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JEE Advanced 2026 · Paper 2, Chemistry question 14

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

CHX3I\ce{CH3I} →3. Red P, BrX24. NHX3 (excess)1. KCN2. HX3OX+, Δ\xrightarrow[\text{3. Red P, }\ce{Br2}\quad\text{4. }\ce{NH3}\text{ (excess)}]{\text{1. KCN}\quad\text{2. }\ce{H3O+},\ \Delta} X

Caprolactam →ΔHX3OX+\xrightarrow[\Delta]{\ce{H3O+}} Y

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

Answer and worked solution

Official answer: 85018

X (glycine):
CHX3I→KCNCHX3CN→HX3OX+CHX3COOH→Red P/BrX2BrCHX2COOH→NHX3HX2NCHX2COOH,\ce{CH3I ->[KCN] CH3CN ->[H3O+] CH3COOH ->[Red P/Br2] BrCH2COOH ->[NH3] H2NCH2COOH},
M=75M=75.

Y (6-aminohexanoic acid): hydrolysis of caprolactam gives HX2N(CHX2)X5COOH\ce{H2N(CH2)5COOH}, M=131M=131.

Z (nylon-2-nylon-6). One linear chain of 1000 monomers has 999 amide links, each releasing one HX2O\ce{H2O}:
500(75)+500(131)−999(18)=37500+65500−17982=85018 g.500(75)+500(131)-999(18)=37500+65500-17982=85018\text{ g}.