A Level Chemistry: Alcohols and Phenol — Revision Notes

Condensed for revision. For the full explanation, use the Phenol: Reactions and Acidity study guide, then test yourself with the practice questions. For the AS reactions of alcohols, see the AS Alcohols revision notes.

Syllabus: Cambridge International AS & A Level Chemistry 9701, 2025–2027, A Level content: subtopics 32.1 Alcohols and 32.2 Phenol.

Alcohols: esters from acyl chlorides (32.1)

An alcohol reacts with an acyl chloride at room temperature to give an ester and HCl. No catalyst or heating is needed.

CH₃COCl + CH₃CH₂OH → CH₃COOCH₂CH₃ + HCl
ethanoyl chloride + ethanol → ethyl ethanoate + hydrogen chloride

	Alcohol + acyl chloride	Alcohol + carboxylic acid
Conditions	room temperature	heat with concentrated H₂SO₄ catalyst
Reaction	vigorous, not reversible	slow, reversible (equilibrium)
By-product	HCl (steamy fumes)	H₂O
Yield	high	limited by the equilibrium

The mechanism is addition–elimination (nucleophilic O of the alcohol attacks the δ+ carbonyl carbon; Cl⁻ is lost). See the Carboxylic Acids and Acyl Chlorides revision notes.

Phenol reacts the same way to give a phenyl ester (phenyl benzoate is the syllabus example in 33.2):

C₆H₅COCl + C₆H₅OH → C₆H₅COOC₆H₅ + HCl

Producing phenol (32.2.1)

Step 1, diazotisation: phenylamine with HNO₂ (made from NaNO₂ and dilute HCl) below 10 °C gives benzenediazonium chloride.

NaNO₂ + HCl → HNO₂ + NaCl
C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O

Step 2: warm the diazonium salt solution with water. Nitrogen gas is given off and phenol forms.

C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl

Below 10 °C keeps the diazonium salt from decomposing while it is made; warming is then used on purpose to decompose it into phenol.

Reactions of phenol (32.2.2)

Reagent and conditions	Product	Observation / note
(a) NaOH(aq)	sodium phenoxide, C₆H₅O⁻Na⁺	phenol dissolves to give a colourless solution
(b) Na(s)	sodium phenoxide + H₂(g)	effervescence
(c) diazonium salt in NaOH(aq), below 10 °C	azo compound	yellow/orange colour (often a precipitate)
(d) dilute HNO₃(aq), room temperature	2-nitrophenol + 4-nitrophenol	no H₂SO₄ needed
(e) Br₂(aq), room temperature	2,4,6-tribromophenol	orange bromine water decolourised; white precipitate

Equations:

(a)  C₆H₅OH + NaOH → C₆H₅ONa + H₂O
(b)  2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂
(c)  C₆H₅N₂⁺ + C₆H₅O⁻ → C₆H₅N=NC₆H₄OH
(d)  C₆H₅OH + HNO₃ → HOC₆H₄NO₂ + H₂O          (2- and 4-isomers)
(e)  C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr

Azo coupling (c): the diazonium ion is the electrophile; it substitutes at the 4-position of the phenoxide ion. The product, 4-hydroxyazobenzene (IUPAC: 4-(phenyldiazenyl)phenol), contains the –N=N– (azo) link between two rings. The extended delocalised system absorbs visible light, so azo compounds are coloured and used as dyes.

Phenol gives no CO₂ with carbonates: it is too weak an acid to release CO₂ from Na₂CO₃ or NaHCO₃, so there is no effervescence. This distinguishes phenol from carboxylic acids.

Acidity of phenol (32.2.3 and 32.2.4)

Phenol is a weak acid; it partly dissociates in water:

C₆H₅OH ⇌ C₆H₅O⁻ + H⁺

Why it is acidic: in the phenoxide ion, a lone pair (the negative charge) on O is in a p orbital that overlaps with the π system of the ring. The negative charge is delocalised over the ring, so the phenoxide ion is stabilised. The equilibrium shifts to the right more than for water or ethanol.

Order of acidity: phenol > water > ethanol

Acid	Anion	Effect on the anion	Acid strength
phenol, C₆H₅OH	phenoxide, C₆H₅O⁻	charge delocalised into the ring: stabilised	strongest of the three
water, H₂O	hydroxide, OH⁻	no delocalisation, no alkyl group	intermediate
ethanol, C₂H₅OH	ethoxide, C₂H₅O⁻	ethyl group is electron-donating; it intensifies the negative charge on O: destabilised	weakest

The more stable the anion, the less readily it accepts H⁺ back, so the stronger the acid.

Evidence: all three react with sodium to give hydrogen, but only phenol (not ethanol) reacts with NaOH(aq).

2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂
C₂H₅OH + NaOH → no significant reaction

Ring reactivity and directing effect (32.2.5 and 32.2.6)

Why phenol reacts under milder conditions than benzene: a lone pair on the O of –OH overlaps with the π system and is partly delocalised into the ring. This increases the electron density of the ring, so it attracts electrophiles more strongly and is activated.

Reaction	Benzene	Phenol
Nitration	concentrated HNO₃ + concentrated H₂SO₄, 25–60 °C	dilute HNO₃(aq), room temperature
Bromination	Br₂ with AlBr₃ catalyst (halogen carrier)	Br₂(aq), room temperature, no catalyst
Degree of substitution	one Br	three Br (2,4,6)

The electron-rich ring can polarise Br₂ by itself, so no halogen carrier is needed, and a weaker electrophile source (dilute HNO₃) is enough for nitration.

Directing effect: the –OH group directs incoming groups to the 2-, 4- and 6-positions.

Other phenolic compounds (32.2.7)

Any compound with –OH bonded directly to an aromatic ring reacts like phenol. Examples:

• Naphthol, C₁₀H₇OH dissolves in NaOH(aq) and couples with a diazonium salt in alkaline solution to give an intensely coloured azo dye.

  C₁₀H₇OH + NaOH → C₁₀H₇ONa + H₂O

• 4-methylphenol with Br₂(aq): the 4-position is already taken, so only the 2- and 6-positions are brominated. The product is 2,6-dibromo-4-methylphenol.

  CH₃C₆H₄OH + 2Br₂ → CH₃C₆H₂Br₂OH + 2HBr

Count the free 2-, 4- and 6-positions to predict how many mol of Br₂ react.

Exam traps

• Diazotisation: below 10 °C. Making phenol from the diazonium salt: warm with water.
• Nitration of phenol uses dilute HNO₃ at room temperature and gives a mixture of 2- and 4-nitrophenol; do not add concentrated H₂SO₄.
• Bromine water gives 2,4,6-tribromophenol: three Br, 3HBr, white precipitate. No catalyst.
• Acidity order is phenol > water > ethanol. Explain with the stability of the anion, not the strength of the O–H bond.
• Ethanol’s alkyl group destabilises ethoxide; it does not “stabilise the ion”.
• Phenol reacts with NaOH but gives no effervescence (no CO₂) with Na₂CO₃ or NaHCO₃; ethanol reacts with neither NaOH nor carbonates.
• Alcohol + acyl chloride: room temperature, no catalyst, by-product HCl (not water).

Self-test

1. Write the equation for the reaction of propan-1-ol with ethanoyl chloride and name the ester formed.
2. Give two advantages of using an acyl chloride rather than a carboxylic acid to make an ester.
3. State the reagents and conditions to convert phenylamine into phenol in two steps.
4. Write the equation for phenol reacting with sodium.
5. Describe what is seen when bromine water is added to phenol, and write the equation.
6. Explain why phenol is more acidic than ethanol.
7. Why does the nitration of phenol not need concentrated sulfuric acid?
8. Predict the organic product when 2-naphthol is added to benzenediazonium chloride in NaOH(aq), and state why it is coloured.
9. How many moles of Br₂ react with one mole of 2-methylphenol? Name the product.
10. 0.940 g of phenol reacts completely with excess bromine water. Calculate the mass of 2,4,6-tribromophenol formed. (Ar: H = 1.0, C = 12.0, O = 16.0, Br = 79.9)

Answers:

1. CH₃COCl + CH₃CH₂CH₂OH → CH₃COOCH₂CH₂CH₃ + HCl; propyl ethanoate.
2. Any two: reacts at room temperature; no catalyst; not reversible, so higher yield; faster.
3. NaNO₂ and dilute HCl (HNO₂) below 10 °C to form the diazonium salt; then warm with water.
4. 2C₆H₅OH + 2Na → 2C₆H₅ONa + H₂.
5. Orange bromine water is decolourised and a white precipitate forms; C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr.
6. In phenoxide the negative charge on O is delocalised into the ring’s π system, stabilising the ion; in ethoxide the electron-donating ethyl group intensifies the charge on O, destabilising it. So phenol dissociates more.
7. A lone pair on O is delocalised into the ring, increasing its electron density; the activated ring reacts with dilute HNO₃ at room temperature.
8. An azo compound (azo dye) with an –N=N– link joining the benzene ring and the naphthol ring. The extended delocalised system absorbs visible light.
9. Two: the 4- and 6-positions are free (position 2 holds CH₃). Product: 2,4-dibromo-6-methylphenol (the same compound as “4,6-dibromo-2-methylphenol” in the question’s numbering; IUPAC gives bromo the lower locants because b comes before m).
10. Mr(C₆H₅OH) = 6(12.0) + 6(1.0) + 16.0 = 94.0, so n = 0.940 / 94.0 = 0.0100 mol. Mr(C₆H₂Br₃OH) = 6(12.0) + 3(1.0) + 3(79.9) + 16.0 = 330.7. 1 : 1 ratio, so mass = 0.0100 × 330.7 = 3.31 g.

These are original notes written for revision. Check the full syllabus wording in the official 9701 syllabus.