JAM Alcohols, Phenols and Ethers — The Reaction Summary
These three functional groups share one oxygen atom and almost nothing else in exam behaviour — an alcohol oxidises, a phenol couples with diazonium salts and reacts without a catalyst towards bromine, and an ether mostly just sits there until you attack it with concentrated HX. This is a compressed, revision-style pass through the reactions that distinguish the three, with the acidity discussion (already covered in depth in the companion resonance article) referenced rather than repeated.
Alcohols — the oxidation ladder and the Lucas test
| Alcohol class | Mild oxidant (PCC) | Strong oxidant (hot KMnO₄ / K₂Cr₂O₇, H⁺) |
|---|---|---|
| 1°, R–CH₂OH | Stops at the aldehyde | Goes on to the carboxylic acid |
| 2°, R₂CHOH | Gives the ketone | Gives the same ketone; resists further oxidation under normal conditions |
| 3°, R₃COH | No reaction | Resists oxidation — there is no hydrogen on the carbinol carbon to remove |
This oxidation-level difference is also the basis of the Lucas test: conc. HCl + anhydrous ZnCl₂ (the Lucas reagent) reacts with alcohols via an SN1-type pathway, so reactivity tracks carbocation stability. A 3° alcohol gives immediate turbidity (a separate alkyl chloride layer) at room temperature; a 2° alcohol takes several minutes, often needing gentle warming; a 1° alcohol shows no visible reaction at room temperature at all.
Worked example 1 — identifying an isomer by the Lucas test. Three isomeric C₄H₁₀O alcohols — butan-1-ol, butan-2-ol and 2-methylpropan-2-ol — are each shaken with Lucas reagent. Which shows turbidity within a minute at room temperature?
2-methylpropan-2-ol (a tertiary alcohol) — its carbocation forms fastest via SN1, so it reacts essentially instantly. Butan-2-ol (secondary) would take several minutes; butan-1-ol (primary) would show no reaction without heating.
Other standard alcohol reactions, briefly: Fischer esterification (acid-catalysed, reversible, driven to completion with excess alcohol or by removing water); acid-catalysed dehydration to the Zaitsev alkene, easiest for 3° alcohols and hardest for 1°, tracking the same carbocation stability that governs the Lucas test.
Phenols — reactions that alcohols simply cannot do
Phenol's acidity and the resonance reasoning behind it are covered fully in the companion article on hybridisation and resonance. What phenol's electron-rich ring adds, beyond simple acidity, is a set of reactions unavailable to an ordinary alcohol:
- Kolbe-Schmitt reaction — sodium phenoxide, treated with CO₂ under pressure and moderate heat, undergoes electrophilic attack by CO₂ at the ring's ortho position (favoured by intramolecular hydrogen bonding in the product), giving salicylic acid after acidification.
- Reimer-Tiemann reaction — phenol with CHCl₃ and NaOH generates dichlorocarbene (:CCl₂) in situ, which attacks the phenoxide ring; hydrolysis of the resulting intermediate gives salicylaldehyde (ortho-hydroxybenzaldehyde).
- Bromination with no catalyst — phenol decolourises bromine water instantly, giving a white precipitate of 2,4,6-tribromophenol, with no Lewis-acid catalyst needed at all. An ordinary benzene ring needs FeBr₃ for bromination to proceed; phenol's ring is activated enough to react on its own, and multiple substitution happens because each new bromine only slightly reduces the ring's very high reactivity.
- Azo coupling — phenol reacts with a diazonium salt (see the companion article on amines) at the position para to –OH, giving a coloured azo dye.
- Ferric chloride test — neutral FeCl₃ solution gives phenols a characteristic violet, blue or green colour via an iron-phenolate complex; simple alcohols give no such colour. This is the standard quick test for distinguishing a phenol from an alcohol in a spot test.
Industrially, phenol is made mainly by the cumene process: cumene (isopropylbenzene) is air-oxidised to cumene hydroperoxide, which then undergoes an acid-catalysed rearrangement to give phenol and acetone together — a route that co-produces a commercially useful second product, which is part of why it is the dominant industrial method.
Ethers — Williamson synthesis and the cleavage rules
This only works cleanly when the alkyl halide is the less hindered of the two possible partners. Using a bulky alkyl halide with a strongly basic alkoxide invites E2 elimination instead of substitution.
Worked example 2 — a Williamson synthesis that fails, and the fix. To make tert-butyl methyl ether, a student reacts sodium methoxide (CH₃O⁻Na⁺) with 2-bromo-2-methylpropane (tert-butyl bromide). Explain why this is a poor choice, and give a route that works.
Methoxide is a small, strongly basic nucleophile facing a bulky tertiary substrate — exactly the conditions that favour E2 elimination over SN2 substitution. The major product is 2-methylpropene, not the ether.
Correct approach: swap which partner supplies the alkyl halide. Sodium tert-butoxide + iodomethane (CH₃I) reacts by SN2 at the unhindered methyl carbon, giving tert-butyl methyl ether cleanly — the bulky group stays as the alkoxide (a spectator to the SN2 step) rather than as the site of attack.
Ether cleavage with excess concentrated HI or HBr depends on which alkyl groups are present. If both are 1°/2°, the reaction proceeds by SN2: iodide attacks the less hindered carbon, giving that alkyl iodide and the alcohol from the other side; with excess HI and heat, that alcohol is also converted to its alkyl iodide. If a 3° group is present, cleavage instead proceeds by SN1 at that carbon, since the more stable carbocation forms preferentially. An aryl-alkyl ether behaves differently again — the Ar–O bond has partial double-bond character from resonance and does not cleave; only the alkyl-oxygen bond breaks, giving the phenol (which then survives unchanged) plus the alkyl halide.
Worked example 3 — ether cleavage stoichiometry. 5.40 g of anisole (methyl phenyl ether, C₇H₈O, M = 108.14 g/mol) is heated with excess HI. Find the mass of methyl iodide (CH₃I, M = 141.94 g/mol) produced, assuming complete reaction.
Moles of anisole = 5.40 ÷ 108.14 = 0.0499 mol.
Only the alkyl-oxygen bond cleaves (phenol survives), so 1 mol anisole gives 1 mol CH₃I.
Mass of CH₃I = 0.0499 × 141.94 = 7.09 g.
Common mistakes
- Using PCC and expecting a carboxylic acid from a 1° alcohol. PCC is a mild oxidant that stops at the aldehyde; only a strong oxidant under acidic, heated conditions goes all the way to the acid.
- Choosing the bulky alkyl halide for a Williamson synthesis. Always make the sterically hindered group the alkoxide, and let the small alkyl halide be the SN2 electrophile.
- Assuming an aryl ether cleaves at the aryl-oxygen bond. It does not — resonance keeps that bond intact, and only the alkyl side is attacked.
- Forgetting phenol needs no Lewis-acid catalyst for bromination. The ring is already activated enough on its own, unlike benzene.
- Confusing the ferric chloride test's purpose. It distinguishes phenols from alcohols; it does not distinguish 1°/2°/3° alcohols from each other — that is what the Lucas test is for.
Exam relevance
| Question style | What to check first |
|---|---|
| Predict oxidation product of an alcohol | Its class (1°/2°/3°) and whether the oxidant is mild (PCC) or strong |
| Identify an alcohol from a distinguishing test | Lucas test speed for alcohols; ferric chloride colour for phenol vs alcohol |
| Predict the product of a phenol-specific named reaction | Kolbe-Schmitt (→ salicylic acid) vs Reimer-Tiemann (→ salicylaldehyde) |
| Plan a Williamson synthesis | Which partner is more hindered — that one must be the alkoxide, not the halide |
| Predict ether cleavage products | Whether any alkyl group is 3° (SN1) and whether an aryl-oxygen bond is present (does not cleave) |
Verify molar masses before a cleavage or yield calculation. The molar mass tool accepts any formula and shows the element-wise breakdown, useful for checking a multi-step alcohol/phenol/ether stoichiometry problem quickly.
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