JEE Advanced 2026 — Paper 2 — CHEM-III solution

JEE Advanced2026Paper 2CHEM-III

In the following reaction sequence, major products $X$ and $Y$ are acyclic monomers.

$500$ mol of $X$ completely reacts with $500$ mol of $Y$ to give $1$ mol of a single biodegradable acyclic copolymer $Z$ as the only product. The amount of $Z$ formed in grams is _____.

Given: Atomic mass (in amu): H: 1, C: 12, N: 14, O: 16, Br: 80.

Your answer

Solution

Identify X. CH $_{3}$ I $KCN$ CH $_{3}$ CN $H_{3} O^{+}$ CH $_{3}$ COOH $Red P, Br_{2}$ BrCH $_{2}$ COOH (Hell-Volhard-Zelinsky $α$ -bromination) $NH_{3} (excess)$ H $_{2}$ N-CH $_{2}$ -COOH. So $X$ = glycine (C $_{2}$ H $_{5}$ NO $_{2}$ , M = 75 g/mol).

Identify Y. Acid hydrolysis of caprolactam opens the ring to give 6-aminohexanoic acid, H $_{2}$ N-(CH $_{2}$ ) $_{5}$ -COOH (C $_{6}$ H $_{13}$ NO $_{2}$ , M = 131 g/mol).

Copolymer Z. Each pair of monomers (one glycine + one 6-aminohexanoic acid) forms one repeat unit by amide condensation; $500 + 500$ monomers form one polymer chain of $1000$ residues with $999$ water molecules lost (each amide bond expels one H $_{2}$ O).

Mass balance. Total monomer mass $= 500 \times 75 + 500 \times 131 = 37500 + 65500 = 103000$ g. Water lost $= 999 \times 18 = 17982$ g.

$∴$ Mass of $Z = 103000 - 17982 = 85018$ g.

← Back to full paper

Practice more CHEM-III

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →