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Polymers · Question 1

JEE Advanced 2026 Paper 2NumericalMedium📈 Topic in 17% papers+4
1

In the following reaction sequence, major products X and Y are acyclic monomers. CH3I [ 3. Red P, Br2\ 4. NH3 (excess)] 1. KCN\ 2. H3O+, X [figure] 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 to 85138.02

Solution

[figure] KCN [figure] H3O+, [figure]

SN2 by CN- gives ethanenitrile; acid hydrolysis of –CN gives ethanoic acid.

[figure] Red P, Br2 [figure] NH3 (excess) [figure] (X)

HVZ reaction brominates the α-carbon; excess NH3 replaces Br by -NH2: X = glycine (75 g mol-1).

[figure] [ ]H3O+ [figure] (Y)

Hydrolysis opens the lactam (cyclic amide): Y = 6-aminohexanoic acid (131 g mol-1).

X + Y Z = nylon-2-nylon-6, [-NH-CH2-CO-NH-(CH2)5-CO-]n: one linear chain of 1000 monomer units joined by 999 amide bonds, each releasing one H2O.

Mass of Z =500 75+500 131-999 18=37500+65500-17982=85018 g

Answer: 85018

JEE Advanced 2026 Paper 2

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2

The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is . Use: Atomic mass of N (in amu) = 14

JEE Advanced 2025 Paper 1
3

The molecular weight of U is .

JEE Advanced 2024 Paper 2

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