Cambridge International AS & A Level Chemistry 9701 · A Level topic 32

Hydroxy compounds (A Level)

What this chapter covers32.1–32.2

An –OH group behaves very differently depending on what it is attached to. On an alkyl chain it makes an alcohol — neutral in water, oxidised by dichromate, dehydrated by hot acid. On a benzene ring it makes a phenol — a weak acid that dissolves in sodium hydroxide, and a ring so reactive that it decolourises bromine water instantly and gives a white precipitate. Phenols are the basis of antiseptics, of many drugs, and, through their coupling with diazonium salts, of the azo dyes.

This chapter completes the chemistry of hydroxy compounds begun at AS Level in topic 16. It adds one new reaction of alcohols — the rapid formation of esters with acyl chlorides — and then develops the chemistry of phenol: how it is made, why it is acidic, and why its ring reacts with electrophiles under far milder conditions than benzene. One idea runs through the whole of the phenol section: a lone pair on the oxygen atom overlaps with the delocalised π system of the ring.

Esters from alcohols and acyl chlorides32.1.1

At AS Level, esters were made by heating an alcohol with a carboxylic acid and a few drops of concentrated sulfuric acid. That reaction is slow and reversible; it reaches an equilibrium in which a large fraction of the reactants remains unreacted. Acyl chlorides, RCOCl, react with alcohols much more readily. Ethanoyl chloride added to ethanol reacts vigorously at room temperature, the mixture warms, and steamy fumes of hydrogen chloride are given off:

The product is ethyl ethanoate. The reaction goes essentially to completion, because the HCl escapes and nothing drives the reaction backwards; no catalyst and no heating are needed.

The reaction is an addition–elimination. The alcohol oxygen uses a lone pair to attack the strongly δ+ carbonyl carbon of the acyl chloride; the C=O then re-forms and a chloride ion is lost, followed by the H⁺ from the oxygen, giving HCl. The full mechanism is developed with the other reactions of acyl chlorides in the next chapter.

Past-paper practice · Set 32A · Alcohols with acyl chlorides

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 32A.1[3]

Tartaric acid is 2,3-dihydroxybutanedioic acid, HO₂CCH(OH)CH(OH)CO₂H.

Answer and marking guidance

With an excess of LiAlH₄: both –CO₂H groups are reduced to –CH₂OH, giving HOCH₂CH(OH)CH(OH)CH₂OH ✔. With an excess of CH₃COCl: both secondary –OH groups are converted to ethanoate esters, –OCOCH₃, giving HO₂CCH(OCOCH₃)CH(OCOCH₃)CO₂H ✔. Types of reaction: reduction; condensation ✔ (both needed). Examiner insight: marks were low. CH(OH)₂CH(OH)₂ was a common wrong product with LiAlH₄, and "electrophilic substitution" a common wrong type for the acyl chloride reaction. Many esters were drawn ambiguously — they could have been methyl esters (wrong) or ethanoate esters (right); showing the ester linkage fully displayed avoids this.

Phenol: structure and properties32.2

Phenol, C₆H₅OH, is a colourless crystalline solid at room temperature (it often turns pink on standing, as a trace is oxidised). Its –OH group forms hydrogen bonds, so phenol melts well above benzene and is moderately soluble in water; the rest of the molecule is a non-polar ring, which limits that solubility. Phenol is toxic and corrosive to skin.

The key to its chemistry is the bond between the ring and the oxygen atom. The oxygen has two lone pairs. One of them occupies a p-type orbital parallel to the p orbitals of the ring carbons, and it overlaps with the delocalised π system (Figure 32.1). The lone pair is partly delocalised into the ring. This one fact has three consequences, each examined:

• the C–O bond gains partial double-bond character and is stronger — phenol does not undergo the substitution reactions of alcohols in which the C–O bond breaks (as with chlorobenzene in chapter 31);
• electron density is drawn away from the O–H bond and the phenoxide ion formed when phenol loses H⁺ is stabilised — phenol is a weak acid;
• the ring becomes more electron-rich than benzene — phenol reacts with electrophiles far more readily, under milder conditions, at the 2-, 4- and 6-positions.

Making phenol from phenylamine32.2.1

Phenol is made in the laboratory in two stages from phenylamine, which itself comes from benzene by nitration and reduction (chapters 30 and 34).

Stage 1: diazotisation, below 10 °C

Phenylamine is dissolved in dilute hydrochloric acid and cooled in ice; a cold solution of sodium nitrite, NaNO₂, is added, keeping the temperature below 10 °C. Sodium nitrite and hydrochloric acid produce nitrous acid, HNO₂, in the mixture:

Nitrous acid converts phenylamine into the benzenediazonium ion, C₆H₅N₂⁺, present as benzenediazonium chloride:

The diazonium ion contains the group –N⁺≡N, with the positive charge on the nitrogen atom bonded to the ring. It is stable enough to use only when cold: above about 10 °C it decomposes, losing nitrogen gas.

Stage 2: warming with water

That decomposition is exactly what is wanted in the second stage. When the solution of the diazonium salt is warmed, nitrogen gas bubbles off and the –N₂⁺ group is replaced by –OH from water:

Why phenol is acidic32.2.3, 32.2.4

Phenol dissociates slightly in water, releasing H⁺ ions:

It is a weak acid — far weaker than a carboxylic acid — but it is much stronger than ethanol or water. The explanation lies in the stability of the ion that is left behind. The stronger the acid, the more stable (lower in energy) its conjugate base.

The phenoxide ion

When phenol loses H⁺, the negative charge left on the oxygen does not stay there. A lone pair on the O⁻ overlaps with the ring's π system, and the negative charge is delocalised over the oxygen and the ring — in particular onto the carbons at the 2-, 4- and 6-positions. Spreading the charge over several atoms stabilises the phenoxide ion, so the equilibrium lies further to the right than for water or ethanol. Delocalisation also weakens the O–H bond in phenol itself, because electron density is drawn from the O–H bond towards the ring.

Water, ethanol and phenol compared

• Water → OH⁻: the charge is on one oxygen atom with nothing to spread it or to intensify it.
• Ethanol → CH₃CH₂O⁻: the ethyl group releases electron density towards the oxygen (positive inductive effect). This increases the negative charge density on the oxygen of the ethoxide ion, destabilising it, so ethanol is a weaker acid than water.
• Phenol → C₆H₅O⁻: the negative charge is delocalised into the ring and spread out; the ion is stabilised and phenol is the strongest acid of the three.

Order of acid strength: phenol > water > ethanol.

Phenol as an acid: sodium hydroxide and sodium32.2.2(a)(b)

With sodium hydroxide

Phenol is only slightly soluble in water, but it dissolves readily in aqueous sodium hydroxide, forming a colourless solution of sodium phenoxide:

Ethanol does not react with NaOH(aq) in this way: it is too weak an acid. The reaction is the simplest test that separates a phenol from an alcohol. Adding a strong acid to the sodium phenoxide solution re-forms phenol, which separates out again.

With sodium metal

Like alcohols and water, phenol reacts with sodium, giving hydrogen gas and sodium phenoxide. Phenol is a solid, so it is melted (or dissolved in an inert solvent) first:

This is a redox reaction: sodium is oxidised to Na⁺, and hydrogen in the O–H group is reduced to H₂. The gas released is H₂, not H⁺.

Past-paper practice · Set 32B · Making phenol, and its acidity

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 32B.1[6]

Parts (b) and (c) of the same question are 32C.1.

Answer and marking guidance

(a) M1: HNO₂, or NaNO₂ + HCl ✔; M2: then warm (T ≥ 10 °C) with water ✔. (d) C₆H₅OH + NaOH → C₆H₅ONa + H₂O ✔ (any equation with phenol acting as an acid). (e) phenol > water > ethanol ✔; two of: the lone pair on oxygen in phenol is delocalised into the ring; the ethyl group in ethanol has a positive inductive (electron-donating) effect; a correct statement about stabilisation of the anion or weakening of the O–H bond; a correct statement about ease of donating H⁺ ✔✔ (two statements per mark). Examiner insight: (a) was found difficult — candidates needed the reagents for the diazonium salt and to recall that it decomposes on warming in water. (d) was answered well. (e) discriminated well: a clear statement was needed for phenol and for ethanol on how and why the anion is stabilised or destabilised (or the O–H bond weakened or strengthened), related to the ease of proton donation.

Question 32B.2[6]

Answer and marking guidance

(i) step 1: concentrated HNO₃ + concentrated H₂SO₄, about 50–55 °C ✔; step 2: Sn + concentrated HCl, heat ✔; step 4: warm with water ✔. (ii) K is the benzenediazonium ion / chloride, C₆H₅N₂⁺Cl⁻ ✔. (iii) electrophilic substitution ✔. (iv) C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O ✔. Examiner insight: (i) was answered well by many; common errors were heating under reflux in step 1, omitting "concentrated" in step 1 or 2, and omitting a temperature or heat in step 4. (iii) was usually fully credited. (iv) was often wrong — a common error was C₆H₅NO₂ + 2[H] → C₆H₅NH₂ + O₂.

Bromination of phenol32.2.2(e), 32.2.5, 32.2.6

When bromine water is added to an aqueous solution of phenol at room temperature, the orange colour disappears at once and a white precipitate forms, smelling of antiseptic. The precipitate is 2,4,6-tribromophenol: three bromine atoms have substituted into the ring, at both positions next to the –OH group and at the position opposite it.

Compare benzene. It does not react with bromine water at all; it needs pure bromine and a halogen carrier such as AlBr₃, and even then only one bromine atom substitutes. Three differences in conditions, and in outcome, all come from the oxygen lone pair:

• No catalyst. Delocalisation of an oxygen lone pair into the ring raises the electron density of the ring. The ring is electron-rich enough to polarise an approaching Br₂ molecule itself, inducing a dipole Br^δ+–Br^δ−; the δ+ end is then attacked by the ring. No halogen carrier is needed to generate Br⁺.
• Aqueous bromine at room temperature. The ring is so much more reactive than benzene that a dilute solution reacts instantly.
• Three substitutions, at 2, 4 and 6. The extra electron density is greatest at the carbons next to and opposite the –OH group, so these are the positions attacked, and each one is substituted.

Nitration of phenol32.2.2(d), 32.2.5

Benzene is nitrated by a mixture of concentrated nitric and sulfuric acids between 25 °C and 60 °C, because only the nitronium ion is a strong enough electrophile to attack its ring. Phenol's ring is so electron-rich that dilute nitric acid at room temperature is enough, and no sulfuric acid is needed. The products are a mixture of 2-nitrophenol and 4-nitrophenol, formed by substitution at the positions activated by –OH:

With concentrated nitric acid the reaction is much more vigorous and further substitution occurs. Using dilute acid limits the reaction to one nitro group.

Coupling with diazonium salts: azo dyes32.2.2(c)

A benzenediazonium ion is a weak electrophile — too weak to attack benzene — but it attacks the very electron-rich ring of the phenoxide ion. When a cold solution of benzenediazonium chloride is added to phenol dissolved in aqueous sodium hydroxide, a yellow-orange precipitate forms at once. It is an azo compound, 4-hydroxyazobenzene (4-(phenylazo)phenol), in which two benzene rings are joined by an azo group, –N=N–:

The reaction is an electrophilic substitution of the phenol ring. The diazonium ion attacks through its terminal nitrogen atom, at the 4-position of the phenol (the 2-position is used if the 4-position is already occupied). Both nitrogen atoms are kept — the product contains –N=N–, not –N≡N. The alkaline conditions convert phenol to phenoxide, whose ring is even more electron-rich, and the mixture is kept cold so that the diazonium salt does not decompose.

The azo group joins two aromatic rings into one large delocalised system, which absorbs visible light, so azo compounds are strongly coloured and are widely used as dyes. Changing the phenol or amine used gives different colours: other azo dyes are made by the same route. The chapter on nitrogen compounds returns to diazonium salts and azo dyes.

Summary of phenol's reactions32.2.2, 32.2.6

Other phenolic compounds: naphthols32.2.7

Any compound with an –OH group bonded directly to an aromatic ring behaves as a phenol, and the syllabus expects you to apply phenol's reactions to unfamiliar examples. The naphthols are the standard example. Naphthalene, C₁₀H₈, consists of two benzene rings fused along one edge; 2-naphthol has an –OH group on one of the rings.

• 2-Naphthol dissolves in NaOH(aq) to form the sodium salt, and reacts with Na to give H₂: it is a weak acid, like phenol.
• Its ring reacts with electrophiles without a catalyst, at positions activated by the –OH group.
• In alkaline solution it couples with benzenediazonium chloride to give a bright orange-red azo dye — the classic laboratory test for a diazonium salt.

The same approach works for any phenolic compound in a question: find the –OH on the ring, then apply acidity, reaction with Na and NaOH, ring substitution at positions 2, 4 and 6 relative to the –OH (if they are free), and coupling in alkaline solution. Other groups already on the ring keep their own chemistry.

Past-paper practice · Set 32C · Reactions of phenol and other phenolic compounds

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 32C.1[5]

Answer and marking guidance

(b) 2-nitrophenol and 4-nitrophenol, each drawn and named ✔✔. (c)(i) 2,4,6-tribromophenol, drawn and named ✔✔. (ii) bromine is decolourised AND a white precipitate forms ✔. Examiner insight: (b) was usually answered well. In (c)(i) the structure was usually right but "tri" was often left out of the name. The observations in (c)(ii) were not well known: effervescence and white fumes were common errors.

Question 32C.2[10]

Answer and marking guidance

(a)(i) M1 curly arrow from inside the hexagon to the N of NO₂⁺ ✔; M2 correct intermediate ✔; M3 second arrow from C–H into the ring AND H⁺ ✔. (ii) HSO₄⁻ + H⁺ → H₂SO₄ ✔. (b) benzoic acid → 3-nitrobenzoic acid ✔; phenol → 2-nitrophenol or 4-nitrophenol ✔. (c) M1 order: phenol > benzene > benzoic acid ✔; for phenol: a lone pair / p orbital on O is delocalised into the ring, and the ring attracts / polarises NO₂⁺ better; for benzoic acid: –COOH / C=O is electron-withdrawing — two such points for one mark, three for two ✔✔; M4 the OH increases (or COOH decreases) the electron density in the ring compared with benzene ✔. Examiner insight: the mechanism errors were the usual ones (first arrow not from inside the ring; partial ring through the sp³ carbon; second arrow from the H; no H⁺). The directing effects of –COOH and –OH were well known, though a minority replaced the original group with NO₂. In (c) most had the right order but many explained it in terms of donating protons or weakening O–H bonds, and some wrote "charge density" for electron density.

Question 32C.3[5]

Part (d)(iii) of this question, comparing the acidity of phenol with a carboxylic acid, is in the next chapter.

Answer and marking guidance

(d)(i) C₆H₅OH + Na → C₆H₅O⁻Na⁺ + ½H₂ ✔. (ii) N is 2,4,6-tribromophenol ✔. (e) a p orbital / lone pair on oxygen overlaps with the ring π system (is delocalised into the ring) ✔; the electron density of the ring increases ✔; so it attracts / polarises the electrophile better ✔. Examiner insight: in (d)(i) many formed H⁺ instead of H₂; (ii) was usually correct. (e) proved difficult: answers lacked precision, for example saying the OH group activates the ring without explaining how — through overlap of the oxygen lone pair with the π system increasing the electron density.

Question 32C.4[7]

Answer and marking guidance

(a) M1 a p orbital / lone pair on O overlaps with (is delocalised into) the ring ✔; M2 greater electron density in the ring, especially at the 2-, 4- and 6-positions ✔; M3 so it polarises electrophiles more easily ✔. (c)(i) C₆H₅OH + Na → C₆H₅O⁻Na⁺ + ½H₂ ✔. (ii) R = 2,4,6-tribromophenol ✔. (iii) reaction 1: dilute (aqueous) HNO₃ ✔; reaction 2: benzenediazonium chloride (C₆H₅N₂⁺Cl⁻) in alkaline solution (NaOH) ✔. Examiner insight: (a) was done well — most recognised the delocalised oxygen lone pair; the effect on electron density was sometimes absent or described as "charge density". In (c)(i) many found the equation difficult, most often forming H⁺ instead of H₂. In (c)(iii) errors were concentrated HNO₃ or adding H₂SO₄ for reaction 1, and leaving out the diazonium salt for reaction 2.

Question 32C.5[6]

Compound R is 4-chloro-3,5-dimethylphenol; its name is question 29C.1 in chapter 29.

Answer and marking guidance

Na: the sodium phenoxide (O⁻Na⁺ in place of OH), redox ✔; CH₃COCl: the ethanoate ester, –OCOCH₃ in place of –OH, condensation ✔; Br₂(aq): bromine at both positions next to the OH (2,6-dibromo; the 4-position is occupied by Cl), (electrophilic) substitution ✔; benzenediazonium chloride: –N=N–C₆H₅ at a position next to the OH, (electrophilic) substitution ✔ — each structure [1] × 4; types of reaction ✔✔. Examiner insight: most had the products with Na and Br₂; the others were less well known. Errors: Friedel–Crafts ring substitution with CH₃COCl; replacing the ring OH with Br; replacing the OH or Cl with –N=N–C₆H₅; keeping N≡N in the azo product. Common wrong reaction types were "acid–base" for Na and "diazotisation" for the coupling.

Question 32C.6[5]

Answer and marking guidance

(a) 1: C₆H₅O⁻Na⁺ ✔; 2: C₆H₅O⁻Na⁺ ✔; 3: the azo compound C₆H₅N=NC₆H₄OH (4-position; –O⁻ also accepted in the alkaline solution) ✔; 4: with an excess of phenol, both acyl chloride groups form phenyl esters: C₆H₅OCOC₆H₄COOC₆H₅ ✔. (b) reactions 2 and 4 (ethanol also reacts with Na and with acyl chlorides; it does not react with NaOH(aq) or couple with diazonium ions) ✔. Examiner insight: (a) was generally answered correctly. Errors were wrong charges on the salts (C₆H₅O⁺Na⁻), losing the –OH from the azo dye, and drawing part of a polymer chain in row 4 instead of the molecule containing two phenol residues. (b) was scored by only a small proportion.

Misconceptions and how the topic is assessed32.1–32.2

Self-test32.1–32.2

Ten questions on the whole chapter. Each gives its reason once you answer.

Definitions to learn32.1–32.2

Summary

• Alcohols react with acyl chlorides at room temperature to give esters and HCl: CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl. The reaction is fast and goes to completion.
• Phenol is made from phenylamine: NaNO₂ and dilute HCl below 10 °C give benzenediazonium chloride; warming with water gives phenol and N₂.
• An oxygen lone pair in phenol is delocalised into the ring π system. This strengthens C–O, weakens O–H and increases the electron density of the ring, especially at the 2-, 4- and 6-positions.
• Phenol is a weak acid: the phenoxide ion is stabilised by delocalisation of its charge. Acid strength: phenol > water > ethanol.
• Phenol reacts with NaOH(aq) and with Na to give sodium phenoxide; it does not release CO₂ from carbonates.
• Bromine water: decolourised, white precipitate of 2,4,6-tribromophenol; no catalyst. Dilute HNO₃, room temperature: 2- and 4-nitrophenol.
• In NaOH(aq) below 10 °C, benzenediazonium ions couple with phenol at the 4-position to give a coloured azo compound.
• The same reactions apply to other phenols, such as the naphthols; ring substitution occurs only at free activated positions.

Examination checklist

• Can I write the equation for an alcohol with an acyl chloride, name the ester, and give the advantages over using a carboxylic acid?
• Can I give both stages, with conditions, for making phenol from phenylamine, and a route from benzene?
• Can I place water, ethanol and phenol in order of acidity and explain each position?
• Can I write equations for phenol with NaOH and with Na, with correct charges and H₂?
• Can I give the reagents, conditions, observations and products for bromination and nitration of phenol, and explain why they differ from benzene?
• Can I draw the azo compound from phenol and a diazonium salt, with –N=N– and the –OH kept?
• Can I predict the products of these reactions for an unfamiliar phenol?

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