Chemistry · JEE Advanced only

Kolbe Reaction of Phenols (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 1.5 to 2 hours. Spend 30 minutes understanding the mechanism and conditions, 30 minutes comparing with Reimer-Tiemann reaction, and 30-60 minutes solving practice problems involving identification of products..

Last updated: 26 October 2026

Direct answer

The Kolbe reaction is an electrophilic aromatic substitution where phenol reacts with carbon dioxide under pressure in the presence of sodium hydroxide to form salicylic acid (ortho-hydroxybenzoic acid).

The Kolbe reaction is an electrophilic aromatic substitution where phenol reacts with carbon dioxide under pressure in the presence of sodium hydroxide to form salicylic acid (ortho-hydroxybenzoic acid). The phenoxide ion attacks CO₂, and after acidification, you get a carboxylic acid group attached to the ring at the ortho position. This reaction is industrially important for making aspirin.

Official syllabus

What the JEE Advanced syllabus says

“Kolbe reaction”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Reimer - Tiemann reaction.

JEE Main covers Reimer-Tiemann reaction (which adds CHO group) but not Kolbe reaction. For Advanced, you need to know both reactions and be able to distinguish when each applies based on reagents.

Key ideas

What you must understand

Phenol is first converted to sodium phenoxide using NaOH
Sodium phenoxide reacts with CO₂ at high temperature (125°C) and pressure (4-7 atm)
The product after acidification is salicylic acid (2-hydroxybenzoic acid)
Ortho position is favored due to the mechanism involving a cyclic intermediate
At higher temperatures (above 200°C), para-hydroxybenzoic acid becomes the major product
This is an example of carboxylation of an aromatic ring

Formulas

Formulas and key facts

C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O
C₆H₅O⁻Na⁺ + CO₂ → o-HOC₆H₄COO⁻Na⁺ (at 125°C, 4-7 atm)
o-HOC₆H₄COO⁻Na⁺ + H⁺ → o-HOC₆H₄COOH (salicylic acid)
Overall: Phenol + CO₂ + NaOH → Salicylic acid (after acidification)
Salicylic acid + Acetic anhydride → Aspirin (acetylsalicylic acid)
Kolbe reaction: adds COOH group at ortho position
Reimer-Tiemann: adds CHO group at ortho position (uses CHCl₃/NaOH)

Avoid

Common mistakes

Confusing Kolbe reaction with Reimer-Tiemann reaction. Kolbe uses CO₂ and gives COOH. Reimer-Tiemann uses CHCl₃ and gives CHO.
Forgetting that phenol must first be converted to phenoxide ion. Neutral phenol does not react with CO₂.
Writing the product as para-isomer at normal conditions. At 125°C, ortho-product dominates. Para-product forms at much higher temperatures.
Not acidifying the final product. The immediate product is sodium salicylate, not salicylic acid.

Worked example

A JEE Advanced-level example

Question. Compound X is obtained when phenol is treated with NaOH followed by CO₂ at 125°C and 4-7 atm pressure, then acidified. Compound X is then treated with acetic anhydride in presence of a few drops of concentrated H₂SO₄. Identify the final product and write its IUPAC name.

Show solution

Step 1: Phenol + NaOH gives sodium phenoxide (C₆H₅O⁻Na⁺)
Step 2: Sodium phenoxide + CO₂ at 125°C, 4-7 atm gives sodium salicylate
Step 3: Acidification of sodium salicylate gives salicylic acid (Compound X = 2-hydroxybenzoic acid)
Step 4: Salicylic acid has both OH and COOH groups. With acetic anhydride and H₂SO₄, the phenolic OH gets acetylated
Step 5: Final product is acetylsalicylic acid (aspirin), IUPAC name: 2-acetoxybenzoic acid

Answer: The final product is Aspirin. IUPAC name: 2-acetoxybenzoic acid (or 2-acetyloxy benzoic acid)

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Kolbe Reaction of Phenols in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Reimer - Tiemann reaction.

How long does Kolbe Reaction of Phenols take to prepare for JEE Advanced?

About 1.5 to 2 hours. Spend 30 minutes understanding the mechanism and conditions, 30 minutes comparing with Reimer-Tiemann reaction, and 30-60 minutes solving practice problems involving identification of products., if you already know the related JEE Main topic.