In Kolbe's reaction, phenol undergoes:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Electrophilic substitution
In Kolbe’s reaction, phenol is first converted to sodium phenoxide by NaOH. The phenoxide ion is even more reactive than phenol towards electrophilic aromatic substitution, so it is attacked by carbon dioxide, a weak electrophile, at the ortho position. Acidification then gives salicylic acid (2-hydroxybenzoic acid):
\( \displaystyle \ce{C6H5OH + NaOH -> C6H5ONa + H2O} \)
\( \displaystyle \ce{C6H5ONa + CO2} \xrightarrow{\text{400 K, 4-7 atm}} \ce{HOC6H4COONa} \)
\( \displaystyle \ce{HOC6H4COONa + H+ -> HOC6H4COOH + Na+} \)
A ring hydrogen is replaced by –COOH and the aromatic ring is kept, so it is a substitution, with \( \ce{CO2} \) as the electrophile.
Option (1) Electrophilic addition — Incorrect. The aromatic ring is kept; nothing adds across a double bond.
Option (3) Nucleophilic addition — Incorrect. This is the typical reaction of C=O compounds; here the ring is the electron-rich partner and \( \ce{CO2} \) is the electrophile.
Option (4) Nucleophilic substitution — Incorrect. No leaving group on the ring is replaced by a nucleophile; a ring hydrogen is replaced by an electrophile.
NCERT (2026-27 edition):
Kolbe’s Reaction
