Identify compound (Z) in the following reaction sequence: C6H5Cl + NaOH 623 K, 300 atm X HCl Y Conc. HNO3 Z

Step 1: Chlorobenzene + NaOH (623 K, 300 atm) → X

This is the Dow process (SNAr on unactivated ring under forcing conditions): $\ce{C6H5Cl + 2NaOH -> C6H5ONa + NaCl + H2O}$ X = sodium phenoxide

Step 2: X + HCl → Y

$\ce{C6H5ONa + HCl -> C6H5OH + NaCl}$ Y = Phenol

Step 3: Phenol + Conc. HNO3 → Z

Phenol undergoes electrophilic aromatic nitration. -OH is a strong ortho/para director.

With concentrated (but not fuming/excess) HNO3:

• First nitration: gives a mixture of o-nitrophenol and p-nitrophenol.
• With conc. HNO3 (moderate excess): second nitration can occur at the remaining activated positions → 2,4-dinitrophenol.

Note: -NO2 at para partially deactivates the ring, but the remaining o/p positions relative to -OH are still reactive.

Z = 2,4-Dinitrophenol ✓

Key Points to Remember:

• Chlorobenzene → SNAr (harsh) → phenol.
• Phenol + HNO3 → primarily ortho/para nitration.
• With conc. HNO3: can get dinitration → 2,4-dinitrophenol.
• With excess fuming HNO3: picric acid (2,4,6-trinitrophenol).