Amount of Barium Sulfate Formed in Carius Method of Sulphur Estimation

To determine the amount of $\text{BaSO}_4$ formed in the Carius method, we first need to determine the structure and molecular formula of compound M.

Step 1: Reaction Sequence Analysis for J and K

1. Friedel-Crafts Acylation: Reaction of $m$-xylene ($1,3\text{-dimethylbenzene}$) with chloroacetyl chloride ($\text{Cl-CH}_2\text{-CO-Cl}$) in the presence of anhydrous $\text{AlCl}_3$ yields $2\text{-chloro-1-}(2,4\text{-dimethylphenyl})\text{ethan-1-one}$: $\text{Ar-H} + \text{Cl-CH}_2\text{-CO-Cl} \xrightarrow{\text{anhyd. AlCl}_3} \text{Ar-CO-CH}_2\text{Cl}$ where $\text{Ar} = 2,4\text{-dimethylphenyl } (\text{C}_8\text{H}_9)$.

2. Finkelstein Reaction: Treatment with $\text{NaI}$ converts the alkyl chloride to the alkyl iodide: $\text{Ar-CO-CH}_2\text{Cl} \xrightarrow{\text{NaI, heat}} \text{Ar-CO-CH}_2\text{I}$

3. Ether Formation: Nucleophilic substitution with sodium $3\text{-nitrophenoxide}$ ($\text{NaO-Ar'}$, where $\text{Ar'} = 3\text{-nitrophenyl}$) yields compound J: $\text{Ar-CO-CH}_2\text{I} + \text{NaO-Ar'} \rightarrow \text{Ar-CO-CH}_2\text{-O-Ar'} \quad (\mathbf{J})$

4. Reduction and Bromination:

   • Reduction of the carbonyl group of J using $\text{NaBH}_4$ yields the secondary alcohol $\text{Ar-CH(OH)-CH}_2\text{-O-Ar'}$.

   • Subsequent bromination using $\text{PBr}_3$ replaces the hydroxyl group with bromine to form compound K: $\mathbf{J} \xrightarrow{1.\ \text{NaBH}_4 \quad 2.\ \text{PBr}_3} \text{Ar-CH(Br)-CH}_2\text{-O-Ar'} \quad (\mathbf{K})$

   • Molecular Formula of K: $\text{C}_{16}\text{H}_{16}\text{BrNO}_3$

   • Molar Mass of K: $M_{\mathbf{K}} = (16 \times 12) + (16 \times 1) + 80 + 14 + (3 \times 16) = 350\text{ g/mol}$ (This matches the given molar mass of $350\text{ g/mol}$).

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Step 2: Identification of Compound M

Compound K reacts with sodium thiophenolate ($\text{PhSNa}$) in a nucleophilic substitution reaction where the bromine atom is replaced by the $-\text{SPh}$ group to yield compound M: $\mathbf{K} \xrightarrow{\text{PhSNa}} \text{Ar-CH(SPh)-CH}_2\text{-O-Ar'} \quad (\mathbf{M})$

• Molecular Formula of M: $\text{C}_{22}\text{H}_{21}\text{NO}_3\text{S}$
• Molar Mass of M: $M_{\mathbf{M}} = (22 \times 12) + (21 \times 1) + 14 + (3 \times 16) + 32 = 379\text{ g/mol}$

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Step 3: Estimation of Sulphur by Carius Method

1. Number of moles of M: $\text{Moles of } \mathbf{M} = \frac{\text{Mass of } \mathbf{M}}{\text{Molar mass of } \mathbf{M}} = \frac{3.79\text{ g}}{379\text{ g/mol}} = 0.01\text{ mol}$

2. Precipitation as $\text{BaSO}_4$: Since $1\text{ mole}$ of M contains $1\text{ mole}$ of sulphur atoms, all sulphur from $0.01\text{ mol}$ of M is quantitatively converted to $\text{BaSO}_4$: $\text{Moles of } \text{BaSO}_4 \text{ formed} = \text{Moles of } \mathbf{M} = 0.01\text{ mol}$

3. Molar Mass of $\text{BaSO}_4$: $M_{\text{BaSO}_4} = 137 + 32 + (4 \times 16) = 233\text{ g/mol}$

4. Mass of $\text{BaSO}_4$ formed: $\text{Mass of } \text{BaSO}_4 = 0.01\text{ mol} \times 233\text{ g/mol} = 2.33\text{ g}$

Final Answer: The amount of $\text{BaSO}_4$ formed is 2.33 $\text{g}$.