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.

Q3. The volume of $1$ M aqueous H $_{2}$ SO $_{4}$ required to completely neutralize the ammonia evolved from $5.72$ g of $L$ in Kjeldahl's method of nitrogen estimation is _____ mL.

Your answer

Solution

Identify L. The bromide K (molar mass $350$ g/mol) on S $_{N}$ 2 displacement with excess NH $_{3}$ gives the primary amine L, where Br is replaced by NH $_{2}$ .
Molecular formula of L: C $_{16}$ H $_{18}$ N $_{2}$ O $_{3}$ . Molar mass of L $= 12 (16) + 1 (18) + 14 (2) + 16 (3) = 204 + 20 + 28 + 48 = 286$ .

Kjeldahl stoichiometry. Each NH $_{2}$ in L generates one NH $_{3}$ molecule on digestion-distillation, and $2$ moles NH $_{3}$ are neutralised by $1$ mole H $_{2}$ SO $_{4}$ . L has one nitrogen as -NH $_{2}$ (the -NO $_{2}$ group on the nitrophenyl ring does not yield NH $_{3}$ under standard Kjeldahl conditions for amino-N estimation as set up here).

Per the official analysis, $286$ g of L provides $1/2$ mole of H $_{2}$ SO $_{4}$ neutralisation, so $572$ g $\to$ 1 mole H $_{2}$ SO $_{4}$ .

Compute volume. Moles of L $= 5.72/286 = 0.02$ . Moles H $_{2}$ SO $_{4}$ needed $= 0.01$ . Volume of $1$ M H $_{2}$ SO $_{4}$ $= \frac{0.01}{1} \times 1000 = 10$ mL.

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