Question

Consider the following sequence of reactions:

4-nitrotoluene
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)}

• A. 301
• B. 208
• C. 228
• D. 146

Hint:

The $-NHCOCH_3$ group is ortho/para directing. Since the para position is occupied by the methyl group, bromine enters the ortho position.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a sequence of functional group transformations: reduction of a nitro group, acetylation of the resulting amine, and electrophilic aromatic substitution (bromination).

Step 2: Key Formula or Approach:
1. Sn/HCl reduces $-NO_2$ to $-NH_2$.
2. $Ac_2O$ converts $-NH_2$ to $-NHCOCH_3$ (A: 4-methylacetanilide).
3. $Br_2/AcOH$ brominates the ring ortho to the activating acetamido group (B: 2-bromo-4-methylacetanilide).
Step 3: Detailed Explanation:
- Molar mass of 4-Nitrotoluene ($C_7H_7NO_2$): $7(12) + 7(1) + 14 + 2(16) = 84 + 7 + 14 + 32 = 137$ g/mol.
- Molar mass of B ($C_9H_{10}NOBr$): $9(12) + 10(1) + 14 + 16 + 80 = 108 + 10 + 14 + 16 + 80 = 228$ g/mol.
- Since the stoichiometry is 1:1 throughout the sequence:
- 137 mg of reactant (1 mmol) will produce 1 mmol of product B.
- Mass of B $= 1 \text{ mmol} \times 228 \text{ mg/mmol} = 228$ mg.
Step 4: Final Answer:
The mass of B produced is 228 mg.