The major product (A) formed in the following reaction sequence is

Solution

Chemistry ma’am · 5 steps

Acetylation tames the amino group so bromine adds only once, at para, and NaOH takes the acetyl off.

1. Reduce first. Sn and HCl reduce the nitro group to an amino group, so nitrobenzene becomes aniline, C₆H₅NH₂. The -NO₂ is gone before any bromine arrives, so its meta-directing effect never comes into play.
2. Protect the amine. Ac₂O with pyridine turns aniline into acetanilide, C₆H₅NHCOCH₃. The nitrogen lone pair is now shared with the C=O group, so the ring is only moderately activated, not strongly activated.
3. Brominate once. With Br₂ in acetic acid, the -NHCOCH₃ group is still an ortho, para director, but it is now gentle enough to allow just one substitution. The bulky acetamido group crowds the ortho positions, so bromine goes para, giving 4-bromoacetanilide.
4. Remove the acetyl. Aqueous NaOH hydrolyses the amide back to a free -NH₂ group. The bromine stays where it is on the ring.
5. So. A is 4-bromoaniline (p-bromoaniline), option (b).

The idea to keep. To get monobromo aniline, protect -NH₂ as an acetanilide first. Acetylation cuts the activation, the bulky group sends Br to para, and hydrolysis gives the amine back.

Answer from the NTA final answer key. Solution by PrepWiser.