JEE Challenger
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ChemHardAdv 2024 • Paper 2

Sum of Oxygen Atoms in Ozonolysis Oxidation Products

Comprehension Passage

An organic compound P\mathbf{P} with molecular formula C9H18O2\text{C}_9\text{H}_{18}\text{O}_2 decolorizes bromine water and also shows positive iodoform test. P\mathbf{P} on ozonolysis followed by treatment with H2O2\text{H}_2\text{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\mathbf{U}.

[Given, atomic mass: H=1\text{H} = 1, C=12\text{C} = 12, O=16\text{O} = 16]

Sum of number of oxygen atoms in S\mathbf{S} and T\mathbf{T} is _____.

Official Numerical Answer2

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesOxidation and Reduction (Organic)Alkanes, Alkenes, Alkynes

Step-by-Step Solution

To determine the sum of the number of oxygen atoms in compounds S\mathbf{S} and T\mathbf{T}, we analyze the reactions step-by-step:

1. Analysis of Compound P\mathbf{P} and its Ozonolysis Products (Q\mathbf{Q} and R\mathbf{R})

  • The molecular formula of P\mathbf{P} is C9H18O2\text{C}_9\text{H}_{18}\text{O}_2.
  • The Degree of Unsaturation (DU) of P\mathbf{P} is: DU=C+1−H2=9+1−182=1\text{DU} = C + 1 - \frac{H}{2} = 9 + 1 - \frac{18}{2} = 1
  • Since P\mathbf{P} decolorizes bromine water, this single degree of unsaturation corresponds to a C=C\text{C}=\text{C} double bond.
  • Oxidative ozonolysis (O3/H2O2\text{O}_3 / \text{H}_2\text{O}_2) of P\mathbf{P} cleaves the double bond to yield two carboxylic acid derivative compounds, Q\mathbf{Q} and R\mathbf{R}.
  • Q\mathbf{Q} and R\mathbf{R} undergo copolymerization (500 moles of Q\mathbf{Q} + 500 moles of R\mathbf{R}) to give a single copolymer U\mathbf{U}, which is the well-known biodegradable polymer PHBV (poly(3-hydroxybutyrate-co-3-hydroxyvalerate)).

Thus, Q\mathbf{Q} and R\mathbf{R} are β\beta-hydroxy carboxylic acids:

  • Q\mathbf{Q} (3-hydroxybutanoic acid): CH3−CH(OH)−CH2−COOH\text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{COOH}
  • R\mathbf{R} (3-hydroxypentanoic acid): CH3−CH2−CH(OH)−CH2−COOH\text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_2-\text{COOH}

Structure of P\mathbf{P}: P=CH3−CH(OH)−CH2−CH=CH−CH2−CH(OH)−CH2−CH3\mathbf{P} = \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_3

  • P\mathbf{P} shows a positive iodoform test due to the presence of the CH3−CH(OH)−\text{CH}_3-\text{CH}(\text{OH})- group.
  • Q\mathbf{Q} contains the CH3−CH(OH)−\text{CH}_3-\text{CH}(\text{OH})- group, so it shows a positive iodoform test.
  • R\mathbf{R} contains CH3−CH2−CH(OH)−\text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})-, so it does not give a positive iodoform test.

2. Oxidation with PCC followed by Heating

When Q\mathbf{Q} and R\mathbf{R} are treated with Pyridinium Chlorochromate (PCC), their secondary alcohol groups (−CH(OH)−-\text{CH}(\text{OH})-) are oxidized to carbonyl groups (−C(=O)−-\text{C}(=\text{O})-), yielding β\beta-keto acids:

  1. Formation of S\mathbf{S} from Q\mathbf{Q}: CH3−CH(OH)−CH2−COOH→PCCCH3−C(=O)−CH2−COOH→ΔCH3−C(=O)−CH3+CO2↑\text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{COOH} \xrightarrow{\text{PCC}} \text{CH}_3-\text{C}(=\text{O})-\text{CH}_2-\text{COOH} \xrightarrow{\Delta} \text{CH}_3-\text{C}(=\text{O})-\text{CH}_3 + \text{CO}_2 \uparrow

    • S\mathbf{S} is propan-2-one (acetone), CH3COCH3\text{CH}_3\text{COCH}_3.
    • S\mathbf{S} shows a positive iodoform test.
    • Number of oxygen atoms in S=1\mathbf{S} = 1.

  2. Formation of T\mathbf{T} from R\mathbf{R}: CH3−CH2−CH(OH)−CH2−COOH→PCCCH3−CH2−C(=O)−CH2−COOH→ΔCH3−CH2−C(=O)−CH3+CO2↑\text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_2-\text{COOH} \xrightarrow{\text{PCC}} \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_2-\text{COOH} \xrightarrow{\Delta} \text{CH}_3-\text{CH}_2-\text{C}(=\text{O})-\text{CH}_3 + \text{CO}_2 \uparrow

    • T\mathbf{T} is butan-2-one, CH3COCH2CH3\text{CH}_3\text{COCH}_2\text{CH}_3.
    • T\mathbf{T} shows a positive iodoform test.
    • Number of oxygen atoms in T=1\mathbf{T} = 1.

3. Conclusion

  • Number of oxygen atoms in S=1\mathbf{S} = 1
  • Number of oxygen atoms in T=1\mathbf{T} = 1

Sum of oxygen atoms=1+1=2\text{Sum of oxygen atoms} = 1 + 1 = 2

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