Chemistry · Aldehydes and Ketones

JEE Advanced 2024 — Paper 2 — Question 43

An organic compound P\mathbf{P} with molecular formula C9H18O2\mathrm{C}_{9} \mathrm{H}_{18} \mathrm{O}_{2} decolorizes bromine water and also shows positive iodoform test. P\mathbf{P} on ozonolysis followed by treatment with H2O2\mathrm{H}_{2} \mathrm{O}_{2} gives Q\mathbf{Q} and R\mathbf{R}. While compound Q\mathbf{Q} shows positive iodoform test, compound R\mathbf{R} does not give positive iodoform test. Q\mathbf{Q} and R\mathbf{R} on oxidation with pyridinium chlorochromate (PCC) followed by heating give S\mathbf{S} and T\mathbf{T}, respectively. Both S\mathbf{S} and T\mathbf{T} show positive iodoform test. Complete copolymerization of 500 moles of Q\mathbf{Q} and 500 moles of R\mathbf{R} gives one mole of a single acyclic copolymer U. [Given, atomic mass: H=1,C=12,O=16\mathrm{H}=1, \mathrm{C}=12, \mathrm{O}=16 ]

The molecular weight of U\mathbf{U} is

Answer: 93018

Numerical answer — enter this value.

Step-by-step solution

Q+R⟶\mathrm{Q}+\mathrm{R} \longrightarrow Polymer (U)(\mathrm{U}) Since moles of U=1U=1. Moles = Weight  Molar mass =\frac{\text { Weight }}{\text { Molar mass }} Weight of ' UU ' == Weight of (Q+R)−(Q+R)- Weight of water Weight of ' QQ ' =500×104=52,000 g=500 \times 104=52,000 \mathrm{~g} Weight of ' R ' =500×118=59,000 g=500 \times 118=59,000 \mathrm{~g} Weight of ' H2O\mathrm{H}_{2} \mathrm{O} ' =999×18=17982 g=999 \times 18=17982 \mathrm{~g} Weight of polymer =52000+59000−17982=52000+59000-17982

=93018 g/mol=93018 \mathrm{~g} / \mathrm{mol}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Chemistry
Chapter
Aldehydes and Ketones
Topic
Chemical properties of Aldehydes & Ketones
An organic compound P with molecular formula C 9 H 18 O 2 decolorizes… | JEE Advanced 2024 PYQ with Solution · DhiX AI