Consider the following reaction:

On estimation of bromine in 1.00 g of R using Carius method, the amount of AgBr formed (in g) is ________.

[Given: Atomic mass of H = 1, C = 12, O = 16, P = 31, Br = 80, Ag = 108]

The correct answer is 1.50.

Step 1: Identify compound R (Major Product)

As shown in the image, the starting material is 4-bromobenzyl alcohol (4-Br-C₆H₄-CH₂OH):

When 4-bromobenzyl alcohol reacts with red phosphorus and bromine ($\text{red P}+{\text{Br}}_2$), phosphorus tribromide (${\text{PBr}}_3$) is formed in situ. This reagent selectively replaces the aliphatic hydroxyl group ($-\text{OH}$) attached to the benzylic carbon with a bromine atom ($-\text{Br}$).

Therefore, the major product R is 4-bromobenzyl bromide (1-bromo-4-(bromomethyl)benzene), which has the molecular formula:

$\text{R}={\text{C}}_7{\text{H}}_6{\text{Br}}_2$

Step 2: Calculate the molar mass of compound R and AgBr

Using the given atomic masses (H = 1, C = 12, Br = 80, Ag = 108):

$\text{Molar mass of R (}{\text{C}}_7{\text{H}}_6{\text{Br}}_2\text{)}=(7\times 12)+(6\times 1)+(2\times 80)=84+6+160=250\text{g/mol}$

$\text{Molar mass of AgBr}=108+80=188\text{g/mol}$

Step 3: Determine the amount of AgBr formed using Carius Method

In the Carius estimation of halogens, all bromine atoms present in the organic compound (compound R) are completely converted into silver bromide (AgBr).

Since 1 mole of ${\text{C}}_7{\text{H}}_6{\text{Br}}_2$ contains 2 moles of bromine atoms, 1 mole of compound R yields 2 moles of AgBr precipitate.

$\text{Moles of R in 1.00 g}=\frac{1.00}{250}\text{mol}$

$\text{Moles of AgBr formed}=2\times \text{Moles of R}=2\times \frac{1.00}{250}=\frac{2.00}{250}\text{mol}$

Step 4: Calculate the mass of AgBr in grams

$\text{Mass of AgBr formed}=\text{Moles of AgBr}\times \text{Molar mass of AgBr}$

$\text{Mass of AgBr formed}=\frac{2.00}{250}\times 188=\frac{376}{250}=1.504\approx 1.50\text{g}$

Thus, the amount of AgBr formed is 1.50 g.