JEE Mains Chemistry · Alcohols, Phenols and Ethers

Preparation of Phenols and Named Reactions

Phenol is made from chlorobenzene, benzenesulphonic acid, a diazonium salt or cumene, and it is recognised by its named reactions: zinc dust gives benzene, chromic acid gives benzoquinone, Reimer–Tiemann gives salicylaldehyde and Kolbe gives salicylic acid.

All Alcohols, Phenols and Ethers notes JEE Mains Chemistry strategy

Why this matters

Eighteen PYQs, sixteen of them multiple choice, and two from 2026. Seven ask how phenol is made, including a step-yield calculation and the cumene process with its hydroperoxide. Eleven test the named reactions: five on Reimer–Tiemann alone (its reagent, its intermediate, ortho against para), four are match lists of reagents and products, one is on zinc dust and one asks why HCl cannot replace the phenolic OH.

Concept 1 of 2: Four ways to make phenol

An OH has to be put on a benzene ring. A chlorine or a sulphonic acid group can be pushed off by hydroxide, but only under harsh conditions. A diazonium group leaves easily as nitrogen gas. Cumene is oxidised by air to a hydroperoxide, which splits in acid into phenol and acetone.

Definition

From chlorobenzene (Dow): fuse with NaOH at 623 K and about 300 atm, then acidify. The C–Cl bond has partial double-bond character, hence the harsh conditions.
From benzenesulphonic acid: benzene + oleum gives $\mathrm{C_6H_5SO_3H}$ ; fusion with NaOH and then acid gives phenol.
From a diazonium salt: aniline with $\mathrm{NaNO_2}$ /HCl at 273–278 K gives $\mathrm{C_6H_5N_2^+Cl^-}$ , which warm water hydrolyses to phenol.
From cumene: air oxidises cumene to cumene hydroperoxide, $\mathrm{C_6H_5C(CH_3)_2OOH}$ ; dilute acid splits it into phenol and propanone.
For a route of several steps, the overall yield is the product of the step yields.

Starting material	Reagents	Intermediate	Products
Chlorobenzene	NaOH, 623 K, about 300 atm; then $\mathrm{H^+}$	Sodium phenoxide	Phenol
Benzenesulphonic acid (benzene + oleum)	Fused NaOH; then $\mathrm{H^+}$	Sodium phenoxide	Phenol
Aniline	$\mathrm{NaNO_2}$ + HCl at 273–278 K; then warm water	Benzenediazonium chloride, $\mathrm{C_6H_5N_2^+Cl^-}$	Phenol, $\mathrm{N_2}$ and HCl
Cumene (isopropylbenzene)	$\mathrm{O_2}$ (air); then dilute acid	Cumene hydroperoxide, $\mathrm{C_6H_5C(CH_3)_2OOH}$	Phenol and propanone (acetone) The intermediate is a hydroperoxide (O–O–H), not an alcohol or an ester; acetone is the by-product.

Carbolic acid is the old name for phenol.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q135Moderate

Example 1 · Alcohols, Phenols and Ethers · Preparation of Phenols and Named Reactions

Hydrolysis of which compound will give carbolic acid?

Hydrolysis of benzal chloride gives benzaldehyde

\mathrm{C_6H_5CHCl_2}

with water gives

\mathrm{C_6H_5CHO}

, not phenol. Cumene does not hydrolyse at all; it must be oxidised to the hydroperoxide first.

Multiply step yields, do not add them

Two steps of 80% and 50% give 40% overall, not 65% or 130%.

Drill 6 more on four ways to make phenol

Concept 2 of 2: Named reactions of phenol

Most phenol reactions start from the phenoxide ion, whose ring is very electron-rich. A weak electrophile such as dichlorocarbene or carbon dioxide then attacks the ortho carbon. Other reactions remove the oxygen (zinc dust) or oxidise the ring (chromic acid).

Definition

Zinc dust: heating phenol with Zn removes the oxygen as ZnO and gives benzene.
Oxidation with $\mathrm{Na_2Cr_2O_7/H_2SO_4}$ gives benzo-p-quinone, a conjugated diketone.
Reimer–Tiemann: $\mathrm{CHCl_3}$ + aqueous NaOH make dichlorocarbene, $\mathrm{:CCl_2}$ . The phenoxide attacks it at the ortho carbon; the intermediate is a substituted benzal chloride, and hydrolysis then acid gives salicylaldehyde (ortho major, para minor).
Kolbe: sodium phenoxide + $\mathrm{CO_2}$ at about 400 K and 4–7 atm, then $\mathrm{H^+}$ , gives salicylic acid (2-hydroxybenzoic acid).
Acetylation: phenol + acetic anhydride gives phenyl acetate; salicylic acid gives aspirin.
Phenol does not give chlorobenzene with HCl: its C–O bond has partial double-bond character.

Reaction	Reagents	Product	Key point
Reduction by zinc dust	Zn, heat	Benzene	Zinc takes the oxygen as ZnO
Oxidation	$\mathrm{Na_2Cr_2O_7/H_2SO_4}$ (chromic acid)	Benzo-p-quinone	A conjugated diketone, O=C at C-1 and C-4
Reimer–Tiemann	$\mathrm{CHCl_3}$ + aq. NaOH, then $\mathrm{H^+}$	Salicylaldehyde (2-hydroxybenzaldehyde), ortho major	Electrophile $\mathrm{:CCl_2}$ ; intermediate is the o-(dichloromethyl)phenoxide, a substituted benzal chloride
Kolbe	NaOH; $\mathrm{CO_2}$ at 400 K, 4–7 atm; then $\mathrm{H^+}$	Salicylic acid (2-hydroxybenzoic acid)	$\mathrm{CO_2}$ is a weak electrophile, so the phenoxide ion is needed
Acetylation	Acetic anhydride or $\mathrm{CH_3COCl}$	Phenyl acetate; salicylic acid gives aspirin	Reaction at the O–H, not on the ring
Nitration with conc. $\mathrm{HNO_3}$	Conc. $\mathrm{HNO_3}$ (best via phenol-2,4-disulphonic acid)	Picric acid (2,4,6-trinitrophenol)	A yellow, strongly acidic solid
With HCl	Conc. HCl	No reaction: no chlorobenzene	The phenolic C–O bond does not break

CHCl₃ puts in a CHO group, CO₂ puts in a COOH group; both go ortho to the OH.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q124Moderate

Example 2 · Alcohols, Phenols and Ethers · Preparation of Phenols and Named Reactions

Match List-I with List-II.

	List-I (Reactants)		List-II (Products)
(A)	Phenol, $Zn/\Delta$	(I)	Salicylaldehyde
(B)	Phenol, $CHCl_{3}, NaOH, HCl$	(II)	Salicylic acid
(C)	Phenol, $CO_{2}, NaOH, HCl$	(III)	Benzene
(D)	Phenol, conc. $HNO_{3}$	(IV)	Picric acid

Reimer–Tiemann gives the ortho aldehyde as the major product

The phenoxide oxygen guides

\mathrm{:CCl_2}

to the ortho carbon. A statement that p-hydroxybenzaldehyde is the major product is false. The two isomers can be separated by steam distillation, because the ortho one is volatile.

Chloroform gives the aldehyde, carbon dioxide gives the acid

\mathrm{CHCl_3}

/NaOH makes salicylaldehyde;

\mathrm{CO_2}

/NaOH makes salicylic acid. Swapping them is the usual wrong option in a match list.

Drill 10 more on named reactions of phenol

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Reference tables (2)

Four ways to make phenol4 rows

Starting material	Reagents	Intermediate	Products
Chlorobenzene	NaOH, 623 K, about 300 atm; then $\mathrm{H^+}$	Sodium phenoxide	Phenol
Benzenesulphonic acid (benzene + oleum)	Fused NaOH; then $\mathrm{H^+}$	Sodium phenoxide	Phenol
Aniline	$\mathrm{NaNO_2}$ + HCl at 273–278 K; then warm water	Benzenediazonium chloride, $\mathrm{C_6H_5N_2^+Cl^-}$	Phenol, $\mathrm{N_2}$ and HCl
Cumene (isopropylbenzene)	$\mathrm{O_2}$ (air); then dilute acid	Cumene hydroperoxide, $\mathrm{C_6H_5C(CH_3)_2OOH}$	Phenol and propanone (acetone) The intermediate is a hydroperoxide (O–O–H), not an alcohol or an ester; acetone is the by-product.

Carbolic acid is the old name for phenol.

Named reactions of phenol7 rows

Reaction	Reagents	Product	Key point
Reduction by zinc dust	Zn, heat	Benzene	Zinc takes the oxygen as ZnO
Oxidation	$\mathrm{Na_2Cr_2O_7/H_2SO_4}$ (chromic acid)	Benzo-p-quinone	A conjugated diketone, O=C at C-1 and C-4
Reimer–Tiemann	$\mathrm{CHCl_3}$ + aq. NaOH, then $\mathrm{H^+}$	Salicylaldehyde (2-hydroxybenzaldehyde), ortho major	Electrophile $\mathrm{:CCl_2}$ ; intermediate is the o-(dichloromethyl)phenoxide, a substituted benzal chloride
Kolbe	NaOH; $\mathrm{CO_2}$ at 400 K, 4–7 atm; then $\mathrm{H^+}$	Salicylic acid (2-hydroxybenzoic acid)	$\mathrm{CO_2}$ is a weak electrophile, so the phenoxide ion is needed
Acetylation	Acetic anhydride or $\mathrm{CH_3COCl}$	Phenyl acetate; salicylic acid gives aspirin	Reaction at the O–H, not on the ring
Nitration with conc. $\mathrm{HNO_3}$	Conc. $\mathrm{HNO_3}$ (best via phenol-2,4-disulphonic acid)	Picric acid (2,4,6-trinitrophenol)	A yellow, strongly acidic solid
With HCl	Conc. HCl	No reaction: no chlorobenzene	The phenolic C–O bond does not break

CHCl₃ puts in a CHO group, CO₂ puts in a COOH group; both go ortho to the OH.

Watch out for (4)

Hydrolysis of benzal chloride gives benzaldehyde
→ Four ways to make phenol
Multiply step yields, do not add them
→ Four ways to make phenol
Reimer–Tiemann gives the ortho aldehyde as the major product
→ Named reactions of phenol
Chloroform gives the aldehyde, carbon dioxide gives the acid
→ Named reactions of phenol

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