Estimation of Bromine in Organic Compound using Carius Method

To determine the amount of $\text{AgBr}$ formed, we first identify the structure and molecular formula of the major product $\mathbf{R}$.

1. Reaction Step: when 4-bromobenzyl alcohol is treated with red phosphorus and bromine ($\text{red phosphorus}/\text{Br}_2$), the primary alcoholic group ($-\text{CH}_2\text{OH}$) is converted into a bromomethyl group ($-\text{CH}_2\text{Br}$). The aromatic bromine atom remains unaffected. $\text{Br}-\text{C}_6\text{H}_4-\text{CH}_2\text{OH} \xrightarrow{\text{red P}/\text{Br}_2} \text{Br}-\text{C}_6\text{H}_4-\text{CH}_2\text{Br} \quad (\mathbf{R})$

2. Molar Mass of Product $\mathbf{R}$ ($\text{C}_7\text{H}_6\text{Br}_2$): $\text{Molar Mass of } \mathbf{R} = (7 \times 12) + (6 \times 1) + (2 \times 80) = 84 + 6 + 160 = 250\text{ g mol}^{-1}$

3. Moles of $\mathbf{R}$ in $1.00\text{ g}$: $\text{Moles of } \mathbf{R} = \frac{1.00\text{ g}}{250\text{ g mol}^{-1}} = 0.004\text{ mol}$

4. Estimation of Bromine using Carius Method: In the Carius method, all bromine atoms present in the organic compound are quantitatively converted into $\text{AgBr}$. Since each molecule of $\mathbf{R}$ contains $2$ bromine atoms: $\text{Moles of } \text{AgBr} \text{ formed} = 2 \times \text{Moles of } \mathbf{R} = 2 \times 0.004\text{ mol} = 0.008\text{ mol}$

5. Molar Mass of $\text{AgBr}$: $\text{Molar Mass of } \text{AgBr} = 108 + 80 = 188\text{ g mol}^{-1}$

6. Mass of $\text{AgBr}$ Formed: $\text{Mass of } \text{AgBr} = 0.008\text{ mol} \times 188\text{ g mol}^{-1} = 1.504\text{ g}$

Rounding to two decimal places, the mass of $\text{AgBr}$ formed is $1.50\text{ g}$ (or $1.504\text{ g}$).