Question

Write the reaction involved in Kolbe's reaction.

Hint:

Kolbe's reaction is an important method for preparing salicylic acid from phenol through the intermediate formation of sodium phenoxide.

Answer 1

Step 1: Convert phenol to sodium phenoxide.
Phenol reacts with NaOH to form the phenoxide ion, which is more reactive toward electrophiles than phenol: $C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O$.
Step 2: Carboxylation under pressure (Kolbe-Schmitt reaction).
Sodium phenoxide is heated with dry $CO_2$ at 373 K and 4-7 atm. $CO_2$ acts as the electrophile and undergoes electrophilic substitution at the ortho position (activated by the phenoxide oxygen): \[ C_6H_5ONa + CO_2 \xrightarrow[4\text{-}7\,atm]{373\,K} o\text{-}HOC_6H_4COONa \text{ (sodium salicylate)} \]
Step 3: Acidify to obtain salicylic acid.
Treatment with dilute HCl liberates salicylic acid from its sodium salt: $o\text{-}HOC_6H_4COONa + HCl \rightarrow o\text{-}HOC_6H_4COOH + NaCl$. \[ \boxed{\text{Product: } o\text{-}HOC_6H_4COOH \text{ (Salicylic acid / 2-Hydroxybenzoic acid)}} \]