Question

Write the reaction involved in Kolbe's reaction.

Hint:

In Kolbe's reaction (Kolbe-Schmitt reaction), sodium phenoxide is heated with carbon dioxide (CO 2 ) under pressure. The CO 2 adds a -COOH group at the ortho position.

Answer 1

Step 1: Note the starting material.
In Kolbe's reaction (Kolbe-Schmitt reaction), we start with sodium phenoxide, made by treating phenol with \(NaOH\).

Step 2: Note the reagent and conditions.
Sodium phenoxide is heated with carbon dioxide (\(CO_2\)) under pressure (about 400 K and 4 to 7 atm).

Step 3: See what gets added.
\(CO_2\) acts as the electrophile and adds a carboxyl-type group at the ortho position of the ring, giving sodium salicylate.

Step 4: Write the reaction.
\[C_6H_5ONa \xrightarrow{CO_2,\ \Delta,\ \text{pressure}} \xrightarrow{H^+} \text{2-hydroxybenzoic acid}\]
Step 5: Name the product.
After acidification the product is salicylic acid, that is 2-hydroxybenzoic acid (\(o\)-\(HO\!-\!C_6H_4\!-\!COOH\)).

Answer: Sodium phenoxide heated with \(CO_2\) under pressure, then acidified, gives salicylic acid (2-hydroxybenzoic acid).