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

To solve this problem, first, we need to analyze the sequence of reactions starting from 4-nitrotoluene and determine the final product (B).

Given the sequence of reactions:

1. The first step involves reducing 4-nitrotoluene with Sn and HCl, followed by neutralization. This converts the nitro group (\(\text{-NO}_2\)) to an amino group (\(\text{-NH}_2\)), forming 4-aminotoluene (A).
2. Next, the acetylation of the amino group using acetic anhydride (\((\text{CH}_3\text{CO})_2\text{O}\)) will convert 4-aminotoluene to N-acetyl-4-aminotoluene.
3. Finally, bromination using \(\text{Br}_2/\text{AcOH}\) takes place selectively at the para position (with respect to the acetylated amino group) due to the activating nature of the acetylamino group, forming compound (B).

Now, to find the mass of \(B\) produced from 137 mg of 4-nitrotoluene:

• Molar mass of 4-nitrotoluene: \((12 \times 7 + 1 \times 7 + 14 + 16 \times 2) = 137 \, \text{g/mol}\)
• Molar mass of (B): \((12 \times 8 + 1 \times 8 + 14 + 16 \times 1 + 80 \times 1 + 12 + 16) = 228 \, \text{g/mol}\)

Since 4-nitrotoluene converts completely to (B) in a 1:1 mole ratio:

The mass of (B) produced from 137 mg of 4-nitrotoluene is:

\(\frac{137 \, \text{mg}}{137} \times 228 = 228 \, \text{mg}\)

Therefore, the correct answer is 228 mg.