A2 32 Hydroxy compounds | A Level Chemistry 9701

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🎯What you need to be able to do

Describe the reaction of alcohols with acyl chlorides to form esters, using ethyl ethanoate.
Recall how phenol is made from phenylamine via a diazonium salt.
Recall the reactions of phenol with NaOH(aq), with Na(s), with diazonium salts in alkali, with dilute HNO₃, and with bromine water.
Explain the acidity of phenol, and the relative acidities of water, phenol and ethanol.
Explain why phenol nitrates and brominates under much milder conditions than benzene, and that the OH group directs to the 2-, 4- and 6-positions.
Apply the chemistry of phenol to other phenolic compounds such as naphthol.

📚The chemistry

32.1 Alcohols with acyl chlorides

At AS, esters were made from an alcohol and a carboxylic acid, a slow, reversible reaction needing a catalyst (topic 18). An acyl chloride does the same job far better:

\[ \mathrm{CH_3COCl + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + HCl} \]

Ethanoyl chloride and ethanol give ethyl ethanoate at room temperature, vigorously, with steamy fumes of HCl. The reaction is not reversible and needs no catalyst, so the yield is high. The mechanism is addition–elimination (topic 33).

32.2 Phenol

Phenol, C₆H₅OH, has an OH group bonded directly to a benzene ring. The ring and the OH change each other’s behaviour: the OH makes the ring far more reactive, and the ring makes the OH more acidic.

Making phenol

From phenylamine, in two stages:

Diazotisation: react phenylamine with nitrous acid, HNO₂, made in situ from NaNO₂ and dilute HCl, below 10 °C. This forms a diazonium salt, benzenediazonium chloride: \[ \mathrm{C_6H_5NH_2 + HNO_2 + HCl \rightarrow C_6H_5N_2^{+}Cl^{-} + 2H_2O} \] The low temperature is essential: the diazonium ion is unstable and decomposes above about 10 °C.
Warm the diazonium salt solution with water: nitrogen is given off and phenol forms. \[ \mathrm{C_6H_5N_2^{+} + H_2O \rightarrow C_6H_5OH + N_2 + H^{+}} \]

A flow chart. Phenylamine with sodium nitrite and dilute hydrochloric acid below 10 degrees Celsius gives benzenediazonium chloride, which decomposes above about 10 degrees. Warming the diazonium salt with water gives phenol and nitrogen gas. Adding phenol in aqueous sodium hydroxide to the diazonium salt instead gives a yellow-orange azo dye, coupled usually at the 4-position. — The diazonium salt is made cold, then either warmed with water to give phenol or coupled with a phenol to give an azo dye.

Phenol as an acid

With NaOH(aq): phenol dissolves, forming sodium phenoxide. \[ \mathrm{C_6H_5OH + NaOH \rightarrow C_6H_5O^{-}Na^{+} + H_2O} \]
With Na(s): fizzes, releasing hydrogen. \[ \mathrm{2C_6H_5OH + 2Na \rightarrow 2C_6H_5O^{-}Na^{+} + H_2} \]

Phenol is a weak acid — stronger than water and ethanol, but too weak to react with carbonates. That last point is a useful test: carboxylic acids fizz with sodium carbonate, phenols do not.

Why phenol is acidic: relative acidity

\[ \text{ethanol} \; < \; \text{water} \; < \; \text{phenol} \quad \text{(increasing acidity)} \]

The strength of an acid depends on how stable its anion is. Compare the three anions:

Phenoxide, C₆H₅O⁻: a lone pair on the oxygen overlaps with the delocalised π system of the ring, so the negative charge is spread over the ring. The charge is less concentrated on oxygen, the ion is more stable, and it has less tendency to recapture H⁺. So phenol dissociates more than water.
Hydroxide, OH⁻: the charge stays on the oxygen, with nothing to spread it.
Ethoxide, C₂H₅O⁻: the ethyl group is electron-donating, pushing extra electron density onto the oxygen and intensifying the charge. It is the least stable, so ethanol is the weakest acid (topic 16).

The more the negative charge is spread, the more stable the anion and the stronger the acid.

The ring reactions: much easier than benzene

The same lone-pair overlap works in the other direction too: it feeds electron density into the ring, which becomes much more attractive to electrophiles. The ring is activated, so phenol reacts under much milder conditions than benzene, and the OH group directs to the 2-, 4- and 6-positions.

Bromination: phenol decolourises bromine water at room temperature, with no catalyst, giving a white precipitate of 2,4,6-tribromophenol. The increased electron density in the ring is enough to polarise Br₂ without a halogen carrier, and all three activated positions react. \[ \mathrm{C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr} \] Benzene needs pure bromine and an AlBr₃ catalyst, and substitutes only once.
Nitration: phenol reacts with dilute nitric acid at room temperature, giving a mixture of 2-nitrophenol and 4-nitrophenol. \[ \mathrm{C_6H_5OH + HNO_3 \rightarrow HOC_6H_4NO_2 + H_2O} \] Benzene needs concentrated HNO₃ and concentrated H₂SO₄ at 25–60 °C.
Coupling with diazonium salts: in NaOH(aq), phenol (as phenoxide) reacts with a diazonium salt to form an azo compound, with the –N=N– group linking two rings, usually at the 4-position of the phenol: \[ \mathrm{C_6H_5N_2^{+} + C_6H_5OH \rightarrow C_6H_5N{=}NC_6H_4OH + H^{+}} \] The product is a brightly coloured (yellow-orange) azo dye; the extended delocalisation across both rings and the N=N link is what makes these compounds coloured.

Other phenols

Any compound with an OH on an aromatic ring behaves the same way. Naphthol (an OH on naphthalene, two fused benzene rings) dissolves in NaOH(aq) to form a naphthoxide, reacts with sodium, decolourises bromine water, and couples with diazonium salts in alkali to give intensely coloured azo dyes — 2-naphthol gives a red-orange dye, and is used as a test for diazonium ions.

✏️Worked example

0.940 g of phenol is dissolved in water, and bromine water is added until no more reacts. (a) Describe what you would see and write the equation. (b) Calculate the mass of organic product and the amount of Br₂ used. [M_r: phenol 94.0; C₆H₂Br₃OH 330.7] (c) Explain why this reaction happens without a catalyst when benzene needs one. (d) Explain whether phenol or ethanol would react with aqueous sodium hydroxide.

(a) The orange bromine water is decolourised immediately and a white precipitate forms (with an antiseptic smell):

\[ \mathrm{C_6H_5OH + 3Br_2 \rightarrow C_6H_2Br_3OH + 3HBr} \]

(b)

\[ n(\text{phenol}) = \frac{0.940}{94.0} = 0.0100\ \mathrm{mol} \qquad m(\text{product}) = 0.0100 \times 330.7 = 3.31\ \mathrm{g} \] \[ n(\mathrm{Br_2}) = 3 \times 0.0100 = 0.0300\ \mathrm{mol} \]

(c) A lone pair on the oxygen of the OH group overlaps with the delocalised π system, increasing the electron density of the ring, especially at the 2-, 4- and 6-positions. This electron-rich ring can polarise Br₂ molecules by itself and attack them, so no halogen carrier is needed to make Br⁺. Benzene’s ring has lower electron density and cannot.

(d) Phenol reacts, forming sodium phenoxide and water, because it is acidic enough: the phenoxide ion is stabilised by delocalising its negative charge into the ring. Ethanol does not react with NaOH(aq): it is a weaker acid than water, because its ethyl group intensifies the negative charge on the ethoxide ion.

Phenol, with its 2, 4 and 6 positions marked as activated by the OH group, reacts with three molecules of bromine in bromine water at room temperature, with no catalyst, to give 2,4,6-tribromophenol, a white precipitate, and three HBr. Amounts: 0.940 grams of phenol is 0.0100 moles, using 0.0300 moles of bromine and giving 3.31 grams of product. — The worked example’s numbers: 0.0100 mol of phenol uses 0.0300 mol of Br₂ and gives 3.31 g of product.

Check it. Three Br replace three H on the ring, and three HBr form, so both Br₂ and HBr must be 3 × the phenol. Mass balance: the product is heavier than the phenol by 3 × (79.9 − 1.0) = 236.7, and 94.0 + 236.7 = 330.7.

Giving monobromophenol, or the benzene conditions. Phenol with bromine water gives the trisubstituted product, because all three activated positions react. And answering “phenol needs AlBr₃” or “phenol nitrates with a nitrating mixture” misses the point of the question: the reagents are milder because the ring is activated.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Compare making ethyl ethanoate from ethanol using (a) ethanoic acid and (b) ethanoyl chloride.

(a) Ethanoic acid needs heat and a concentrated H₂SO₄ catalyst; the reaction is slow and reversible, so the yield is limited by equilibrium; by-product water. (b) Ethanoyl chloride reacts rapidly at room temperature without a catalyst; the reaction is not reversible, so the yield is high; by-product HCl (steamy fumes). CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl.

2. Give the reagents and conditions for converting phenylamine into phenol, and explain why the first stage is kept cold.

Stage 1: NaNO₂ and dilute HCl (making HNO₂) below 10 °C → benzenediazonium chloride. Stage 2: warm the solution → phenol + N₂. The first stage is kept below 10 °C because the diazonium ion is unstable: at higher temperatures it decomposes before it can be used (in stage 2 that decomposition is exactly what is wanted, which is why it is then warmed).

3. Arrange water, ethanol and phenol in order of acidity and explain the position of phenol.

Acidity: ethanol < water < phenol. When phenol loses H⁺, the phenoxide ion forms. A lone pair on its oxygen overlaps with the delocalised π system, so the negative charge is delocalised over the ring. This makes phenoxide more stable than hydroxide (whose charge is concentrated on one oxygen), so phenol loses H⁺ more readily than water.

4. Describe a chemical test to distinguish phenol from benzoic acid.

Add sodium carbonate (or hydrogencarbonate) solution. Benzoic acid, a carboxylic acid, gives effervescence of CO₂. Phenol is too weak an acid to react with carbonate: no fizzing. (Alternatively, bromine water: phenol decolourises it with a white precipitate; benzoic acid does not react.)

5. Name the products of nitrating phenol, give the conditions, and explain the difference from benzene.

2-nitrophenol and 4-nitrophenol, with dilute nitric acid at room temperature. Benzene needs concentrated nitric acid with concentrated sulfuric acid at 25–60 °C. The difference is because the OH group activates the ring: a lone pair on its oxygen feeds electron density into the π system, so the ring is attacked by electrophiles much more readily, at the 2- and 4-positions.

6. 2-naphthol is added to a solution of benzenediazonium chloride in alkali. Predict what you would see, and name the type of product.

A brightly coloured red-orange precipitate of an azo compound (azo dye) forms. The naphthol, like phenol, is deprotonated in alkali and its activated ring couples with the diazonium ion, forming an –N=N– link between the two aromatic systems. The extended delocalisation makes it strongly coloured. This reaction is used to detect diazonium ions.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Chemguide (Jim Clark) — the phenol pages, including its acidity and ring reactions
Royal Society of Chemistry — practical guide for making an azo dye