Animated Solution for Chemistry - Organic Chemistry: Consider the following reaction. On estimation of bromine in $1.00\text{ g}$ of $\mathbf{R}$ using Carius method, the amount of $\text{AgBr}$ formed (in $\text{g}$) is ______. [Given : Atomic mass of $\text{H}=1$, $\text{C}=12$, $\text{O}=16$, $\text{P}=31$, $\text{Br}=80$, $\text{Ag}=108$]

Let's embark on a fascinating journey that bridges organic synthesis with quantitative physical chemistry. We are given a starting material that features a benzene ring. Attached to this ring are two distinct groups: a bromine atom at the para position, and a hydroxymethyl group ($-{\text{CH}}_2\text{OH}$) at the top. This molecule is known as p-bromobenzyl alcohol.

Our reagent of choice is a mixture of red phosphorus and bromine ($\text{Red P}/{\text{Br}}_2$). This is a classic, highly reliable reagent used to convert aliphatic alcohols into their corresponding alkyl bromides.

Here is where the magic of chemical selectivity comes into play. Does the reagent attack the bromine already on the benzene ring? Absolutely not! Halogens directly attached to an aromatic ring are stabilized by resonance, giving the carbon-halogen bond a partial double-bond character. Therefore, they are highly resistant to nucleophilic substitution under these conditions.

The $\text{Red P}/{\text{Br}}_2$ reagent selectively targets the aliphatic $-\text{OH}$ group. The hydroxyl group is cleanly substituted by a bromine atom, transforming our starting material into p-bromobenzyl bromide (Product $\mathbf{R}$).

Let's determine the molecular formula of our newly formed product $\mathbf{R}$. It contains a benzene ring (${\text{C}}_6{\text{H}}_4$), a $-{\text{CH}}_2\text{Br}$ group, and another $-\text{Br}$ atom. Summing it all up, the molecular formula is ${\text{C}}_7{\text{H}}_6{\text{Br}}_2$.

Now, the problem shifts gears into analytical chemistry. We are asked to estimate the amount of bromine in $1.00\text{ g}$ of product $\mathbf{R}$ using the Carius method.

The Carius method is a robust technique used to estimate the total halogen content in an organic compound. The compound is heated with fuming nitric acid in the presence of silver nitrate. This aggressive treatment completely breaks down the organic framework, converting every single halogen atom present in the molecule into a precipitate of silver halide.

This is the critical catch: because our product $\mathbf{R}$ (${\text{C}}_7{\text{H}}_6{\text{Br}}_2$) contains two bromine atoms per molecule, the stoichiometry dictates that $1\text{ mole}$ of $\mathbf{R}$ will yield exactly $2\text{ moles}$ of silver bromide ($\text{AgBr}$).

First, we need the exact molar mass of our product $\mathbf{R}$.

Next, we calculate the number of moles of $\mathbf{R}$ present in our $1.00\text{ g}$ sample:

Because of the $1:2$ stoichiometry we established earlier, the moles of $\text{AgBr}$ formed will be twice the moles of $\mathbf{R}$:

Finally, we convert these moles back into mass. The molar mass of $\text{AgBr}$ is the sum of the atomic masses of silver ($108$) and bromine ($80$), which equals $188\text{ g/mol}$.

Rounding to appropriate significant figures, the amount of $\text{AgBr}$ formed is $1.50\text{ g}$.