The final major product of the following reaction is :

Solution

Chemistry ma’am · 4 steps

Protect the NH₂ as an amide, let the bulky NHCOCH₃ send Br para, then take the acetyl off.

1. Protect the amine. Ac₂O/pyridine turns -NH₂ into -NHCOCH₃. A free -NH₂ would bind the Lewis acid FeCl₃ and get brominated many times over. The amide is only moderately activating, so you get one bromine.
2. Who directs. -NHCOCH₃ is a stronger ortho/para director than -CH₃, so it decides where Br goes. Number the ring with NHCOCH₃ at C1 and CH₃ at C3. The open sites it activates are C2, C4 and C6.

   Site	Relative to NHCOCH₃	Relative to CH₃	Verdict
   C2	ortho	ortho	squeezed between both groups, too crowded
   C4	para	ortho	both groups agree, no crowding from the bulky amide: major
   C6	ortho	para	next to the bulky amide, minor
   C5	meta	meta	not activated by either group

3. Remove the acetyl. OH⁻ with heat hydrolyses the amide back to -NH₂. The Br and CH₃ do not move.
4. So. The product is 4-bromo-3-methylaniline: Br para to NH₂ and ortho to CH₃. That is option (b).

The idea to keep. Acetylating an aniline before ring substitution gives you one substituent, not three, and the bulky amide pushes the new group para to itself. Hydrolysis then gives back the free amine.

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