Br2 → C6H2Br3OH + 3HBr. A single substitution happens only under milder conditions, not with excess bromine water. Bromine does not replace the –OH group, so bromobenzene does not form. The ring keeps its delocalised system: this is substitution, not addition.
The O lone pair increases electron density most at the 2-, 4- and 6-positions, so –OH is 2,4-directing (6 is equivalent to 2). Positions 3 and 5 are where electron-withdrawing groups like –NO2 direct. Position 1 already carries the OH. Position 4 alone is wrong: the 2-position also reacts, which is why nitration gives both 2- and 4-nitrophenol.
Ethanoic acid is strong enough to release CO2 from carbonate; phenol is not. Both react with sodium to give H2. Both are neutralised by NaOH, which is why phenol dissolves in it. Phenol is the weaker acid, so if anything ethanoic acid turns indicator more red.
n(phenol) = 0.0200 mol, giving 0.0200 × 330.7 = 6.61 g of product. Each ring takes 3Br2 (one Br onto the ring, one into HBr), so 0.0600 mol Br2. 0.0200 mol Br2 treats it as a 1 : 1 reaction. 2.20 g divides the mass by three. 19.8 g multiplies by three, as if three separate products formed.
n(phenylamine) = 4.65 ÷ 93.0 = 0.0500 mol, giving 0.0500 mol C6H5N2+ and so 0.0500 mol N2 = 0.0500 × 24.0 = 1.20 dm3. 2.40 dm3 doubles the moles, as if N2 contained one N per molecule. 0.600 dm3 halves them. 1.12 dm3 uses 22.4 dm3 mol−1, the volume at s.t.p., not room conditions.
Delocalising an O lone pair into the π system makes the ring electron-rich (especially at 2, 4 and 6), so it is attacked by weaker electrophiles under milder conditions; the products are 2- and 4-nitrophenol. Ionising the O–H bond is not the cause: the phenoxide route is not how nitration works, and nitric acid already supplies H+. Phenol (pKa 10.0) is a far weaker acid than nitric acid. An electron-withdrawing group would deactivate the ring, not activate it.
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.