An organic compound A, molecular formula C6H6O, gives a violet colour with neutral FeCl3 solution. A on treatment with CO2 and NaOH at 400 K under pressure gives B, which on acidification gives a compound C. Compound C reacts with acetyl chloride to give D, which is a popular painkiller.
In the reaction sequence given in the passage, what is the compound (C)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Salicylic acid
(A), \( \ce{C6H6O} \), is phenol. With NaOH it gives sodium phenoxide, which takes up \( \ce{CO2} \) at the ortho position under pressure (Kolbe’s reaction) to give sodium salicylate (B):
\( \displaystyle \ce{C6H5OH + NaOH -> C6H5ONa + H2O} \)
\( \displaystyle \ce{C6H5ONa + CO2} \xrightarrow[\text{pressure}]{410\,\mathrm{K}} \ce{HOC6H4COONa} \)
Acidification replaces Na by H and gives 2-hydroxybenzoic acid, salicylic acid (C):
\( \displaystyle \ce{HOC6H4COONa + HCl -> HOC6H4COOH + NaCl} \)
Acetylation of its –OH group by acetyl chloride then gives aspirin (D), which confirms (C).
Why the other options are incorrect
Option (1) Benzoic acid — Incorrect. \( \ce{C6H5COOH} \) has no –OH group; the phenolic –OH of (A) stays in (C), and it is this –OH that is acetylated to give aspirin.
Option (3) Phthalic acid — Incorrect. Phthalic acid, \( \ce{C6H4(COOH)2} \), has two –COOH groups and no –OH; Kolbe’s reaction adds only one –COOH and keeps the –OH.
NCERT (2026-27 edition):
2. Kolbe’s reaction
2. Esterification
