Mass of AgBr Precipitate Formed From Carius Analysis of Organic Product

To determine the mass of $\text{AgBr}$ precipitate formed during the Carius analysis of product $(\text{P})$, we first trace the chemical transformations step-by-step:

1. Reduction step (i): The starting material is 4-nitrotoluene (1-methyl-4-nitrobenzene). Reaction with $\text{Sn/HCl}$ followed by base ($\text{OH}^-$) reduces the nitro group ($-\text{NO}_2$) to an amino group ($-\text{NH}_2$), yielding 4-methylaniline (p-toluidine).

2. Acetylation step (ii): Reaction of 4-methylaniline with acetic anhydride, $(\text{CH}_3\text{CO})_2\text{O}$, protects the amino group to form $N$-(4-methylphenyl)acetamide ($4$-methylacetanilide).

3. Bromination step (iii): Electrophilic aromatic substitution with $\text{Br}_2 / \text{AlBr}_3$ introduces a bromine atom onto the aromatic ring. Since $-\text{NHCOCH}_3$ is a stronger activating group than $-\text{CH}_3$, and the position para to $-\text{NHCOCH}_3$ is occupied by $-\text{CH}_3$, bromination occurs at the position ortho to the $-\text{NHCOCH}_3$ group, giving $N$-(2-bromo-4-methylphenyl)acetamide.

4. Hydrolysis step (iv): Acidic hydrolysis ($\text{H}_3\text{O}^+$) hydrolyzes the amide group back to the primary amine, yielding the major product $(\text{P})$, which is 2-bromo-4-methylaniline.

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Molar Mass Calculation:

• Product $(\text{P})$: 2-bromo-4-methylaniline

  • Formula: $\text{C}_7\text{H}_8\text{BrN}$
  • Molar mass ($M_{\text{P}}$): $M_{\text{P}} = (7 \times 12) + (8 \times 1) + (1 \times 14) + (1 \times 80) = 84 + 8 + 14 + 80 = 186 \text{ g mol}^{-1}$

• Precipitate: Silver Bromide ($\text{AgBr}$)

  • Formula: $\text{AgBr}$
  • Molar mass ($M_{\text{AgBr}}$): $M_{\text{AgBr}} = 108 + 80 = 188 \text{ g mol}^{-1}$

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Mass of Precipitate Formed:

According to Carius analysis, each mole of product $(\text{P})$ containing 1 atom of bromine produces 1 mole of $\text{AgBr}$:

$\text{Moles of }(\text{P}) = \frac{1.0 \text{ g}}{186 \text{ g mol}^{-1}}$

$\text{Mass of AgBr precipitate} = \text{Moles of }(\text{P}) \times M_{\text{AgBr}} = \frac{1.0}{186} \times 188 \approx 1.01075 \text{ g}$

Rounding off to the nearest integer gives 1.