Consider the following sequence of reactions. [figure] Assuming that the reaction proceeds to completion, then 137 mg of 4-nitrotoluene will produce mg of B. (Given molar mass in g mol-1 H : 1, C : 12, N : 14, O : 16, Br : 80)
- 208
- 228
- 146
- 301
Answer
(2) 228
Solution
[figure]
Sn/HCl reduces -NO2 to -NH2; neutralisation gives A = 4-methylaniline (p-toluidine).
Acetylation gives N-(4-methylphenyl)acetamide; bromination in AcOH gives monobromination ortho to -NHCOCH3 (the para position is blocked by CH3).
B = N-(2-bromo-4-methylphenyl)acetamide, C9H10BrNO
Molar mass of 4-nitrotoluene (C7H7NO2) = 84 + 7 + 14 + 32 = 137 g mol-1
137 mg = 1 mmol 1 mmol of B
Molar mass of B = 108 + 10 + 80 + 14 + 16 = 228 g mol-1
mass of B = 1 228 = 228 mg
Answer: (2)