Preparation of Alcohols and Phenols

Ask a chemist for a bottle of ethanol or a bottle of phenol and both are reached by a short reaction, but the routes that get you there look nothing alike. Alcohols come from friendly, familiar chemistry — an alkene soaking up water, a haloalkane swapping halogen for hydroxyl, a carbonyl compound accepting hydride. Phenol resists all of that. Its ring locks the C–OH-precursor bond into place with resonance, so the laboratory has to reach for industrial-strength conditions — fused sodium hydroxide at high pressure, or a two-step oxidation of cumene that quietly manufactures acetone as a bonus. This article walks through both families of routes, alcohols and phenols side by side, and pins down exactly why phenol's chemistry has to be so much more forceful than an alcohol's.

The core insight

Alcohols form by straightforward addition, substitution or reduction reactions at an sp3 carbon, while phenol requires harsher conditions (fused NaOH, or the cumene process) because resonance delocalises the aryl carbon–leaving-group bond and blocks ordinary nucleophilic substitution.

The core idea

Alcohols from alkenes. Acid-catalysed hydration follows Markovnikov's rule: $\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{OH}$ , with OH landing on the more substituted carbon for unsymmetrical alkenes. Hydroboration-oxidation reverses this regiochemistry: alkene $\xrightarrow{\text{B}_2\text{H}_6}$ trialkylborane $\xrightarrow{\text{H}_2\text{O}_2/\text{OH}^-}$ alcohol, putting OH on the less substituted carbon (anti-Markovnikov).

Alcohols from haloalkanes and carbonyls. Nucleophilic substitution converts a haloalkane cleanly: $\text{R–X} + \text{aq. KOH} \xrightarrow{\Delta} \text{R–OH} + \text{KX}$ . Reduction of a carbonyl compound gives an alcohol whose class depends on the starting material — aldehydes give 1° alcohols, ketones give 2° alcohols — using $\text{LiAlH}_4$ , $\text{NaBH}_4$ , or catalytic $\text{H}_2/\text{Ni}$ . A Grignard reagent extends the same idea one carbon further: R–MgX with methanal gives a 1° alcohol, with any other aldehyde a 2° alcohol, and with a ketone a 3° alcohol, after acidic workup.

Phenol, three named routes. The cumene process is the industrial workhorse: benzene and propene combine under acid catalysis to give cumene, air oxidises cumene to cumene hydroperoxide, and dilute acid rearranges that peroxide into phenol plus acetone. The Dow process fuses chlorobenzene with NaOH at about $623\ \text{K}$ and roughly $300$ atm, then acidifies to release phenol. Diazonium salt hydrolysis is the mildest route on paper: $\text{C}_6\text{H}_5\text{N}_2^+ + \text{H}_2\text{O} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{H}^+$ , simply warming the salt in water. A fourth, older route fuses benzenesulphonic acid with NaOH and then acidifies — chemically similar to the Dow process but rarely used today.

\text{C}_6\text{H}_5\text{Cl} + 2\text{NaOH} \xrightarrow[300\ \text{atm}]{623\ \text{K}} \text{C}_6\text{H}_5\text{ONa} + \text{NaCl} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{OH}

where: the first arrow is the forcing nucleophilic substitution step (fused NaOH, high temperature and pressure) that only works on an aryl halide under these extreme conditions, and the second arrow is a simple acid-base step that liberates the free phenol from its sodium salt.

Figure

Flow diagram of the cumene process from benzene and propene through cumene and cumene hydroperoxide to phenol and acetone

Why it works

There is no single yield or rate formula examined here, so the payoff of this topic is understanding why phenol needs harsher conditions than a plain alcohol — a reasoning chain, not a calculation, but one that appears in almost every board question on this subsection.

Step 1. In a haloalkane such as bromoethane, the C–Br bond is an ordinary, fully single, polar bond between an sp3 carbon and a halogen. Hydroxide can attack that carbon directly in an $S_N2$ -type step, so mild aqueous NaOH or KOH with gentle heating is enough to convert R–Br into R–OH.

Step 2. In an aryl halide such as bromobenzene, the halogen sits on an sp2 ring carbon whose lone pair is delocalised into the aromatic π system. This resonance donates partial double-bond character to the C–Br bond, shortening and strengthening it while simultaneously making the ring carbon less electrophilic and less accessible to a nucleophile. Ordinary aqueous base at mild temperature essentially does nothing to it.

Step 3. To force the same net substitution on an aryl halide, the reaction has to be pushed with fused (molten) NaOH at high temperature (about $623\ \text{K}$ ) and high pressure (about $300$ atm) — the Dow process. Only this much thermal and mechanical energy overcomes the resonance stabilisation and drives the substitution to sodium phenoxide, which acidification then converts to free phenol.

Step 4. The cumene process sidesteps the resonance problem altogether by never breaking an aryl carbon–halogen or carbon–sulphonate bond at all. Instead it builds a peroxide (–OOH) on a side-chain carbon attached to the ring, and dilute acid triggers a 1,2-aryl shift that migrates the ring from carbon to oxygen, cleaving the weak O–O bond and releasing phenol together with acetone. Because this route needs only mild acid rather than a fused-alkali melt at $623\ \text{K}$ , and it turns a valuable by-product (acetone) into revenue, it has displaced the Dow process as the dominant industrial method — even though the Dow process remains the clearer teaching example of why haloarenes resist substitution.

Worked example

From bromobenzene, obtain phenol, and explain why the mild conditions that hydrolyse bromoethane fail here; then compare with the cumene route.

Step 1. Bromoethane hydrolyses readily with aqueous NaOH or KOH under gentle heating because its C–Br bond is a normal, fully single, polar bond, easily attacked by $\text{OH}^-$ in an $S_N2$ -type substitution.

Step 2. Bromobenzene's C–Br bond is different: the ring's π system delocalises into the bond, giving it partial double-bond character. This strengthens and shortens the bond and lowers the electrophilicity of the ring carbon, so aqueous NaOH at mild temperature fails to displace Br.

Step 3. Forcing the substitution requires fusing bromobenzene with NaOH at about $623\ \text{K}$ and about $300$ atm, giving sodium phenoxide; acidifying with dilute HCl then liberates phenol: $\text{C}_6\text{H}_5\text{ONa} + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaCl}$ .

Step 4. By contrast, the cumene route reaches the same product — phenol — from benzene and propene via cumene and cumene hydroperoxide, using only mild dilute acid for the final cleavage step, and it additionally yields acetone. Answer: phenol is obtained from bromobenzene only via the forcing Dow conditions (fused NaOH, 623 K, 300 atm, then acidification), while the cumene process reaches the same product under far milder conditions and with a valuable by-product — which is why industry prefers cumene over Dow.

Competitive stretch

If asked to also make phenol from benzenediazonium chloride, the answer is shorter still: simply warm the diazonium salt with water, $\text{C}_6\text{H}_5\text{N}_2^+ + \text{H}_2\text{O} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{H}^+$ , with no fused alkali and no high pressure needed at all — because the leaving group here is dinitrogen gas, an excellent leaving group that departs on its own once warmed, not a resonance-stabilised aryl-halogen bond.

Common traps

Treating a haloarene like a haloalkane and expecting mild aqueous NaOH to hydrolyse it — aryl halides need fused NaOH at high temperature and pressure, not gentle aqueous base.
Getting Markovnikov's rule direction backwards: acid-catalysed hydration puts OH on the more substituted carbon, while hydroboration-oxidation is anti-Markovnikov and puts OH on the less substituted carbon.
Assuming carbonyl reduction always gives the same class of alcohol — aldehydes give 1° alcohols, ketones give 2° alcohols, regardless of which reducing agent is used.
Naming only phenol as the cumene process product and forgetting acetone — the process always gives both, and acetone is the reason the route is economically preferred.

Ask the tutor: unsure which preparation route fits a given starting material, or want the resonance argument for haloarene resistance walked through step by step? Bring it to the AI tutor.

Why can diazonium salt hydrolysis happen so much more easily than the Dow process, even though both start from a benzene ring

The difference is the leaving group, not the ring. In bromobenzene, the leaving group is bromide, held on by a resonance-stabilised, partial-double-bond C–Br bond that resists substitution. In a diazonium salt, the leaving group is dinitrogen gas, which departs spontaneously on warming because forming stable N2 is highly favourable and the C–N bond, unlike C–Br, does not benefit from the same aromatic stabilisation once the reaction gets underway. That is why warm water alone is enough for the diazonium route, while bromobenzene needs fused NaOH at 623 K and 300 atm.

Is the Dow process still used industrially, given the cumene process exists

Not as the primary route. The cumene process dominates modern industrial phenol production because it runs under milder conditions, uses cheap and abundant benzene, propene and air, and produces acetone as a saleable co-product that offsets cost. The Dow process is still taught because it cleanly demonstrates the resonance argument for why aryl halides resist nucleophilic substitution, and it remains a viable laboratory-scale or historical route, but for large-scale manufacture cumene has been the industry standard for decades.

Do hydroboration-oxidation and acid-catalysed hydration ever give the same alcohol from the same alkene

Only for a symmetrical alkene, where there is no distinction between "more substituted" and "less substituted" carbon, so both routes place OH on chemically equivalent positions and give the same product. For an unsymmetrical alkene the two methods are deliberately complementary: acid-catalysed hydration follows Markovnikov's rule and puts OH on the more substituted carbon, while hydroboration-oxidation is anti-Markovnikov and puts OH on the less substituted carbon, so a chemist chooses the reagent based on which regiochemistry the target alcohol requires.