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 g mol ^ -1 H : 1, C : 12, ~N : 14, O : 16, Br : 80)

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Correct answer: C. 228

Starting with 137 mg of 4-nitrotoluene (C₇H₉NO₂, MW = 137). Step 1: Reduction with Sn/HCl converts NO₂ to NH₂, producing 4-aminotoluene (A). Step 2: Acetylation with (CH₃CO)₂O produces the acetamide intermediate, then bromination with Br₂/AcOH adds bromine to the para position (ortho-para directing due to acetamide group), yielding 4-bromo-N-(4-methylphenyl)acetamide (B) with molecular formula C₉H₁₀BrNO and MW = 228 g/mol. Moles of starting material = 137 mg ÷ 137 g/mol = 1 mmol. Assuming complete reaction, moles of B produced = 1 mmol. Mass of B = 1 mmol × 228 mg/mmol = 228 mg.

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