Preparation of phenols

Phenol looks like a simple molecule — a benzene ring with one –OH group — but that –OH refuses to form the easy way alcohols do. The aromatic ring will not simply swap a hydrogen or a leaving group for –OH under mild conditions, so chemists had to find indirect back doors into the ring. This article walks through the four routes the ISC syllabus names: hydrolysing a diazonium salt, forcing chlorobenzene to react under extreme heat and pressure (Dow's process), fusing benzene sulphonic acid with alkali, and — the route that actually supplies most of the world's phenol — oxidising cumene. Each route solves the same stubborn problem in a different way, and seeing why each one works, rather than just memorising the equations, is what separates a shaky answer from a confident one.

The core insight

Because the aromatic C–H, C–Cl and C–SO₃H bonds resist direct substitution by –OH, phenol is made industrially and in the laboratory only through four indirect routes — diazonium hydrolysis, Dow's process, benzene sulphonic acid fusion, and the cumene process — each of which supplies the driving force the ring itself will not.

The core idea

Route 1 — hydrolysis of a diazonium salt. Benzenediazonium chloride, made by diazotising aniline with $NaNO_2/HCl$ at $0$ – $5\,^\circ C$ , loses nitrogen gas on warming with water and leaves phenol behind:

C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl

where: the diazonium ion is the intermediate, $\Delta$ marks gentle warming (not the harsh conditions of the other routes), and $N_2\uparrow$ is the gas that escapes, leaving the C–OH bond behind.

Route 2 — Dow's process (from chlorobenzene). Chlorobenzene is fused with concentrated aqueous NaOH at roughly $623\ K$ and $300\ \text{atm}$ , then the resulting salt is acidified:

C_6H_5Cl + 2NaOH \xrightarrow[623\,K,\ 300\ atm]{} C_6H_5ONa + NaCl + H_2O \xrightarrow{H^+} C_6H_5OH

where: $623\ K$ and $300\ \text{atm}$ are the forcing conditions needed because the aryl C–Cl bond is unusually strong, and the acidification step ( $H^+$ ) is what converts the intermediate sodium phenoxide into free phenol.

Route 3 — from benzene sulphonic acid. Fusing the sulphonic acid with solid NaOH at high temperature, then acidifying, gives phenol the same way:

C_6H_5SO_3H + 2NaOH \xrightarrow{\Delta,\ \text{fusion}} C_6H_5ONa + Na_2SO_3 + H_2O \xrightarrow{H^+} C_6H_5OH

where: fusion means heating the solids together without solvent, and $Na_2SO_3$ (sodium sulphite) is the by-product carrying away the sulphonic acid group.

Route 4 — the cumene process (industrial manufacture). Cumene, made by Friedel–Crafts alkylation of benzene with propene, is air-oxidised to a hydroperoxide, which acid then cleaves into phenol and acetone:

C_6H_5CH(CH_3)_2 \xrightarrow{O_2} C_6H_5C(CH_3)_2OOH \xrightarrow{H^+,\ H_2O} C_6H_5OH + (CH_3)_2C=O

where: $O_2$ drives slow autoxidation of the isopropyl side chain (not the ring), and the final acid-catalysed cleavage step is the only place a new C–O bond forms at the ring carbon.

Figure

Four arrows from different starting materials converging on phenol, one via diazonium salt hydrolysis, one via Dows process on chlorobenzene, one via benzene sulphonic acid fusion, and one via the cumene process, showing four distinct chemical routes to the same product

Why it works

The stumbling block behind all four routes is the same: the aromatic ring's C–H, C–Cl and C–SO₃H bonds are all strong and resist the kind of easy nucleophilic substitution that converts, say, an alkyl halide into an alcohol. Each named route supplies a different driving force to overcome that resistance.

The diazonium route wins on leaving-group strength. Nitrogen gas is an exceptionally stable, exceptionally good leaving group — once it departs there is no going back — so hydrolysis proceeds on gentle warming with no forcing conditions needed at all. This is also why the route has no aliphatic counterpart: aliphatic diazonium salts are too unstable to isolate, while the aromatic ring's ability to delocalise the positive charge keeps the aryl diazonium salt stable enough to handle at $0$ – $5\,^\circ C$ before it is deliberately hydrolysed.

Dow's process and the sulphonic-acid route win by brute force. Neither the aryl C–Cl bond nor the aryl C–SO₃H bond is a good leaving group under normal conditions, so both routes simply supply enough heat (and, for Dow's process, extreme pressure) to force the substitution through. The direct product of fusion in both cases is the sodium salt — sodium phenoxide, alongside NaCl or Na₂SO₃ — never free phenol; a separate acidification step is always required to liberate the –OH.

The cumene process wins by avoiding the ring altogether. No substitution ever happens at a ring carbon. Oxidation and rearrangement occur entirely in the isopropyl side chain, and the new C–O bond is created only in the final acid-catalysed cleavage of the hydroperoxide. Because the ring itself is never attacked, this route needs no forcing conditions, and it has the further advantage of yielding acetone as a valuable co-product from the same cheap feedstocks (benzene and propene) — which is exactly why it is the dominant industrial method for phenol manufacture today.

Worked example

Starting from benzene, write every equation for a complete synthesis of phenol via the diazonium-salt route, and state why the key intermediate is stable enough to isolate.

Step 1. Nitrate benzene to nitrobenzene. C_6H_6 + HNO_3 \xrightarrow{conc.\ H_2SO_4} C_6H_5NO_2 + H_2O

Step 2. Reduce nitrobenzene to aniline. C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O

Step 3. Diazotise aniline at low temperature. C_6H_5NH_2 + NaNO_2 + HCl \xrightarrow{0\text{–}5\,^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

Step 4. Hydrolyse the diazonium salt by warming with water. C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl

Step 5. Explain the stability. Aromatic diazonium salts, unlike aliphatic ones, survive at $0$ – $5\,^\circ C$ because the ring delocalises the positive charge on the $-N_2^+$ group; but that same salt hydrolyses readily on warming because loss of $N_2$ gas is such a strong driving force. The four-step route benzene to nitrobenzene to aniline to the diazonium salt to phenol is the complete answer, and the low-temperature stability followed by warm hydrolysis is the reasoning that earns full marks.

Competitive stretch

The same benzene-to-phenol conversion can be done in three fewer isolable steps using the cumene process: alkylate benzene with propene ( $C_6H_6 + CH_2=CHCH_3 \xrightarrow{H_3PO_4} C_6H_5CH(CH_3)_2$ ), air-oxidise to the hydroperoxide, then cleave with acid to give phenol plus acetone. Compare the two: the diazonium route needs four separate transformations and generates no useful by-product, while the cumene route needs three and yields commercially valuable acetone alongside phenol — the atom-economy argument examiners look for when asked why industry prefers cumene.

Common traps

Writing Dow's process as though chlorobenzene reacts with NaOH as easily as an alkyl halide would — the aryl C–Cl bond is strong, and the fused NaOH at $623\ K$ , $300\ \text{atm}$ must be stated explicitly.
Stopping at the sodium phenoxide (or sulphite) stage in the chlorobenzene and sulphonic-acid routes and forgetting the final acidification step that actually releases free phenol.
Assuming the diazonium route has an aliphatic analogue — it does not; aliphatic diazonium salts decompose instantly, and only the aromatic ring's charge delocalisation makes the low-temperature salt stable enough to isolate.
Writing the cumene process cleavage as giving phenol alone — the balanced equation always produces acetone as a co-product, and omitting it loses marks.

Ask the tutor: want to work through the full benzene-to-phenol synthesis by another route, or check your own equations for any of these four methods?

Why can't phenol just be made by directly replacing a hydrogen on benzene with an OH group

Because the aromatic ring's C–H bond is strong and the ring is electron-rich, so it resists the kind of direct nucleophilic attack that would install –OH in one step; there is no simple, direct analogue of the way alcohols form from alkyl halides. Every named route instead goes through an intermediate — a diazonium salt, a chlorinated or sulphonated ring forced to react under extreme conditions, or a side-chain oxidation as in the cumene process — because the ring itself only yields the C–OH bond indirectly.

Why does the diazonium salt need to be kept at 0 to 5 degrees Celsius before hydrolysis

At that low temperature the aromatic diazonium salt is stable enough to prepare and briefly store, because the ring delocalises the positive charge on the $-N_2^+$ group, unlike an aliphatic diazonium salt which decomposes on formation. Warming it above this range is exactly what triggers hydrolysis — the same instability that would destroy an aliphatic diazonium salt immediately is deliberately switched on by heating the aromatic one, releasing $N_2$ gas and giving phenol.

Is the cumene process asked about in as much detail as the other three routes

No. The ISC syllabus treats cumene manufacture at "outline of method, equations required, detail not required" level, matching how industrial processes are generally examined in this unit. You should be able to write the three-step sequence (alkylation, air oxidation, acid cleavage) and name acetone as the co-product, but you are not expected to justify reaction mechanisms or conditions in the same depth as for the diazonium, Dow's, and sulphonic-acid routes.