Question

Write the reaction involved in the following:

Kolbe's reaction

Hint:

Kolbe's reaction introduces a \(-COOH\) group at the ortho position of phenol through its phenoxide ion reacting with carbon dioxide.

Answer 1

Step 1: Aim of the reaction.
Kolbe's reaction (Kolbe-Schmitt reaction) puts a \(-COOH\) (carboxylic acid) group onto the phenol ring at the ortho position, giving salicylic acid.

Step 2: First make the phenoxide.
Phenol is treated with sodium hydroxide \((NaOH)\) to form sodium phenoxide. The phenoxide ion is much more reactive towards electrophiles than phenol itself, because of its higher electron density on the ring.

Step 3: React with carbon dioxide.
The sodium phenoxide is treated with carbon dioxide \((CO_2)\) under pressure at about 400 K. \(CO_2\) acts as the weak electrophile here.

Step 4: Position of attack.
The \(CO_2\) attacks the reactive ortho position of the ring, forming sodium salicylate.

Step 5: Acidify.
On treating sodium salicylate with acid, salicylic acid (2-hydroxybenzoic acid) is obtained.

Step 6: Overall reaction.
\[ C_6H_5ONa + CO_2 \xrightarrow{400\,K,\ \text{pressure}} o\text{-}HOC_6H_4COONa \xrightarrow{H^+} o\text{-}HOC_6H_4COOH \]

Answer: Phenol is converted to sodium phenoxide, which reacts with \(CO_2\) under pressure and is then acidified to give salicylic acid (2-hydroxybenzoic acid). This is Kolbe's reaction.