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The KOLBE reaction converts phenol (in the form of sodium phenoxide) into salicylic acid. The conditions and key reagent are

A$\mathrm{CO_2}$ under pressure
B$\mathrm{CHCl_3}$ + NaOH
C$\mathrm{Br_2}$ in $\mathrm{CS_2}$
D$\mathrm{H_2SO_4}$ at $443\,\text{K}$
Answer & Solution
Correct answer: A. $\mathrm{CO_2}$ under pressure
1. NCERT §7.7 (Phenol reactions): KOLBE reaction (also Kolbe-Schmitt) introduces a carboxylic acid group onto the phenoxide. 2. PROCEDURE: sodium phenoxide is heated with $\mathrm{CO_2}$ under PRESSURE (~125°C, ~4 atm), then ACIDIFIED. 3. MECHANISM: the phenoxide ion (activated nucleophile, with negative charge delocalised to ortho/para) attacks CO$_2$ at the ortho position. After workup with acid, the product is ORTHO-hydroxybenzoic acid (SALICYLIC ACID). 4. PRODUCT: salicylic acid is the precursor of ASPIRIN (acetylsalicylic acid) — historically important industrial chemistry. 5. Option B describes the REIMER-TIEMANN reaction (gives aldehyde, not acid). Option C is mild bromination (gives bromophenol). Option D is dehydration of alcohol, unrelated. _Source: NCERT Class 12 Chemistry Part 2, Ch 7, §7.7 (Kolbe reaction), p. 18–19._
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