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← JEE Advanced 2024 · Paper 2
ChemistryNumerical+3Question 49 of 51

JEE Advanced 2024 · Paper 2, Chemistry question 15

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PARAGRAPH I

An organic compound P with molecular formula CX9HX18OX2\ce{C9H18O2} decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with HX2OX2\ce{H2O2} gives Q and R. While compound Q shows positive iodoform test, compound R does not give positive iodoform test. Q and R on oxidation with pyridinium chlorochromate (PCC) followed by heating give S and T, respectively. Both S and T show positive iodoform test.

Complete copolymerization of 500 moles of Q and 500 moles of R gives one mole of a single acyclic copolymer U.

[Given, atomic mass: H = 1, C = 12, O = 16]

The molecular weight of U is _____.

Answer and worked solution

Official answer: 93018

Monomers. Q and R are hydroxy acids:

  • Q = CX4HX8OX3\ce{C4H8O3}, M=104M=104;
  • R = CX5HX10OX3\ce{C5H10O3}, M=118M=118.

Copolymer. Condensation polymerisation forms ester links. A single acyclic chain of 10001000 monomers has 999999 ester bonds, and each one releases a water molecule:
MU=500×104+500×118−999×18=52000+59000−17982=93018.M_U=500\times104+500\times118-999\times18=52000+59000-17982=93018.