Question 18

Identify product M: Compound $K$ ($C_{16}H_{16}BrNO_3$) reacts with $PhSNa$ to replace the $-Br$ atom with a $-SPh$ group via $S_N2$ substitution. The resulting compound $M$ has the formula $C_{22}H_{21}NO_3S$.

Molar Mass of M: $Mass(M) = Mass(K) - Mass(Br) + Mass(S) + Mass(Ph) = 350 - 80 + 32 + 77 = 379$ g/mol.

Carius Method: In this method, all sulphur in the organic compound is converted to sulphate ions and precipitated as $BaSO_4$. Each mole of $M$ contains $1$ mole of Sulphur, which yields $1$ mole of $BaSO_4$.

Calculation: Moles of $M = \frac{3.79}{379} = 0.01$ mol. Moles of $BaSO_4$ formed = $0.01$ mol. Molar mass of $BaSO_4 = 137 (Ba) + 32 (S) + 64 (O_4) = 233$ g/mol. Mass of $BaSO_4 = 0.01 \times 233 = 2.33$ g.