Identify A and B in the following reaction sequence. Bromobenzene Conc. HNO3 (excess) A (i) NaOH; (ii) HCl B

Step 1: Bromobenzene + conc. HNO3 (excess) → A

Bromobenzene undergoes electrophilic nitration. With excess concentrated $\ce{HNO3}$, multiple nitrations occur:

• Br is an ortho/para director (despite being weakly deactivating).
• With excess reagent: 3 nitro groups are introduced at positions 2, 4, and 6 (all ortho/para to Br).
• A = 1-Bromo-2,4,6-trinitrobenzene (picryl bromide)

Step 2: A + (i) NaOH, (ii) HCl → B

1-Bromo-2,4,6-trinitrobenzene has three $\ce{-NO2}$ groups at positions ortho and para to the Br. This activates the ring enormously toward nucleophilic aromatic substitution (SNAr):

• $\ce{OH-}$ (from NaOH) attacks the C bearing Br → Meisenheimer complex forms.
• $\ce{Br-}$ is displaced as leaving group.
• Product after acid workup: 2,4,6-trinitrophenol (picric acid)

B = 2,4,6-Trinitrophenol (picric acid)

Step 3: Key insight

The three $\ce{-NO2}$ groups at ortho/para positions to Br make the C-Br carbon highly electrophilic, enabling easy SNAr. This is the same principle as picryl chloride reacting even with water.

Key Points to Remember:

• Bromobenzene + excess conc. HNO3 → trinitration at 2,4,6 (o/p to Br).
• 1-Bromo-2,4,6-trinitrobenzene undergoes facile SNAr with NaOH → picric acid.
• More NO2 groups at o/p to leaving group → faster SNAr.