In the following reaction sequence, major products $\mathbf{X}$ and $\mathbf{Y}$ are acyclic monomers. $\text{CH}_3\text{I} \xrightarrow{\text{1. KCN}} \xrightarrow{\text{2. } \text{H}_3\text{O}^+, \Delta} \xrightarrow{\text{3. Red P, Br}_2} \xrightarrow{\text{4. } \text{NH}_3 \text{ (excess)}} \mathbf{X}$ Caprolactam $\xrightarrow[\Delta]{\text{H}_3\text{O}^+} \mathbf{Y}$ 500 mol of $\mathbf{X}$ completely reacts with 500 mol of $\mathbf{Y}$ to give 1 mol of a single biodegradable acyclic copolymer $\mathbf{Z}$ as the only product. The amount of $\mathbf{Z}$ formed in grams is ____. Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, Br : 80

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Polymers

In the following reaction sequence, major products

X

and

Y

are acyclic monomers.

CH_{3} I 1. KCN 2. H_{3} O^{+}, Δ 3. Red P, Br_{2} 4. NH_{3} (excess) X

Caprolactam

H_{3} O^{+} Δ 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: 85018

Solution

The first sequence builds glycine,

H_{2} NC H_{2} COOH

(molar mass

75

), as

X

: methyl iodide to acetonitrile, hydrolysis to acetic acid, Hell-Volhard-Zelinsky bromination to bromoacetic acid, then reaction with excess ammonia. Acid hydrolysis of caprolactam gives 6-aminohexanoic acid,

H_{2} N (C H_{2})_{5} COOH

(molar mass

131

), as

Y

. Joining

500

units of each into a single chain of

1000

residues forms

999

amide links, and each link releases one water molecule. The copolymer mass is therefore

500 \times 75 + 500 \times 131 - 999 \times 18 = 37500 + 65500 - 17982 = 85018

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Polymers

Solution

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