Class 12 Alcohols, Phenols and Ethers — The Key Reactions
Three families share one feature — an oxygen atom joined to carbon — and yet they behave completely differently. An alcohol is neutral enough to drink in dilute form; phenol turns blue litmus red; an ether is so unreactive it is used as a solvent for reactions that would destroy the other two. This chapter is really a study of why that oxygen behaves differently in each case. Get that idea, and the reaction list stops being memory work.
The three structures side by side
Phenol: Ar–O–H (–OH directly on an aromatic ring)
Ether: R–O–R′ (no O–H at all)
- R — alkyl group; Ar — aryl (benzene ring).
- Alcohols are classified 1°, 2° or 3° by how many carbon atoms are attached to the carbon carrying the –OH. This single label decides oxidation products, dehydration rate and the Lucas test result.
- An ether has no O–H, so it cannot hydrogen bond to itself. That is why diethyl ether boils at a much lower temperature than butan-1-ol even though their molar masses are almost the same (74.12 g/mol for both).
How each one is made
| Target | Route | Point to remember |
|---|---|---|
| Alcohol | Alkene + H₂O, acid catalysed | Markovnikov addition — OH goes to the more substituted carbon |
| Alcohol | Alkene + B₂H₆, then H₂O₂/OH⁻ | Hydroboration–oxidation: anti-Markovnikov, and syn addition |
| Alcohol | Aldehyde or ketone + NaBH₄ or LiAlH₄ | Aldehyde → 1° alcohol; ketone → 2° alcohol |
| Alcohol | Grignard reagent + carbonyl, then H₃O⁺ | HCHO → 1°, other aldehydes → 2°, ketones → 3° |
| Phenol | Chlorobenzene + NaOH, 623 K, 300 atm | Dow process — drastic conditions are the point |
| Phenol | Benzenesulphonic acid fused with NaOH | Then acidify |
| Phenol | Benzenediazonium salt + warm water | Cleanest laboratory route |
| Phenol | Cumene + O₂, then dilute acid | Industrial; gives propanone as the co-product |
| Ether | Williamson synthesis: R–X + NaOR′ | Use a primary halide, or elimination wins |
| Ether | Alcohol + conc. H₂SO₄ at 413 K | At 443 K the same mixture gives an alkene instead |
The last row is worth underlining. Ethanol with concentrated sulphuric acid gives diethyl ether at about 413 K and ethene at about 443 K. If a question quotes a temperature, it is handing you the product.
Acidity — the heart of the chapter
All three release the O–H proton in principle, but only phenol does it to any measurable extent in water. The order you should be able to defend is:
The reason is what happens after the proton leaves. The phenoxide ion spreads its negative charge into the benzene ring by resonance, so it is comparatively stable. An alkoxide ion has nowhere to put the charge, and the alkyl group's +I effect pushes even more electron density onto the already negative oxygen, making it worse.
| Compound | Approximate pKa in water at 25 °C | Practical consequence |
|---|---|---|
| Ethanoic acid | 4.76 | Dissolves in NaHCO₃ with CO₂ effervescence |
| Phenol | ≈ 10.0 | Dissolves in NaOH but not in NaHCO₃ |
| Water | 14.0 | Reference point |
| Ethanol | ≈ 16 | Neutral to litmus; needs sodium metal to deprotonate |
Values are rounded and differ slightly between textbooks and temperatures, so quote them as approximate. Substituents shift phenol's acidity in a predictable direction: electron-withdrawing groups such as –NO₂ at the ortho and para positions increase acidity (2,4,6-trinitrophenol, picric acid, is a strong acid), while electron-donating groups such as –CH₃ or –OCH₃ decrease it.
Worked example 1 — the pH of a phenol solution
Question. Calculate the pH of 0.10 M phenol. Take Ka = 1.0 × 10⁻¹⁰.
Step 1 — the equilibrium. C₆H₅OH ⇌ C₆H₅O⁻ + H⁺, so Ka = [C₆H₅O⁻][H⁺] ÷ [C₆H₅OH].
Step 2 — set up. Let [H⁺] = x. Then x² ÷ (0.10 − x) = 1.0 × 10⁻¹⁰. Because Ka is tiny, x is far smaller than 0.10 and (0.10 − x) ≈ 0.10.
Step 3 — solve.
x² = 1.0 × 10⁻¹⁰ × 0.10 = 1.0 × 10⁻¹¹
x = √(1.0 × 10⁻¹¹) = 3.16 × 10⁻⁶ mol/L
Step 4 — pH.
pH = −log(3.16 × 10⁻⁶) = 6 − log 3.16 = 6 − 0.50 = 5.50
Check the approximation. Degree of dissociation α = 3.16 × 10⁻⁶ ÷ 0.10 = 3.16 × 10⁻⁵, i.e. about 0.003%. Ignoring x against 0.10 was clearly safe. Note what this means physically: 0.10 M phenol has a pH of only 5.5 — mildly acidic, far weaker than 0.10 M ethanoic acid, which comes out near pH 2.9.
The same sum for ethanol is meaningless: with Ka around 10⁻¹⁶, the H⁺ from water's own ionisation completely dominates, and the solution is neutral for all practical purposes.
Worked example 2 — hydrogen gas from the sodium test
Question. 4.60 g of ethanol reacts completely with excess sodium metal. What volume of hydrogen is produced at STP?
Step 1 — balanced equation.
2 C₂H₅OH + 2 Na → 2 C₂H₅ONa + H₂↑
Left: 4 C, 12 H, 2 O, 2 Na. Right: 4 C, 10 H, 2 O, 2 Na, plus H₂ = 2 H. Totals match.
Step 2 — molar mass of ethanol.
2(12.011) + 6(1.008) + 15.999 = 24.022 + 6.048 + 15.999 = 46.07 g/mol
Step 3 — moles.
n(ethanol) = 4.60 ÷ 46.07 = 0.0999 mol
n(H₂) = 0.0999 ÷ 2 = 0.0499 mol
Step 4 — volume. Two answers, and both are correct depending on the definition your
book uses:
Older convention (273.15 K, 1 atm), molar volume 22.4 L/mol:
0.0499 × 22.4 = 1.12 L
Current IUPAC STP (273.15 K, 100 kPa), molar volume 22.7 L/mol:
0.0499 × 22.7 = 1.13 L
Write down which convention you used. Examiners accept either when it is stated; they cannot accept a bare number that matches neither.
Distinguishing the three in the laboratory
| Test | Alcohol | Phenol | Ether |
|---|---|---|---|
| Neutral FeCl₃ solution | No colour | Violet / purple colour | No colour |
| NaOH solution | Insoluble (does not react) | Dissolves — forms phenoxide | Insoluble |
| NaHCO₃ solution | No reaction | No effervescence | No reaction |
| Sodium metal | H₂ gas evolved | H₂ gas evolved | No reaction |
| Bromine water | No reaction | White precipitate of 2,4,6-tribromophenol | No reaction |
| Lucas reagent (conc. HCl + anhydrous ZnCl₂) | 3° turbid at once, 2° in minutes, 1° only on heating | Not applicable | No reaction |
The NaHCO₃ row is the one examiners love: phenol dissolves in NaOH but gives no carbon dioxide with sodium bicarbonate, whereas a carboxylic acid fizzes. That single test separates phenol from benzoic acid.
Worked example 3 — bromination of phenol
Question. 0.94 g of phenol is treated with excess bromine water. Calculate the maximum mass of 2,4,6-tribromophenol obtainable.
Step 1 — balanced equation.
C₆H₅OH + 3 Br₂ → C₆H₂Br₃OH + 3 HBr
Left: 6 C, 6 H, 1 O, 6 Br. Right: 6 C, (2 + 1) + 3 = 6 H, 1 O, 3 + 3 = 6 Br. Balanced.
Step 2 — molar masses (C 12.011, H 1.008, O 15.999, Br 79.904):
M(C₆H₅OH) = 72.066 + 6.048 + 15.999 = 94.11 g/mol
M(C₆H₂Br₃OH) = 72.066 + 3.024 + 239.712 + 15.999 = 330.80 g/mol
Step 3 — moles and mass.
n(phenol) = 0.94 ÷ 94.11 = 0.00999 mol → 0.00999 mol of product
mass = 0.00999 × 330.80 = 3.30 g
Why three bromines? The –OH group is strongly activating and ortho/para directing. In water the ring is activated so strongly that all three available positions react. Run the same reaction in carbon disulphide at low temperature and you get mainly the mono-substituted para product instead — solvent and temperature control the outcome.
Two named reactions of phenol you must not confuse
- Kolbe's reaction. Sodium phenoxide + CO₂ under pressure, then acidify → salicylic acid (2-hydroxybenzoic acid). Carbon dioxide is the electrophile, so a –COOH group arrives.
- Reimer–Tiemann reaction. Phenol + CHCl₃ + aqueous NaOH, then acidify → salicylaldehyde (2-hydroxybenzaldehyde). Chloroform in alkali makes dichlorocarbene, so a –CHO group arrives.
One-line memory hook: CO₂ gives the acid, CHCl₃ gives the aldehyde.
Ether cleavage — the rule that decides the products
Ethers are unreactive, but hot concentrated HI or HBr breaks the C–O bond. Which side breaks is a standard two-mark question.
For an alkyl aryl ether the halide always attacks the alkyl side, because the aryl C–O bond has partial double-bond character from resonance with the ring. So you get an alkyl halide and a phenol. For a dialkyl ether, the halide attacks the less hindered (usually smaller, primary) group by an SN2 path — unless one group is tertiary, in which case a stable carbocation forms and the halide goes to the tertiary carbon instead.
Common mistakes that cost marks
- Saying phenol is a strong acid. It is a weak acid — as worked example 1 shows, 0.10 M phenol is only pH 5.5.
- Claiming phenol liberates CO₂ from NaHCO₃. It does not. That is precisely the test used to tell phenol from a carboxylic acid.
- Mixing up 413 K and 443 K for ethanol with conc. H₂SO₄ — ether at the lower temperature, alkene at the higher one.
- Forgetting anti-Markovnikov in hydroboration–oxidation. Acid-catalysed hydration and hydroboration give different alcohols from the same alkene.
- Writing bromobenzene as an ether-cleavage product. Anisole with HI gives phenol and iodomethane, never iodobenzene.
- Using Williamson synthesis with a tertiary halide. The alkoxide is a strong base as well as a nucleophile, so elimination takes over and you get an alkene.
- Assuming an ether is inert to everything. Old bottles of diethyl ether form explosive peroxides on standing in air — a genuine laboratory hazard, and a fair viva question.
Why this chapter matters
| Where it appears | What is actually tested |
|---|---|
| CBSE / ICSE Class 12 organic unit | Acidity comparisons with reasons, product prediction, named reactions |
| Conversion questions | Alkene → alcohol → aldehyde/acid, and phenol → salicylic acid |
| Practical and viva | FeCl₃ test, Lucas test, sodium metal test, functional group identification |
| NEET / JEE foundation | Markovnikov vs anti-Markovnikov, oxidation product of each alcohol class |
| IIT-JAM / CUET-PG later on | Mechanisms, substituent effects on pKa, selective protection of –OH |
Do not plan revision around remembered "weightages" — syllabus documents are revised, so check the one your board has issued for your own session.
Practise the acidity numbers, not just the order. The pH / pOH calculator takes a concentration and a Ka (or a pH you already have) and returns the rest, so you can see for yourself how a Ka of 10⁻¹⁰ turns into a pH of 5.5 — and how much further a carboxylic acid moves the needle.
Open the pH / pOH Calculator →Want this chapter taught properly rather than memorised? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes for students right across India — details at abcchemistry.in.