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Most phenol preparations in the textbook need harsh conditions or expensive starting materials. The one used in industry needs a hydrocarbon, air and dilute acid, and it makes a second useful product at the same time. This post explains the steps, the mechanism behind the second one, the mass balance and the traps in exam questions.
Where is this in the syllabus?
The NCERT Class 12 chapter Alcohols, Phenols and Ethers (2026-27 reprint) lists four preparations of phenol, the fourth "From cumene", with a margin note: "Most of the worldwide production of phenol is from cumene." The CBSE 2026-27 Chemistry curriculum includes this unit. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list general methods of preparation for oxygen compounds, and the JEE (Advanced) 2026 brochure lists the preparation of phenols. Check the current year's documents.
What happens in each step?
Cumene itself is isopropylbenzene. OpenStax shows it formed by Friedel–Crafts alkylation of benzene with 2-chloropropane; our post on electrophilic substitution in benzene covers that reaction.
Step 1, air oxidation: NCERT says cumene is oxidised in the presence of air to cumene hydroperoxide, C₆H₅C(CH₃)₂–O–O–H. OpenStax explains that this is a radical oxidation at the benzylic carbon, the one attached to the ring.
Step 2, dilute acid: the hydroperoxide splits into phenol and acetone.
The chemistry has a long industrial history. In 1957 H. Hock and H. Kropf published a review of hydrocarbon autoxidation and the cumene–phenol synthesis that covered the acid cleavage of hydroperoxides and the industrial process of the time.
Why does a phenyl group move?
OpenStax gives the mechanism for step 2:
Acid protonates the terminal oxygen of the hydroperoxide.
The phenyl group shifts from carbon to the neighbouring oxygen as water leaves.
Water adds to the resulting cation, giving a hemiacetal (one carbon carrying both an –OR and an –OH).
The hemiacetal breaks down to phenol and acetone.
Step 2 is a 1,2-shift onto an electron-poor oxygen, the same logic as the hydride and methyl shifts in our post on carbocation rearrangements. The ring ends up on oxygen, which is why the product is phenol, and the three side-chain carbons become acetone.
Worked example: how much acetone per tonne of phenol?
Overall: C₉H₁₂ + O₂ → C₆H₅OH + (CH₃)₂CO.
Check the balance: C₉H₁₂O₂ on each side. Molar masses: cumene 120.19, O₂ 32.00, phenol 94.11 and acetone 58.08 g/mol, and 120.19 + 32.00 = 94.11 + 58.08 = 152.19.
For 1000 kg of phenol:
Moles: 1000 ÷ 94.11 = 10.63 kmol.
Acetone made alongside: 10.63 × 58.08 ≈ 617 kg.
Cumene needed, at 100% yield: 10.63 × 120.19 ≈ 1277 kg.
Oxygen used: 10.63 × 32.00 ≈ 340 kg.
Atom economy, counting only phenol as useful, is 94.11 ÷ 152.19 × 100 ≈ 61.8%. If the acetone is sold too, every atom in the idealised equation ends up in a product. OpenStax calls the process efficient for this reason: two valuable chemicals come out at once. The arithmetic also has a consequence: this route cannot make a tonne of phenol without also making about 0.62 tonne of acetone.
How does it compare with the other routes?
NCERT's other three preparations:
From chlorobenzene: fused with NaOH at 623 K and 320 atm, then acidified.
From benzenesulphonic acid: benzene sulphonated with oleum, heated with molten NaOH, then acidified.
From diazonium salts: aniline with NaNO₂ and HCl at 273–278 K, then warmed with water.
OpenStax notes that phenol was made for many years by the Dow process, from chlorobenzene and NaOH, before the cumene route took over.
Which mistakes cost marks?
Starting step 1 from benzene. Air oxidises cumene, not benzene.
Drawing a methyl shift in step 2. The phenyl group moves.
Getting the by-product wrong. It is acetone (propanone), from the isopropyl group, not acetaldehyde or acetic acid.
Confusing it with KMnO₄ oxidation. The JEE Advanced brochure lists making benzoic acid from alkylbenzenes; a strong oxidant cuts the side chain to –COOH. Air followed by acid gives phenol.
Forgetting the intermediate's name, cumene hydroperoxide, which questions often ask for.
For phenol's acidity and tests, see our post on alcohols, phenols and ethers.
A practice routine
Write NCERT's four phenol preparations from memory, with conditions.
Draw the four-step mechanism of step 2 with curved arrows.
Redo the mass balance for 250 kg of phenol. You should get about 154 kg of acetone.
Answer NCERT Exercise 7.9: the equations for preparing phenol from cumene.
When a mechanism step will not come out right, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, draw cumene, colour its three side-chain carbons, and follow them into acetone.
