Preparation of Phenol

ChemistryAlcohols, Phenols And EthersFor JEE aspirants

The preparation of phenol uses four routes: hydrolysis of chlorobenzene with NaOH at 623 K and 300 atm (Dow process), alkali fusion of sodium benzenesulphonate, warming a benzenediazonium salt with water, and the cumene process, which makes phenol and acetone together from benzene and propene. In every case the key idea of the preparation of phenol is the same: put an $- OH$ group on an $s p^{2}$ ring carbon, usually as phenoxide first. Expect one direct question in NEET and JEE Main.

On this page1Four routes2From haloarenes3From sulphonic acid4From diazonium salts5Cumene process6Choose a route7Examples

Key Formulas - Quick Reference

★ Must learnDow: $C_{6} H_{5} Cl$ + NaOH $623 K, 300 atm$ $C_{6} H_{5} ONa$ $H^{+}$ $C_{6} H_{5} OH$
★ Must learnSulphonate: $C_{6} H_{5} S O_{3} Na$ + 2NaOH $\to$ $C_{6} H_{5} ONa$ + $N a_{2} S O_{3}$ + $H_{2} O$ (fusion), then $H^{+}$
★ Must learnDiazonium: $C_{6} H_{5} N_{2}^{+} C l^{-}$ + $H_{2} O$ $warm$ $C_{6} H_{5} OH$ + $N_{2}$ + HCl
Diazotisation needs 273-278 K; the salt decomposes if warmed before water is added
★ Must learnCumene: $C_{6} H_{5} CH (C H_{3})_{2}$ $O_{2}$ hydroperoxide $H_{3} O^{+}$ $C_{6} H_{5} OH$ + $C H_{3} COC H_{3}$
★ Must learno/p $N O_{2}$ groups make haloarene hydrolysis easier: 623 K (none), 443 K (one), 368 K (two), 328 K (three)
Phenol always forms as phenoxide in alkali; acidify to isolate it

1. Four Routes to Phenol

Phenol, $C_{6} H_{5} OH$ , cannot be made the way alcohols are. You cannot hydrate a benzene ring or reduce a carbonyl group to it. Instead, every laboratory route replaces a group that is already on the ring by $- OH$ , and the industrial route builds the molecule from benzene and propene.

Figure 1: Four starting points, one product. Three routes replace a group already on the ring (Cl, SO

_{3}

H, N

_{2}^{+}

) by OH; the cumene route builds phenol from benzene and propene and gives acetone as a valuable by-product.

Route	Starting material	Reagent and conditions	Where used
Dow process	chlorobenzene	NaOH, 623 K, 300 atm; then dil. HCl	industry (older plants)
Alkali fusion	benzenesulphonic acid	oleum; then fused NaOH; then $H^{+}$	older industrial route
Diazonium hydrolysis	aniline (via diazonium salt)	$NaN O_{2}$ + HCl at 273-278 K; warm with water	laboratory
Cumene process	benzene + propene	$H_{3} PO_{4}$ ; air; dil. acid	about 90% of world phenol

2. From Haloarenes: the Dow Process

Chlorobenzene is fused with sodium hydroxide at 623 K under 300 atm. The product is sodium phenoxide, because phenol is acidic enough to be neutralised by the excess alkali. Dilute hydrochloric acid then liberates phenol.

C_{6} H_{5} Cl + 2NaOH 623 K, 300 atm C_{6} H_{5} ONa + NaCl + H_{2} O

C_{6} H_{5} ONa + HCl C_{6} H_{5} OH + NaCl

Figure 2: Dow process. The aryl C-Cl bond has partial double-bond character, so the swap needs 623 K and 300 atm; the product is the phenoxide salt, freed as phenol by acid.

Why such drastic conditions? The C-Cl bond in chlorobenzene has partial double-bond character (a lone pair of chlorine is delocalised into the ring), the carbon is $s p^{2}$ and the $π$ cloud repels an incoming $OH^{-}$ . An alkyl chloride is hydrolysed by aqueous NaOH on warming; chlorobenzene needs 623 K.

Chloroethane,

C_{2} H_{5} Cl

C-Cl on an

s p^{3}

carbon
single bond, 177 pm
aq. NaOH, warm: ethanol

Chlorobenzene,

C_{6} H_{5} Cl

C-Cl on an

s p^{2}

carbon
partial double bond, 169 pm
NaOH, 623 K, 300 atm: phenol

Electron-withdrawing nitro groups at the ortho and para positions stabilise the negative intermediate formed when $OH^{-}$ adds, so the hydrolysis gets steadily easier.

Figure 3: Each nitro group at an ortho or para position lowers the temperature needed, from 623 K to 328 K, because it stabilises the negative intermediate (Meisenheimer complex).

Exam Trick

"Six-Four-Three-Three." The temperatures fall 623, 443, 368, 328 K as nitro groups are added at 2, 4 and 6. Only ortho and para nitro groups help; a meta nitro group cannot take the negative charge by resonance.

JEE Advanced

Under the Dow conditions most of the phenol is formed by elimination-addition through benzyne. $OH^{-}$ removes the proton next to Cl, chloride leaves, and the strained ring alkyne adds water. The proof: chlorobenzene labelled with $^{14}$ C at C-1 gives phenol with the label at C-1 and at C-2 in roughly equal amounts, which a direct substitution could never do.

Figure 4: At 623 K most of the reaction goes by elimination-addition through benzyne.

OH^{-}

adds to either end of the strained triple bond, which is why a labelled C-1 is found at C-1 and C-2 in roughly equal amounts.

Key idea

A haloarene gives phenol only under forcing conditions, unless nitro groups at o/p positions pull electron density out of the ring.

3. From Benzenesulphonic Acid

Benzene is sulphonated with oleum (fuming sulphuric acid). The benzenesulphonic acid is converted to its sodium salt and fused with solid sodium hydroxide at about 623 K. The melt contains sodium phenoxide, and acidification gives phenol.

C_{6} H_{6} + H_{2} S O_{4} oleum C_{6} H_{5} S O_{3} H + H_{2} O

C_{6} H_{5} S O_{3} Na + 2NaOH fused, 623 K C_{6} H_{5} ONa + N a_{2} S O_{3} + H_{2} O

C_{6} H_{5} ONa + HCl C_{6} H_{5} OH + NaCl

Figure 5: Sulphonate route. Oleum puts SO

_{3}

H on the ring; fusion with NaOH at about 623 K replaces it by O

^{-}

; acid then frees phenol.

Two moles of NaOH are used per mole of sulphonate: one replaces

- SO_{3} Na

(leaving as sulphite) and one neutralises the phenol formed. The same fusion turns naphthalene-2-sulphonic acid into 2-naphthol.

4. From Diazonium Salts

An aromatic primary amine is treated with sodium nitrite and hydrochloric acid at 273-278 K (diazotisation). The benzenediazonium chloride formed is unstable above this temperature. When its solution is warmed with water, or steam distilled with dilute acid, nitrogen escapes and phenol forms.

C_{6} H_{5} N H_{2} + NaN O_{2} + 2HCl 273-278 K C_{6} H_{5} N_{2}^{+} C l^{-} + NaCl + 2 H_{2} O

C_{6} H_{5} N_{2}^{+} C l^{-} + H_{2} O warm C_{6} H_{5} OH + N_{2} ↑ + HCl

Figure 6: Diazonium route. Diazotise in the cold (273-278 K), then warm with water: N

_{2}

leaves as a gas and OH takes its exact position, so any substitution pattern of the amine is kept.

The great advantage is precision: the $- OH$ group appears exactly where the $- NH_{2}$ group was. A modern variant uses copper(I) oxide in a solution of copper(II) nitrate, which lets the hydrolysis run at room temperature and avoids side reactions (the source's 2-bromo-4-methylphenol example uses it).

Exam Trick

"Cold to make, warm to break." Diazotise at 273-278 K, then warm with water. If a question gives $NaN O_{2}$ + HCl followed by warming, the answer is the phenol with OH where the $NH_{2}$ was.

Quick Recall: tap to check

Temperature for diazotisation?

273-278 K (0-5 °C).

Which gas is evolved when benzenediazonium chloride is warmed with water?

Nitrogen,

N_{2}

Why is 3-bromophenol made through the diazonium salt and not by brominating phenol?

OH directs ortho and para; the amine route puts OH exactly where

NH_{2}

was (meta to Br).

5. From Cumene: the Industrial Route

Most of the world's phenol is made from cumene (isopropylbenzene). Benzene and propene give cumene over a phosphoric acid catalyst; air oxidises cumene at its benzylic C-H to cumene hydroperoxide; dilute acid then cleaves the hydroperoxide into phenol and acetone.

C_{6} H_{6} + C H_{3} CH = C H_{2} H_{3} P O_{4}, 523 K C_{6} H_{5} CH (C H_{3})_{2}

C_{6} H_{5} CH (C H_{3})_{2} + O_{2} air C_{6} H_{5} C (C H_{3})_{2} OOH

C_{6} H_{5} C (C H_{3})_{2} OOH H_{3} O^{+} C_{6} H_{5} OH + C H_{3} COC H_{3}

Figure 7: Cumene process, three steps. Friedel-Crafts alkylation, air oxidation at the benzylic C-H, and acid cleavage. Each mole of phenol comes with one mole of acetone.

The acid step is a rearrangement. After the terminal oxygen is protonated, the phenyl group migrates from carbon to oxygen as water leaves. Water then adds to the cation, and the hemiketal splits into the two products.

Figure 8: Why the product is phenol and not a benzene alcohol: the phenyl group migrates from carbon to the electron-poor oxygen as water leaves (a 1,2-shift to oxygen), so the O-C

_{6}

_{5}

bond is made before the molecule splits into phenol and acetone.

Key idea

Cumene process: three cheap inputs (benzene, propene, air) give two valuable outputs (phenol and acetone) in a 1:1 molar ratio.

6. Choosing a Route

Questions usually give a starting material and ask for the reagent, or give the reagents and ask for the product. The flowchart turns the four routes into four questions; the mind map after it holds every condition on one screen.

Figure 9: Flowchart: pick the route from what you start with. Only the diazonium route puts OH exactly where you want it on any ring.

Quick Recall: tap to check

Starting material and by-product of the cumene process?

Benzene and propene; acetone is the by-product.

Conditions for the Dow process?

NaOH, 623 K, 300 atm, then acidification.

Reagent that converts sodium benzenesulphonate into sodium phenoxide?

Fused sodium hydroxide.

Figure 10: Mind map: four routes, their conditions and the two mechanisms behind them.

7. Solved Examples

Solved Example 1

Which of these gives phenol and acetone as the two products of one process?
(A) cumene hydroperoxide with dil. acid
(B) chlorobenzene with NaOH at 623 K
(C) benzenediazonium chloride with water
(D) sodium benzenesulphonate with fused NaOH

Solution:

Answer: (A). Acid cleavage of cumene hydroperoxide gives $C_{6} H_{5} OH$ and $C H_{3} COC H_{3}$ in a 1:1 ratio. (B) and (D) give phenoxide (then phenol) with NaCl or $N a_{2} S O_{3}$ ; (C) gives phenol, $N_{2}$ and HCl.

Solved Example 2

Hydrolysis of which compound needs the mildest conditions?
(A) chlorobenzene
(B) 4-chloronitrobenzene
(C) 1-chloro-2,4-dinitrobenzene
(D) 2-chloro-1,3,5-trinitrobenzene

Solution:

Answer: (D). Three nitro groups at the ortho and para positions to Cl stabilise the negative intermediate so well that warm water alone (about 328 K) replaces Cl by OH, giving picric acid.

Solved Example 3

How much phenol and acetone can 1.00 tonne of cumene give at 100% yield?

Solution:

One mole of cumene ( $M = 120.2$ g mol $^{- 1}$ ) gives one mole of phenol (94.1) and one mole of acetone (58.1). Phenol $= 1000 \times 94.1/120.2 = 783$ kg; acetone $= 1000 \times 58.1/120.2 = 483$ kg. The two masses add up to 1266 kg because 1 mol of $O_{2}$ (32 g) is taken up per mole.

Solved Example 4

Suggest a preparation of 3-bromophenol from 3-bromoaniline, and explain why brominating phenol does not work.

Solution:

Diazotise 3-bromoaniline with $NaN O_{2}$ + HCl at 273-278 K and warm the diazonium salt with water: the OH takes the place of $NH_{2}$ , giving 3-bromophenol. Brominating phenol cannot work: OH is an ortho/para director, so bromine would enter at 2, 4 and 6, never at 3.

Solved Example 5

Why is the product of the Dow process sodium phenoxide and not phenol, and what is used to release phenol?

Solution:

Phenol ( $p K_{a} \approx 10$ ) is a much stronger acid than water, so in the hot alkali it is deprotonated to $C_{6} H_{5} O^{-}$ . A stronger acid, dilute HCl (or $C O_{2}$ in water), protonates the phenoxide and releases phenol.

Solved Example 6

Chlorobenzene labelled with

^{14}

C at C-1 is hydrolysed under Dow conditions. Where is the label in the phenol?

Solution:

At C-1 in about half the molecules and at C-2 in the other half. The reaction goes through benzyne, whose two triple-bond carbons are C-1 and C-2; $OH^{-}$ adds to either with nearly equal probability.

Solved Example 7

Identify A and B: aniline

NaNO_{2}, HCl, 273 K

H_{2} O, warm

Solution:

A is benzenediazonium chloride, $C_{6} H_{5} N_{2}^{+} C l^{-}$ ; B is phenol, $C_{6} H_{5} OH$ (with $N_{2}$ and HCl).

Solved Example 8

10.0 g of chlorobenzene is converted to phenol by the Dow process. What is the maximum mass of phenol?

Solution:

$M (C_{6} H_{5} Cl) = 112.6$ g mol $^{- 1}$ and $M (C_{6} H_{5} OH) = 94.1$ g mol $^{- 1}$ , one to one. Mass of phenol $= 10.0 \times 94.1/112.6 = 8.36$ g.

Practice Questions

Name the by-product of the cumene process.Answer: acetone (propanone).
Give the conditions of the Dow process.Answer: NaOH, 623 K, 300 atm, then dil. HCl.
Write the product when benzenediazonium chloride is warmed with water.Answer: phenol, with $N_{2}$ and HCl.
Which reagent converts benzene into benzenesulphonic acid?Answer: oleum (fuming $H_{2} SO_{4}$ ).
Why is 4-nitrochlorobenzene hydrolysed more easily than chlorobenzene?Answer: the para nitro group stabilises the negative intermediate by resonance.
Suggest a route from 4-methylaniline to 4-methylphenol (p-cresol).Answer: diazotise at 273-278 K, then warm with water.
How many moles of NaOH does the alkali fusion of one mole of sodium benzenesulphonate use?Answer: two.

Common Mistakes to Avoid

Watch out

Writing phenol as the direct product of the Dow process. Alkali gives sodium phenoxide; acid is needed afterwards.
Using aqueous NaOH at room temperature for chlorobenzene. Plain haloarenes need 623 K and 300 atm.
Diazotising at room temperature. The diazonium salt must be made at 273-278 K and only then warmed with water.
Forgetting acetone in the cumene process. Every mole of phenol comes with one mole of acetone.
Writing oxidation of cumene at a ring carbon. Air attacks the benzylic C-H of the isopropyl group.
Thinking a meta nitro group helps hydrolysis. Only ortho and para nitro groups stabilise the intermediate.
Writing only one mole of NaOH for the sulphonate fusion. Two are used: one for the swap, one to neutralise phenol.
Using direct bromination of phenol to make 3-bromophenol. OH directs to 2, 4, 6; use the diazonium route.

Frequently Asked Questions

How is phenol prepared in the laboratory?

The usual laboratory method starts from aniline. It is diazotised with sodium nitrite and hydrochloric acid at 273-278 K, and the benzenediazonium chloride solution is warmed with water. Nitrogen gas escapes and phenol forms, with the OH group exactly where the amino group was.

What is the Dow process for phenol?

It is the hydrolysis of chlorobenzene by sodium hydroxide at 623 K and 300 atm. The product is sodium phenoxide, which is acidified with dilute hydrochloric acid to give phenol. The harsh conditions are needed because the aryl C-Cl bond has partial double-bond character.

Why is the cumene process the main industrial route to phenol?

It uses cheap benzene, propene and air, runs under mild conditions and gives acetone, another valuable chemical, as a co-product in a 1:1 molar ratio. About 90 per cent of the world's phenol is made this way.

What is the by-product in the manufacture of phenol from cumene?

Acetone. Acid cleavage of cumene hydroperoxide gives one mole of phenol and one mole of acetone. One tonne of cumene can give at most about 783 kg of phenol and 483 kg of acetone.

Why do nitro groups make chlorobenzene easier to hydrolyse?

A nitro group at the ortho or para position withdraws electron density and stabilises, by resonance, the negative intermediate formed when hydroxide adds. The required temperature falls from 623 K to 443 K with one nitro group and to about 328 K with three.

How is phenol obtained from benzenesulphonic acid?

Benzene is sulphonated with oleum. The sodium salt of benzenesulphonic acid is fused with sodium hydroxide at about 623 K, which gives sodium phenoxide and sodium sulphite. Acidifying the melt then liberates phenol.

Which preparation of phenol questions are common in NEET?

NEET usually asks for the by-product of the cumene process, the reagents and temperature for converting aniline to phenol through the diazonium salt, or the conditions of the Dow process. Remembering acetone, 273-278 K and 623 K with 300 atm answers most of them.

How does JEE Main test the preparation of phenol?

JEE Main links the routes to other chapters: a nitro-substituted haloarene hydrolysed under mild conditions, a substituted aniline taken through its diazonium salt to a specific phenol isomer, or the migration step in the cumene hydroperoxide rearrangement. Track where each substituent ends up.

Previous year questions on Preparation of Phenol

6 questions from past papers, each with a step-by-step solution.

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