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 )

Step-by-step solution

Starting with 137 mg of 4-nitrotoluene (C₇H₉NO₂, MW = 137). Reduction with Sn/HCl converts NO₂ to NH₂, producing 4-aminotoluene (A). 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