Alcohols, Phenols and Ethers · Lesson 4 of 12
Preparing Phenols
“Compare four routes to the same aromatic hydroxyl group and explain why each needs its own activation and work-up.”
• Write balanced preparation equations from chlorobenzene, sulphonates and diazonium salts. • Explain why acidification follows alkaline phenoxide formation. • Trace oxygen and carbon through the cumene process. • Select a synthesis that preserves the required aromatic skeleton.
A phenolic C–O bond requires an aromatic route
Replacing Cl in CH₃CH₂Cl by OH using aqueous alkali is relatively straightforward. Replacing Cl directly on benzene is not the same problem. Chlorobenzene has a shorter, conjugated aryl C–Cl bond and an sp² reaction centre that does not permit ordinary backside SN2 attack. It also does not readily ionise to an unstable phenyl carbocation. Phenol preparations therefore use strong conditions, an activated aromatic precursor, or a different rearrangement strategy. The product identity is shared, but the reaction pathways are not interchangeable.
In the following equations C₆H₅ is the aromatic fragment that remains bonded to oxygen. Sodium phenoxide, C₆H₅ONa, represents an ionic pair of Na⁺ and C₆H₅O⁻. It is a salt rather than a phenol with an extra sodium covalent substituent on the ring. Under strongly alkaline conditions it is the expected species; acid must be added later to recover neutral phenol.
Chlorobenzene: severe alkaline conditions followed by acidification
The two equivalents of NaOH in the first net equation account for both substitution and deprotonation under the alkaline conditions. Writing just one NaOH while giving sodium phenoxide, NaCl and water would fail atom balance. In the second operation, proton transfer changes O⁻ to OH without changing the ring skeleton. This route illustrates why a reaction scheme may omit spectator salts but a claimed balanced equation may not.
Problem
After heating chlorobenzene with concentrated alkali, a student says the mixture already contains isolated neutral phenol. Identify the missing operation.
- 1.The strongly alkaline medium favours phenoxide, the conjugate base of phenol. The C–O bond may have formed, but the protonation state is not yet that of the neutral target.
- 2.Acidify after the conversion. C₆H₅O⁻ accepts H⁺ and becomes C₆H₅OH. Separate inorganic salts are not part of the neutral organic product.
- 3.This is a proton-transfer work-up, not a second aromatic substitution. Keeping those roles separate prevents errors in product formulas.
Sulphonation followed by alkaline fusion
Benzene reacts with oleum to form benzenesulphonic acid. Oleum contains SO₃ in sulphuric acid; the sulphonation can be written with SO₃ as the reacting component. Neutralise to the sodium sulphonate, heat with molten NaOH, then acidify the resulting phenoxide. A sulphonate substituent is thus installed and subsequently replaced by oxygen. Concentrated acid and molten alkali belong to separate stages, since mixing them would merely neutralise the reagents.
Sodium sulphite contains the sulphur removed from the aromatic substituent. Checking sulphur and sodium as well as carbon is useful because compressed ring diagrams often hide the inorganic coproduct. The original benzene carbon count remains six throughout; sulphonation introduces no carbon atom. This route can therefore be planned from benzene, sulphuric acid/oleum and alkali without inventing an extra organic starting material.
Diazonium hydrolysis exchanges a nitrogen group for OH
Aniline is an aromatic primary amine. Treating it with nitrous acid generated from NaNO₂ and HCl at 273–278 K forms benzenediazonium chloride. The cold conditions are important because diazonium salts can decompose; the subsequent conversion to phenol uses warming with water or dilute acid. Do not collapse the cold preparation and warm hydrolysis into one unspecified operation.
The aromatic carbon carrying NH₂ ultimately carries OH. The nitrogen atoms leave as N₂; no nitrogen remains in the phenol product. The hydrolysis should not be rationalised as formation of a stable free phenyl carbocation analogous to a tertiary alkyl cation. At this level the useful conclusions are precursor identity, temperature sequence, gas evolution and correct replacement of the diazonium group.
Cumene makes two useful organic products
Cumene is isopropylbenzene, C₆H₅CH(CH₃)₂. Oxidation with air gives cumene hydroperoxide, C₆H₅C(CH₃)₂OOH. The O–O bond identifies a hydroperoxide; it must not be drawn as a simple alcohol. Acid treatment causes rearrangement and cleavage to phenol and propanone. The ring migrates to oxygen during the rearrangement, so the final phenolic oxygen is directly attached to the ring. The three-carbon side-chain fragment becomes acetone.
For carbon accounting, cumene has nine carbons. Phenol receives six and acetone three. For oxygen accounting, the hydroperoxide has two oxygen atoms and the final products each have one. This industrial route is attractive because the coproduct acetone is valuable; it does not discard the entire side chain as waste. A detailed radical oxidation or rearrangement mechanism is beyond the required scope, but the hydroperoxide connectivity and two-stage sequence are essential.
Problem
If 0.50 mol of cumene hydroperoxide undergoes the ideal cleavage, what amounts of phenol and acetone are predicted?
- 1.The balanced cleavage has coefficient 1 for hydroperoxide, phenol and acetone. One mole of intermediate therefore gives one mole of each product.
- 2.For 0.50 mol intermediate, the theoretical amounts are 0.50 mol phenol and 0.50 mol acetone. Carbon balance is 9×0.50 = 6×0.50 + 3×0.50 = 4.50 mol of carbon atoms.
- 3.This is a theoretical stoichiometric prediction. An actual isolated yield would require conversion and recovery data, which must not be invented.
Quick Quiz
Quiz
What species is favoured after alkaline preparation of phenol, before acidification?
Explanation: Strong base removes the phenolic proton, so acid work-up is needed.
Which conditions belong to diazonium formation from aniline?
Explanation: Nitrous acid generated cold converts the aromatic primary amine into a diazonium salt.
What organic coproduct accompanies phenol in cumene hydroperoxide cleavage?
Explanation: The three-carbon isopropyl fragment becomes acetone.
Which coproduct balances alkaline fusion of sodium benzenesulphonate?
Explanation: The departing sulphur-containing group appears as sulphite in the balanced equation.
Why is ordinary primary-alkyl SN2 reasoning unsuitable for chlorobenzene?
Explanation: Aryl attachment changes geometry and bond character; severe conditions are needed for this preparation.
Practice Problems
- Write the balanced chlorobenzene-to-phenoxide equation and the acid work-up.
- Plan phenol synthesis from benzene through a sulphonic acid.
- Explain the temperature sequence and observable gas in the aniline route.
- Write the structures and atom balance for the cumene route.
- Can dilute aqueous NaOH convert both ethyl chloride and chlorobenzene under identical ordinary conditions? Explain.
- What is the theoretical phenol amount from 1.20 mol hydroperoxide at 75% product yield?
C₆H₅Cl + 2NaOH → C₆H₅ONa + NaCl + H₂O under severe heated, high-pressure conditions. Then C₆H₅ONa + HCl → C₆H₅OH + NaCl. Two NaOH units are necessary for the stated first products; the second step restores the oxygen-bound proton.
Key Takeaways
• Phenol synthesis requires routes suited to aromatic attachment. • Strongly alkaline preparations give phenoxide; later acidification releases phenol. • Chlorobenzene conversion needs severe heat and pressure. • Sulphonation followed by alkaline fusion replaces a sulphonic substituent with oxygen. • Diazonium salts are prepared cold and hydrolysed by warming. • Cumene is first oxidised to a hydroperoxide, then cleaved with acid. • Cumene cleavage produces phenol and acetone with a 1:1 molar relationship.