Question

Consider the following reaction sequence in which \(J, K, L\) and \(M\) are the major products.

Given:
Atomic mass (in amu): H : \(1\), C : \(12\), N : \(14\), O : \(16\), S : \(32\), Br : \(80\), Ba : \(137\)

In sulphur estimation by Carius method, the amount of \(\text{BaSO}_4\) formed from \(3.79\,\text{g}\) of \(M\) is _____ g.

Hint:

In the Carius method, the organic compound is heated with fuming nitric acid in the presence of silver nitrate or barium chloride. For sulphur estimation, the sulphur is oxidized to sulfuric acid and then precipitated as barium sulfate. Use the stoichiometry 1 mole of $\text{S} \rightarrow 1$ mole of $\text{BaSO}_4$ for calculations.

Answer 1

Correct Answer: 2.33

Step 1: Understanding the reaction to find compound M.
Compound M is formed by the reaction of K with sodium thiophenolate ($\text{PhSNa}$).
The $\text{PhS}^-$ ion undergoes nucleophilic substitution replacing the bromide in K.
Structure M: $\text{C}_6\text{H}_3(\text{CH}_3)_2-\text{CH(SPh)}-\text{CH}_2\text{-O-C}_6\text{H}_4\text{-NO}_2$.
Formula of M: $\text{C}_{16}\text{H}_{16}\text{NO}_3(\text{SC}_6\text{H}_5) = \text{C}_{22}\text{H}_{21}\text{NO}_3\text{S}$.

Step 2: Calculating the molar mass of M and BaSO$_4$.
Molar mass of M = $(22 \times 12) + (21 \times 1) + 14 + (3 \times 16) + 32$
$= 264 + 21 + 14 + 48 + 32 = 379 \text{ g/mol.}$
Molar mass of $\text{BaSO}_4$ = $137 + 32 + (4 \times 16) = 137 + 32 + 64 = 233 \text{ g/mol.}$

Step 3: Quantitative estimation of sulphur.
Moles of M used = $\frac{3.79 \text{ g}}{379 \text{ g/mol}} = 0.01 \text{ mol}$.
Since each molecule of M contains one sulphur atom, 1 mole of M will yield 1 mole of $\text{BaSO}_4$ in the Carius method.
Moles of $\text{BaSO}_4$ formed = Moles of sulphur in M = $0.01 \text{ mol}$.

Step 4: Final calculation of the mass of $\text{BaSO_4$.}
\[ \text{Mass of } \text{BaSO}_4 = \text{moles} \times \text{molar mass} = 0.01 \text{ mol} \times 233 \text{ g/mol} = 2.33 \text{ g.} \] Final Answer: 2.33