Separating phenol, benzyl alcohol and anisole - and two Williamson proposals that will not both work
A 15.0 g mixture of phenol, benzyl alcohol and anisole (roughly equal masses) must be separated and each component identified using only: 10% NaOH, dilute HCl, neutral FeCl3 solution, bromine water, Lucas reagent (ZnCl2 in conc. HCl) and a fractional distillation set-up with a 20 cm column. Boiling points: anisole 154 °C, phenol 182 °C, benzyl alcohol 205 °C. The recovered phenol is then to be converted to anisole, and a separate request asks for methyl tert-butyl ether made from methanol and 2-bromo-2-methylpropane, both of which are in stock. A junior chemist proposes simply heating phenol with methyl iodide, and separately proposes treating sodium methoxide with 2-bromo-2-methylpropane. Design the separation and identification sequence, and judge whether the two proposals will give the required ethers.
1. Give a step-by-step separation of the three components, stating what is in each layer at every stage and how each component is finally isolated.
Read the worked answer
Step 1 - acid–base extraction. Shake the mixture with 10% NaOH. Phenol is converted to sodium phenoxide and passes into the aqueous layer; benzyl alcohol and anisole remain in the organic layer because neither is acidic enough to be deprotonated by NaOH (benzyl alcohol pKa ≈ 15, anisole has no acidic O–H). Separate the layers.
Step 2 - recover phenol. Acidify the aqueous layer with dilute HCl until Congo red/universal indicator shows acidic: C6H5O−Na+ + HCl → C6H5OH + NaCl. Phenol separates as an oil or solid; extract with ether, dry over anhydrous Na2SO4, remove ether, and distil (bp 182 °C).
Step 3 - separate the neutral pair. Wash the organic layer with water, dry over anhydrous Na2SO4, and fractionally distil through the 20 cm column. Anisole (154 °C) distils first, benzyl alcohol (205 °C) remains; a 51 °C gap is sufficient for clean separation with a reasonable column, though the first and last fractions should be redistilled.
Step 4 - confirm identities. Phenol: violet colour with FeCl3, and white 2,4,6-tribromophenol with bromine water. Anisole: no FeCl3 colour, decolourises bromine water without a precipitate, bp 154 °C. Benzyl alcohol: no FeCl3 colour, immediate cloudiness with Lucas reagent, bp 205 °C.
2. Which two simple tests would distinguish all three compounds unambiguously, and what would you see in each case?
Read the worked answer
No single test distinguishes all three, so use two orthogonal tests.
Test 1 - neutral FeCl3 solution. Phenol gives an intense violet colour (formation of the phenoxide–Fe(III) complex). Benzyl alcohol and anisole give no colour. This immediately identifies phenol.
Test 2 - Lucas reagent (ZnCl2/conc. HCl), at room temperature. Benzyl alcohol turns cloudy almost at once, because it ionises through a resonance-stabilised benzylic carbocation (the primary benzylic cation is far more stable than a simple primary cation). Anisole does not dissolve and does not give the cloudiness of an alkyl chloride; it floats as a separate layer. Phenol is already accounted for, but for completeness it does not give a chloride layer - the aromatic C–O bond is not broken under these conditions.
Cross-check with bromine water. Phenol gives a white precipitate of 2,4,6-tribromophenol; anisole decolourises bromine water by ring substitution (4-bromoanisole) without a heavy precipitate; benzyl alcohol gives no rapid reaction. Boiling points (154, 182, 205 °C) provide a final confirmation after distillation. Relying on FeCl3 alone would confuse 'no reaction' for anisole and benzyl alcohol, so two tests are essential.
3. Judge the two Williamson proposals: (i) phenol + methyl iodide by heating alone; (ii) sodium methoxide + 2-bromo-2-methylpropane. Give the correct route for each target ether.
Read the worked answer
Proposal (i) fails. Phenol is a poor nucleophile because the oxygen lone pair is delocalised into the ring, and methyl iodide is a weak electrophile; simply heating gives negligible anisole. The correct Williamson route first makes the nucleophile: phenol + NaOH → sodium phenoxide, then phenoxide attacks CH3I (or dimethyl sulphate) by SN2 on the unhindered methyl carbon: C6H5O−Na+ + CH3I → C6H5OCH3 + NaI. Methyl iodide is essential - a secondary or tertiary halide with phenoxide (a strong base/nucleophile) would give mainly elimination.
Proposal (ii) also fails. 2-Bromo-2-methylpropane is a tertiary halide and methoxide is a strong base; the dominant pathway is E2 elimination to 2-methylpropene, not substitution. The correct pairing reverses the roles: use potassium tert-butoxide, (CH3)3C–O−K+, with methyl iodide, (CH3)3C–O−K+ + CH3I → (CH3)3C–O–CH3 + KI. Methyl iodide cannot undergo elimination and is unhindered, so SN2 gives the ether in good yield.
General rule: in Williamson ether synthesis, put the bulky part on the alkoxide and the small part on the halide.