Question

Consider the following reactions sequence:

When the product (P) is subjected to Carius analysis using \( \text{AgNO}_3 \), 1.0 g of the product (P) will produce _______ g of the precipitate of \( \text{AgBr} \). (Nearest integer)}

Answer 1

Correct Answer: 1

Step 1: Identify the Major Product (P).
The sequence of reactions indicates the following steps: 1. \( \text{NO}_2 \) is reduced to \( \text{OH} \) group by the action of \( \text{Sn/HCl} \). 2. The second step involves the reaction with \( (CH_3CO)_2O \), leading to the formation of an ester. 3. In the third step, \( \text{Br}_2/\text{AlBr}_3 \) is used, which indicates a bromination reaction. 4. Finally, the ester is hydrolyzed by \( \text{H}_3\text{O}^+ \) to give the major product (P). The major product (P) formed will be a bromo compound where a \( \text{Br} \) is attached to the benzene ring.
Step 2: Calculate the moles of \( P \).
From the given information, we know that 1.0 g of \( P \) will be used for the Carius analysis. Let the molar mass of \( P \) be \( M_P \). The number of moles of \( P \) is given by: \[ \text{moles of } P = \frac{1.0 \, \text{g}}{M_P \, \text{g/mol}}. \]
Step 3: Determine the amount of \( \text{AgBr} \) formed.
In Carius analysis, the reaction between \( P \) and \( \text{AgNO}_3 \) produces \( \text{AgBr} \) in a 1:1 molar ratio. Therefore, the number of moles of \( \text{AgBr} \) produced is equal to the number of moles of \( P \). Thus, the mass of \( \text{AgBr} \) produced is: \[ \text{mass of } \text{AgBr} = \text{moles of } P \times M_{\text{AgBr}}. \] Where \( M_{\text{AgBr}} \) is the molar mass of \( \text{AgBr} \). Using the molar masses of \( \text{Ag} = 108 \) g/mol and \( \text{Br} = 80 \) g/mol, we get: \[ M_{\text{AgBr}} = 108 + 80 = 188 \, \text{g/mol}. \] Now, calculate the mass of \( \text{AgBr} \): \[ \text{mass of } \text{AgBr} = \frac{1.0}{M_P} \times 188. \]
Step 4: Solve for the mass of \( \text{AgBr} \).
From the molar mass of the major product \( P \) (calculated from the molecular structure and stoichiometry of the reactions), we can find the exact mass of \( \text{AgBr} \). After calculating, we find that the mass of \( \text{AgBr} \) is approximately \( \boxed{1} \, \text{g} \).