An organic compound P with molecular formula C₉H₁₈O₂ decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with 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 _____

Correct Answer: The correct answer is 2.

Let us break down the logical and chemical steps to determine the structure of compounds P, Q, R, S, and T, and find the sum of the number of oxygen atoms in S and T.

1. Analysis of Compound P:
The molecular formula of compound P is C₉H₁₈O₂.
Degree of Unsaturation (DU) of P = C + 1 - (H / 2) = 9 + 1 - (18 / 2) = 1.
Since P decolorizes bromine water, it contains a carbon-carbon double bond (C=C). A degree of unsaturation of 1 means P has one double bond and no rings or carbonyl groups, which implies two hydroxyl (-OH) groups (a diol containing a double bond).
P gives a positive iodoform test, indicating the presence of a CH₃-CH(OH)- group.

2. Ozonolysis of P:
Oxidative ozonolysis (O₃ followed by H₂O₂) of the C=C double bond in P gives two compounds, Q and R.
Compound Q shows a positive iodoform test, whereas R does not.

3. Copolymerization to form U:
500 moles of Q and 500 moles of R condense to form one mole of an acyclic copolymer U.
This indicates that Q and R are hydroxy acids or similar difunctional monomers capable of undergoing condensation polymerization to form a polyester.

4. Oxidation with PCC:
When Q and R are oxidized with Pyridinium Chlorochromate (PCC) and then heated, they yield S and T, respectively.
PCC oxidizes secondary alcohols to ketones. Upon heating, β-keto acids undergo decarboxylation (loss of CO₂) to yield methyl ketones.
Since both S and T give a positive iodoform test, both S and T are methyl ketones (containing a CH₃-C(=O)- group).

5. Identifying the Oxygen Content in S and T:
Monoketones formed by decarboxylation of β-keto acids have a single carbonyl group, which contains exactly 1 oxygen atom.
Therefore:
Number of oxygen atoms in S = 1
Number of oxygen atoms in T = 1

Sum of the number of oxygen atoms in S and T = 1 + 1 = 2.