Alcohols, Phenols and Ethers
Three families built on oxygen — the versatile alcohols, the surprisingly acidic phenols, and the quietly unreactive ethers
- How alcohols, phenols and ethers differ in structure and reactivity.
- Preparing alcohols by hydration, hydroboration–oxidation, carbonyl reduction and Grignard routes.
- Alcohol reactions — esterification, dehydration, oxidation — and the Lucas test.
- Why phenol is far more acidic than an alcohol (phenoxide resonance).
- Electrophilic substitution of phenol, the Kolbe and Reimer–Tiemann reactions.
- The Williamson synthesis and the cleavage of ethers by \(\ce{HI}\).
Three Oxygen Families
All three classes share an oxygen atom but differ in what surrounds it. An alcohol has \(\ce{-OH}\) on an \(sp^3\) carbon; a phenol has \(\ce{-OH}\) directly on an aromatic ring; an ether has oxygen bridging two carbon groups (\(\ce{R-O-R'}\)). That one structural difference makes alcohols versatile, phenols acidic, and ethers largely inert.
| Class | Group | Example | Defining trait |
|---|---|---|---|
| Alcohol | \(\ce{-OH}\) on \(sp^3\) C | ethanol | versatile, weakly acidic |
| Phenol | \(\ce{-OH}\) on ring | phenol | distinctly acidic |
| Ether | \(\ce{C-O-C}\) | diethyl ether | inert, good solvent |
Preparation of Alcohols
Alcohols are reached from alkenes, carbonyls and halides. The choice of method even controls which carbon bears the \(\ce{-OH}\) — acid hydration follows Markovnikov, while hydroboration–oxidation goes anti-Markovnikov.
From the same alkene you can place the \(\ce{-OH}\) on either carbon by choosing the route. Grignard reagents extend the toolkit: \(\ce{HCHO}\) gives a \(1^\circ\) alcohol, an aldehyde gives \(2^\circ\), and a ketone gives \(3^\circ\).
| Method | Reaction | Gives |
|---|---|---|
| Acid hydration | \(\ce{C=C + H2O ->[H+] }\) alcohol | Markovnikov |
| Reduction | \(\ce{RCHO ->[LiAlH4] RCH2OH}\) | 1° alcohol |
| Grignard + ketone | \(\ce{R'2C=O + RMgX -> R'2C(OH)R}\) | 3° alcohol |
| From haloalkane | \(\ce{R-X + aq.\ KOH -> R-OH}\) | any |
Reactions of Alcohols
The \(\ce{-OH}\) group reacts in two ways — losing the acidic \(\ce{O-H}\) hydrogen, or losing the whole \(\ce{-OH}\) as the C–O bond breaks.
| Reaction | Product |
|---|---|
| \(\ce{+ Na}\) | sodium alkoxide \(+\ \ce{H2}\) |
| \(\ce{+ RCOOH}\) (\(\ce{H+}\)) | ester (esterification) |
| conc. \(\ce{H2SO4}\), 443 K | alkene (dehydration) |
| conc. \(\ce{H2SO4}\), 413 K | ether |
| \(\ce{+ HX}\) (Lucas) | haloalkane |
The Lucas Test & Oxidation
Two reactions tell \(1^\circ\), \(2^\circ\) and \(3^\circ\) alcohols apart. The Lucas test (conc. \(\ce{HCl}\) + \(\ce{ZnCl2}\)) measures how fast a cloudy haloalkane forms; oxidation gives different products depending on the class.
The test works because the reaction runs by \(\ce{SN1}\): a \(3^\circ\) alcohol forms its stable carbocation instantly, a \(2^\circ\) more slowly, and a \(1^\circ\) not at all under these mild conditions.
Preparation of Phenol
Phenol is made industrially and in the lab by several routes. The dominant industrial method is the cumene process, which neatly yields acetone as a valuable by-product.
| Route | Reaction / steps |
|---|---|
| Cumene process | cumene \(\to\) cumene hydroperoxide \(\to\) phenol + acetone |
| Dow process | \(\ce{C6H5Cl + NaOH ->[623\,K][300\,atm] C6H5OH}\) |
| From sulphonic acid | \(\ce{C6H5SO3H}\) fused with \(\ce{NaOH}\) |
| From diazonium salt | \(\ce{C6H5N2+ + H2O -> C6H5OH + N2}\) |
Acidity of Phenol
Phenol (\(\mathrm{p}K_a \approx 10\)) is far more acidic than an alcohol (\(\mathrm{p}K_a \approx 16\)). The reason is the stability of the ion it leaves behind: the phenoxide ion spreads its negative charge over the ring by resonance, whereas an alkoxide cannot.
Electron-withdrawing groups (\(\ce{-NO2}\)) stabilise the phenoxide further and raise acidity; the \(2,4,6\)-trinitrophenol (picric acid) is strongly acidic. Electron-donating groups (\(\ce{-CH3}\)) lower it. Note phenol is still too weak to react with \(\ce{NaHCO3}\) — a useful way to tell it from a carboxylic acid.
Reactions of Phenol
The \(\ce{-OH}\) group is strongly activating and ortho/para-directing, so phenol undergoes electrophilic substitution far more readily than benzene — sometimes too readily. Two named reactions install useful groups at the ortho position.
| Reaction | Conditions | Product |
|---|---|---|
| Bromination | \(\ce{Br2}\) water | 2,4,6-tribromophenol (white ppt) |
| Nitration | conc. \(\ce{HNO3}\) | picric acid (2,4,6-trinitrophenol) |
| Kolbe reaction | \(\ce{NaOH}\), then \(\ce{CO2}\) | salicylic acid |
| Reimer–Tiemann | \(\ce{CHCl3} + \ce{NaOH}\) | salicylaldehyde |
| \(\ce{FeCl3}\) test | neutral \(\ce{FeCl3}\) | violet colour (detects phenol) |
Ethers: Preparation & Cleavage
Ethers are the quiet members of the family — no \(\ce{O-H}\), so no hydrogen bonding among themselves, low boiling points, and little reactivity (which makes them excellent solvents). The key synthesis is the Williamson ether synthesis, an \(\ce{SN2}\) reaction.
Hot \(\ce{HI}\) splits ethers at the C–O bond. For an aryl alkyl ether like anisole, only the alkyl–O bond breaks (the aryl–O bond is strengthened by resonance), giving phenol plus the alkyl iodide.
Putting It to Work
Problem. Propene is treated by (a) \(\ce{H2O}/\ce{H+}\) and (b) \(\ce{B2H6}\) then \(\ce{H2O2}/\ce{OH-}\). Give the products.
Solution. Acid hydration is Markovnikov; hydroboration is anti-Markovnikov:
Problem. Which alcohol results when \(\ce{CH3MgBr}\) reacts with acetone, then water?
Solution. A ketone + Grignard gives a tertiary alcohol:
Problem. Three alcohols give turbidity (i) at once, (ii) in 5 min, (iii) not at all. Identify each.
Solution. Speed tracks carbocation stability (SN1):
Problem. Which is more acidic, ethanol or phenol? Why?
Solution. The phenoxide ion is resonance-stabilised; the ethoxide is not:
Problem. To make tert-butyl methyl ether, which halide–alkoxide pair should you choose?
Solution. The 1° partner must be the halide (a 3° halide would eliminate):
Problem. What product forms when phenol is treated with excess bromine water?
Solution. The strongly activated ring brominates at all three o/p sites:
Chapter Summary
Alcohols (sp³ C–OH), phenols (ring C–OH), ethers (C–O–C) — set by what holds the O.
Hydration (Markovnikov) vs hydroboration (anti); reduction; Grignard (1°/2°/3°).
Esterify, dehydrate, oxidise; Lucas test & oxidation distinguish 1°/2°/3°.
Phenoxide resonance makes phenol far more acidic than alcohols; \(\ce{-NO2}\) raises it.
Strongly o/p-activating; Kolbe → salicylic acid; Reimer–Tiemann → salicylaldehyde.
Williamson (SN2) synthesis; inert solvents; \(\ce{HI}\) cleaves the C–O bond.
Problems
For each item, first decide which family it concerns, then apply the relevant preparation, reaction or acidity principle. Difficulty rises down the list.
- Distinguish an alcohol, a phenol and an ether by structure.
- How would you make propan-1-ol and propan-2-ol from the same alkene?
- Which alcohol forms when methylmagnesium bromide reacts with formaldehyde?
- Describe the Lucas test and the basis on which it distinguishes 1°, 2° and 3° alcohols.
- Give the oxidation products of a primary, a secondary and a tertiary alcohol.
- Outline the cumene process for manufacturing phenol.
- Explain why phenol is more acidic than ethanol, with the help of resonance.
- Arrange phenol, p-nitrophenol, p-cresol and picric acid by acidity.
- Write the Kolbe and Reimer–Tiemann reactions of phenol.
- Why does phenol give 2,4,6-tribromophenol with bromine water?
- Write the Williamson synthesis of ethyl methyl ether and justify the choice of halide.
- Give the products when anisole is heated with \(\ce{HI}\), and explain which bond breaks.