JEE Advanced 2026 — Paper 2 — CHEM-IV solution

JEE Advanced2026Paper 2CHEM-IV

Question Stem (Q3-Q4). 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.

Q4. In sulphur estimation by Carius method, the amount of BaSO $_{4}$ formed from $3.79$ g of $M$ is _____ g.

Your answer

Solution

Identify M. K + PhSNa (S $_{N}$ 2 with thiolate) gives the thioether M (Br replaced by SPh). Molecular formula of M: C $_{22}$ H $_{21}$ NO $_{3}$ S, molar mass $= 12 (22) + 1 (21) + 14 + 16 (3) + 32 = 264 + 21 + 14 + 48 + 32 = 379$ g/mol.

Carius method stoichiometry. Each sulphur atom in M ends up as one BaSO $_{4}$ . M contains $1$ sulphur atom per molecule.

Calculate. Moles of M $= 3.79/379 = 0.01$ . Moles BaSO $_{4}$ $= 0.01$ . Mass BaSO $_{4}$ $= 0.01 \times 233 = 2.33$ g (M $_{BaSO_{4}} = 137 + 32 + 64 = 233$ ).

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