JEE Advanced 2022 Paper 1, Chemistry Section 1 Q7: Preparation of Phenol

Step 1 (Dow process): chlorobenzene + NaOH (623 K, 300 atm) gives sodium phenoxide, which on acid work-up gives phenol.

Step 2: phenol is first sulphonated by conc. H${}_2$SO${}_4$ at the ortho and para positions (the SO${}_3$H groups act as blocking and activating positions, also help control product distribution), then nitrated by conc. HNO${}_3$. The sulphonic groups are displaced during nitration, yielding 2,4,6-trinitrophenol (picric acid) as the major product Q.

Composition of picric acid C${}_6$H${}_3$N${}_3$O${}_7$:

Molar mass = $6(12)+3(1)+3(14)+7(16)=72+3+42+112=229$ g/mol.

Weight percent of H = $\frac{3}{229}\times 100=1.31\%$.