JEE Main 2026 · 23 January · morning · Single correct · +4 / −1
Consider the following sequence of reactions.
Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce
____ mg of B.
(Given molar mass in gmol⁻¹H:1,C:12, N:14,O:16,Br:80 )
- (a) 208
- (b) 228
- (c) 146
- (d) 301
Answer and solution
Answer (b) 228
Solution
Chemistry ma’am · 4 steps
137 mg is exactly 1 mmol, so work out B's molar mass and you have your answer in mg.
- Moles of start. 4-Nitrotoluene is C₇H₇NO₂: 84 + 7 + 14 + 32 = 137 g mol⁻¹. So 137 mg is exactly 1 mmol, and every step is 1 : 1.
- Make A. Sn/HCl reduces -NO₂ to -NH₃⁺, and neutralising frees the amine. A is p-toluidine, CH₃C₆H₄NH₂.
- Protect, then brominate. Acetic anhydride turns -NH₂ into -NHCOCH₃. This group is a milder activator, so Br₂/AcOH adds only one Br. The para position is taken by CH₃, so Br goes ortho to -NHCOCH₃. B is 2-bromo-4-methylacetanilide, C₉H₁₀BrNO.
M_B = 9(12) + 10(1) + 80 + 14 + 16 = 228 g mol⁻¹
- So. 1 mmol of B weighs 228 mg, which is option (b).
The idea to keep. Always check whether the given mass is a neat number of millimoles. Acetylating an aniline before bromination is how you stop at a single Br.
Answer from the NTA final answer key. Solution by PrepWiser.