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JEE Advanced 2024 · Paper 2, Chemistry question 14

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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]

Sum of number of oxygen atoms in S and T is _____.

Answer and worked solution

Official answer: 2

P. CX9HX18OX2\ce{C9H18O2} has one degree of unsaturation. It decolourises bromine water (one C=C\ce{C=C}) and gives the iodoform test, so it contains CHX3CH(OH)X−\ce{CH3CH(OH)-}. The second O is another OH. So P is
CHX3CH(OH)CHX2CH=CHCHX2CH(OH)CHX2CHX3.\ce{CH3CH(OH)CH2CH=CHCH2CH(OH)CH2CH3}.

Q and R. Oxidative ozonolysis (OX3\ce{O3}, then HX2OX2\ce{H2O2}) splits P into two hydroxy acids:

  • Q = CHX3CH(OH)CHX2COOH\ce{CH3CH(OH)CH2COOH}, iodoform positive;
  • R = CHX3CHX2CH(OH)CHX2COOH\ce{CH3CH2CH(OH)CH2COOH}, iodoform negative.

S and T. PCC oxidises each to a β-keto acid, which loses COX2\ce{CO2} on heating:

  • S = acetone, CHX3COCHX3\ce{CH3COCH3};
  • T = butanone, CHX3COCHX2CHX3\ce{CH3COCH2CH3}.

Both are methyl ketones and give the iodoform test.

Answer. Oxygen atoms: 1+1=21+1=2. The key also accepts 4, which is the count if S and T are taken as the β-keto acids before decarboxylation.

Topics:Oxidation and reductionAlkenesCarboxylic acidsDetection of functional groups
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