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 $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{Br}: 80$ )

Question

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

\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16, \mathrm{Br}: 80

)

Quantrex Academy · Accepted Answer

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.

Consider the following sequence of reactions. Assuming that the...

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