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Notes/Chemistry/Paper 4/Hydroxy Compounds
CAIEA2 Level9701§32

Hydroxy Compounds

The one group that is a weak acid, a nucleophile towards acyl chlorides and the strongest ring activator on the course — from diazonium routes to azo dyes and drug-molecule phenols.

120 min read 7 sub-topics
55
question parts
2021–2025 · 21 papers
6 marks
per paper
≈ 6% of the paper
2.1/3
avg difficulty
moderate
#13
most examined
of 15 topics by marks

Phenol is the busiest small molecule on Paper 4. Over 2021–2025 it anchored 55 examined parts worth 119 marks across 21 papers, and nearly every part leans on the same opening move: find where the −OH-\text{OH} sits. On a saturated carbon it is an ordinary alcohol — barely acidic, sluggish towards electrophiles. Directly ON the benzene ring it becomes a weak acid strong enough to dissolve in NaOH\text{NaOH}, a nucleophile that attacks acyl chlorides, and the strongest ring activator on the course — one lone pair, borrowed by the ring, doing all three jobs at once. The route through: §01 what the direct ring attachment changes, §02 esters from acyl chlorides — ethanol's reaction extended to phenol, §03 producing phenol from a diazonium salt, §04 the acidity ladder phenol > water > ethanol and the NaOH\text{NaOH}/Na\text{Na} reactions that prove it, §05 nitration and bromination under mild conditions with the 2,4,6-directing rule, §06 azo coupling — phenol meets a diazonium salt, and §07 running the whole playbook on real molecules: tyrosine, capsaicin and salbutamol.

Before you start you should be able to

  • Benzene's delocalised π system, the electrophilic-substitution mechanism and directing effects — including the mild-condition contrast with methylbenzene (this subject's A2 Hydrocarbons note, §§01–02 and §07)

  • Alcohols: the O–H group's weak acidity, reaction with sodium, and ester formation (this subject's AS Hydroxy Compounds note)

  • Acyl chlorides: the electron-deficient carbonyl carbon and addition–elimination reactions (this subject's A2 Carboxylic Acids note)

  • Lone-pair delocalisation into an aromatic ring strengthening a bond — met for the aryl C–X bond (this subject's A2 Halogen Compounds note, §03)

  • Relative acid strength as position of equilibrium: the stronger the acid, the more stable its conjugate base (AS Chemical Equilibria)

By the end of this page you can

  • Explain what makes phenol different from an alcohol — the oxygen lone pair delocalised into the ring π system — and state its consequences: a weakened O–H bond with a stabilised phenoxide ion, a strengthened C–O bond, and an activated ring donating density to the 2-, 4- and 6-positions

  • Describe the reaction of acyl chlorides to form esters using ethyl ethanoate as the example, extend it to phenyl ethanoate from ethanoyl chloride and phenol, and write both equations with the HCl co-product

  • Recall the production of phenol: phenylamine with HNO₂ or NaNO₂ and dilute acid below 10 °C to give the diazonium salt, then warming the salt with water

  • Explain the acidity of phenol, rank water, ethanol and phenol (and place carboxylic acids above them), and write the NaOH(aq) and Na(s) equations — knowing that Na reacts with all of them but NaOH rejects alcohols and Na₂CO₃ rejects phenol

  • State the mild conditions for nitrating (dilute HNO₃(aq), room temperature) and brominating (Br₂(aq)) phenol, explain WHY they are milder than benzene's using the delocalisation chain, and predict products: the 2-/4-nitrophenol mixture and 2,4,6-tribromophenol with its decolourisation-plus-white-precipitate observations

  • Give the reagents and conditions for coupling a diazonium salt with phenol in NaOH(aq) at 0–10 °C, and draw azo dye products joined through −N=N− para to the −OH

  • Apply all of the above to substituted phenols such as naphthol, tyrosine, capsaicin and salbutamol — counting free 2/4/6 positions and letting excess Br₂(aq) hit every one of them

01

What makes phenol special

Syllabus requirement · §32

“

explain the acidity of phenol · describe and explain the relative acidities of water, phenol and ethanol · explain why the reagents and conditions for the nitration and bromination of phenol are different from those for benzene — every one of these rests on the single structural idea built in this section (syllabus 32.2, learning outcomes 3, 4 and 5)

”

One group, three personalities

Ethanol is mildly annoying to oxidise and barely acidic. Phenol — same −OH-\text{OH} tag, different address — dissolves in NaOH\text{NaOH}(aq) as freely as a carboxylic acid's cousin, hands electrons into its own benzene ring so generously that bromine water needs no catalyst, and holds onto its ring carbon so tightly that its C–O bond behaves more like a double bond. Three behaviours, one structural difference: in phenol the oxygen is welded directly to a ring carbon.

This section builds that difference properly, because §§04–06 are just its consequences being cashed in. If you can explain the overlap to yourself now, the rest of the topic turns into bookkeeping.

First, the address test: phenol versus alcohol

Both families carry −OH-\text{OH}. The classifier is WHERE the oxygen sits:

  • Alcohol (alkanol): the O is bonded to an sp³ carbon — ethanol CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}, propan-1-ol, phenylmethanol C6H5CH2OH\text{C}_6\text{H}_5\text{CH}_2\text{OH} (note the −CH2−-\text{CH}_2- spacer: the ring is one whole bond away and plays no part). The oxygen's lone pairs are localised on the oxygen and belong to nobody else.
  • Phenol: the O is bonded directly to a flat sp² ring carbon of the delocalised π system — phenol C6H5OH\text{C}_6\text{H}_5\text{OH}, the cresols (methylphenols), naphthol. One of oxygen's lone pairs occupies a p orbital aligned sideways-on with the ring's π cloud, and orbitals that overlap SHARE electrons.

That last sentence is the entire topic. Everything else in this note is what follows.

phenol — one oxygen lone pair rewrites the whole chemistryTHE OVERLAP — O lone pair meets the π cloudOHthe O lone pair delocalises INTO the π system(a p orbital on O overlaps the ring's p orbitals)drawn as the two gold lobes feeding the dashed cloudring ACTIVATEDπ density up — especially at C-2, C-4, C-6aryl C–O bond STRONGERpartial double-bond character — hard to breakO–H more polarised than in alcoholsH⁺ lost more readily ⇒ phenol is ACIDIC (§04)phenol's acidity (§04), its activated ring (§05) and its dye chemistry (§06)all trace back to this one lone-pair overlap

Two panels compared. Left: ethanol — the O–H group on an sp³ carbon; both oxygen lone pairs stay localised, nothing is shared with anything. Right: phenol — the O–H on a ring carbon; a p orbital holding one oxygen lone pair overlaps sideways with the ring's delocalised π system (shaded overlap region between O and the ring). Three consequence chips hang off the phenol panel: ring activated (density donated in, greatest at the 2-, 4- and 6-positions) · O–H weakened, phenoxide stabilised (acidity) · C–O bond strengthened.

The donation, and its three consequences

When the oxygen lone pair delocalises into the ring, electron density leaves the oxygen's neighbourhood and arrives in the ring — unevenly. Because of how the overlapping orbitals combine, the donated density appears greatest at the 2-, 4- and 6-positions (two ortho sites, one para site relative to the −OH-\text{OH}). Follow the electrons and three consequences drop out:

  1. The O–H bond weakens and the phenoxide ion stabilises — acidity. Electron density pulled away from oxygen leaves the H more exposed as δ+\delta+ and the O–H bond easier to break; and IF the proton leaves, the leftover negative charge on oxygen can spread into the ring (to those same 2,4,6 positions) instead of sitting alone. A stabilised conjugate base means the ionisation equilibrium sits further right than water's. This is §§04's whole job.
  2. The ring activates and directs to 2,4,6. Extra electron density makes the ring greedier towards electrophiles — phenol polarises Br2\text{Br}_2 or generates conditions for NO2+\text{NO}_2^+ far more gently than benzene can — and the density is thickest exactly at the 2-, 4- and 6-positions, so attack lands there. §§05–06 cash this in.
  3. The C–O bond strengthens. Sharing lone-pair density between O and the ring carbon gives the C–O linkage partial double-bond character — shorter, stronger, unwilling to break. Exactly the argument you met for chlorine in chlorobenzene (A2 Halogen Compounds note, §03): phenol resists cleavage of its C–O bond for the same reason chlorobenzene resists cleavage of its C–Cl bond.
Explaining ANY phenol behaviour, in three moves
  1. 1

    Name the cause: a lone pair (p orbital) on the oxygen overlaps with / is delocalised into the ring's π system.

    This exact clause opens virtually every published mark scheme answer for this topic — acidity, activation, directing, all of them.

  2. 2

    Pick the consequence the question asks for. Acidity questions want: O–H bond weakened OR the phenoxide/conjugate-base anion stabilised. Reactivity questions want: electron density in the ring increased (especially at 2,4,6) → electrophiles attracted/polarised more easily.

    Quoting the right half of the ledger matters: 'stabilised anion' earns nothing in a bromination question, and 'more reactive ring' earns nothing in an acidity question.

  3. 3

    Land the observable: the equation, the product positions, or the conditions change that follows.

    Examiners pay the final mark for the concrete outcome — never stop at 'so it is more reactive'; say what happens, where, and with which reagent.

Worked demo — sort the −OH families before any theory

For each compound, classify the −OH-\text{OH} group and state ONE chemical consequence of your classification:

A=C6H5OHB=C6H5CH2OHC=CH3CH(OH)CH3\text{A} = \text{C}_6\text{H}_5\text{OH} \qquad \text{B} = \text{C}_6\text{H}_5\text{CH}_2\text{OH} \qquad \text{C} = \text{CH}_3\text{CH(OH)CH}_3

Show full working
  1. 1

    A: the O is bonded directly to a ring carbon → phenol. Consequence: the O lone pair feeds the π system, so A is acidic enough to react with NaOH\text{NaOH}(aq).

    No spacer atoms: the subscript pattern C₆H₅–OH means O-on-ring. This is the compound the rest of the note is about.

  2. 2

    B: the O sits on the −CH2−-\text{CH}_2- carbon OUTSIDE the ring → alcohol (phenylmethanol). Consequence: localised lone pairs, so B is barely acidic — no reaction with NaOH\text{NaOH}(aq), only with sodium metal.

    The classic trap: a benzene ring in the picture does not make something a phenol. Count the atoms between O and ring — here there is a full CH₂ in between.

  3. 3

    C: O on an sp³ carbon again → alcohol (propan-2-ol). Consequence: same family as B — the electron-donating alkyl groups actually make its O–H the LEAST acidic of the trio.

    Preview of §04's ladder: alkyl groups push electrons TOWARDS oxygen, the opposite of the ring's pull.

Answer

A phenol — reacts with NaOH(aq) · B, C alcohols — no reaction with NaOH(aq), Na(s) only

Address first, behaviour second: trace the bond from O to its carbon before predicting anything.

Your turn

Classification drills — invented, so the reflex is yours before the past-paper questions arrive.

  1. 1

    Sort these four oxygen-containing aromatics into phenols and non-phenols, and name the non-phenol that would be MOST easily mistaken for a phenol:

    (a) 4-methylphenol (b) C6H5COOH\text{C}_6\text{H}_5\text{COOH} (c) C6H5CH(OH)CH3\text{C}_6\text{H}_5\text{CH(OH)CH}_3 (d) naphthalen-2-ol (naphthol, two fused rings, O on a ring carbon)

    Stuck? Show hint

    Ask one question of each structure: is THIS oxygen bonded to a ring carbon?

    Show solution
    1. 1

      (a) Phenol — the −OH is on a ring carbon (the methyl rides elsewhere on the ring).

      Substituents like −CH₃ do not change which atom the oxygen is bonded to; (a) is a substituted phenol.

    2. 2

      (b) Not a phenol — the O–H belongs to a carboxylic acid group, −COOH; neither of its oxygens is bonded to a ring carbon.

      Benzoic acid contains no C(ring)–O bond at all. Its acidity is a §33 story — one notch ABOVE phenol on the ladder.

    3. 3

      (c) Not a phenol — the O sits on the side-chain sp³ carbon; the ring is one bond removed.

      This is the B-style decoy: aromatic molecule, ordinary alcohol behaviour.

    4. 4

      (d) Phenol — the O is bonded directly to a ring carbon of the fused system; naphthol behaves chemically like phenol (syllabus 32.2 asks you to apply phenol's chemistry to it).

      'Ring carbon' counts in any aromatic system, fused rings included — §06 exercises a naphthol coupling.

    Answer

    Phenols: (a) and (d) · non-phenols: (b) carboxylic acid, (c) side-chain alcohol — (c) is the great imposter

  2. 2

    Complete each explanation frame with the missing clause:

    (i) The O–H bond in phenol is weaker than in ethanol because …

    (ii) Bromination of phenol proceeds without a catalyst because …

    (iii) The C–O bond in phenol is stronger than in cyclohexanol because …

    Stuck? Show hint

    All three start from the same five-word cause. What differs is WHICH consequence you cash in.

    Show solution
    1. 1

      (i) … the oxygen lone pair is delocalised into the ring, pulling electron density away from the O–H bond (and any resulting phenoxide negative charge is stabilised in the ring).

      Either half — weakened bond OR stabilised anion — is creditable; quoting both is safer.

    2. 2

      (ii) … the donated lone pair raises the ring's π electron density (most at 2,4,6), so the ring can polarise Br2\text{Br}_2 to Br+\text{Br}^+-character WITHOUT an AlBr3\text{AlBr}_3 carrier.

      The catalyst's job for benzene is exactly this polarisation — phenol's own electrons do it for free.

    3. 3

      (iii) … the shared lone-pair density gives the C–O bond partial double-bond character, making it shorter and stronger than cyclohexanol's pure single bond.

      Same architecture as the aryl C–X argument in the Halogen Compounds note — reuse it deliberately.

    Answer

    (i) LP delocalised into ring → O–H weakened/anion stabilised · (ii) higher ring π density polarises Br₂ unaided · (iii) partial double-bond character in C–O

Practise spotting phenol units and their consequences in unfamiliar moleculesReal past-paper questions · Phenol: preparation and reactions

The rest of this note

Checking your access…

Can you do all of these?

  • I can look at any structure and say within seconds whether an −OH is a phenol (on the ring) or an alcohol (on an sp³ carbon), and predict differently because of it

  • I can write the equations for phenol with NaOH(aq), with Na(s) and with ethanoyl chloride, naming every organic product and co-product

  • I can reproduce the examiner's acidity chain word-for-word: lone pair delocalises into ring → O–H weakened OR phenoxide stabilised; and the ethanol counter-leg: positive inductive alkyl push

  • I can complete the nine-box tick table: Na reacts with acid, phenol AND alcohol; NaOH(aq) with acid and phenol only; Na₂CO₃(aq) with acid only

  • I can give reagents and conditions for making phenol from phenylamine — cold (≤10 °C) diazotisation, then warming the salt with water

  • I can run the three-mark activation chain: lone pair overlaps ring → greater π electron density at 2,4,6 → polarises Br₂/NO₂⁺ without a catalyst

  • I can state the mild conditions for nitrating and brominating phenol, contrast them with benzene's, and justify the difference

  • I can draw 2-nitrophenol, 4-nitrophenol and 2,4,6-tribromophenol, and give BOTH observations for the bromine-water reaction

  • I can couple a diazonium salt with phenol in cold NaOH(aq) and draw the azo dye with the −N=N− link para to the −OH (or at the 2-position if 4 is blocked)

  • I can apply the playbook to unfamiliar phenols: find the −OH, count free 2/4/6 sites, let excess Br₂(aq) take them ALL, and remember an alkene side chain adds Br₂ too

Now do the questions
55 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes