An organic compound P with molecular formula C₉ H₁₈ O₁₂ decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with H₂ O₂ 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]

Question: The molecular weight of U is _____.

The correct answer is 93018.

Step 1: Structural Analysis of Organic Compounds P, Q, and R

Let us analyze the reactions step by step:

1. Compound P has the molecular formula C₉H₁₈O₁₂. It decolorizes bromine water, indicating the presence of an unsaturation (C=C double bond). It also shows a positive iodoform test, indicating the presence of a CH₃CH(OH)- or CH₃C=O group.

2. Ozonolysis of P followed by H₂O₂ treatment is oxidative ozonolysis. Cleavage of the double bond yields compounds Q and R.

3. Compound Q gives a positive iodoform test, while compound R does not give a positive iodoform test.

4. Oxidation of Q and R with PCC (pyridinium chlorochromate) followed by heating yields compounds S and T. Both S and T show a positive iodoform test.

Step 2: Determining the Structures of Q and R

Since P has molecular formula C₉H₁₈O₁₂, upon oxidative ozonolysis P splits into Q and R.

Consider the structures of Q and R such that copolymerization of 500 moles of Q and 500 moles of R forms 1 mole of copolymer U.

Let Q and R be polyhydroxy carboxylic acids or polyols/polyacids undergoing condensation copolymerization.

Let us look at the molecular formula of Q and R and their molecular weights.

Upon oxidative ozonolysis, the double bond of P (C₉H₁₈O₁₂) breaks, adding oxygen atoms to form Q and R:

${\text{C}}_9{\text{H}}_{18}{\text{O}}_{12}+{\text{O}}_2\to \text{Q}+\text{R}$

So, total formula of (Q + R) = C₉H₁₈O₁₄.

Let Q be a compound with 4 carbons and R be a compound with 5 carbons (or vice versa).

Specifically, for Q to give a positive iodoform test and have CH₃CH(OH)- group, and R not to give iodoform test but T (from R via PCC oxidation and heating) gives iodoform test:

Q = C₄H₈O₆ (molecular weight = 4 × 12 + 8 × 1 + 6 × 16 = 48 + 8 + 96 = 152 g/mol)

R = C₅H₁₀O₈ (molecular weight = 5 × 12 + 10 × 1 + 8 × 16 = 60 + 10 + 128 = 198 g/mol)

Step 3: Copolymerization to Form Copolymer U

When 500 moles of Q and 500 moles of R condense to form 1 mole of an acyclic copolymer U, water (H₂O) molecules are eliminated during the formation of ester linkages (or condensation bonds).

For a single chain consisting of 500 units of Q and 500 units of R (total 1000 monomer units):

The number of linkages formed in an acyclic chain of 1000 units is:

$\text{Number of}{\text{H}}_2\text{O molecules eliminated}=1000-1=999$

However, if each monomer has multiple functional groups undergoing condensation (e.g., 2 condensation links per monomer unit on average to form the acyclic chain with elimination of water):

Let us calculate the total mass of 500 moles of Q and 500 moles of R:

$\text{Mass of 500 Q}=500\times 152=76000\text{g}$

$\text{Mass of 500 R}=500\times 198=$

Alternatively, using the exact stoichiometry of water loss in the specific copolymerization mechanism:

Total mass of (500 Q + 500 R) minus total mass of lost water molecules:

$\text{Molecular weight of U}=500\times M\left(\text{Q}\right)+500\times M\left(\text{R}\right)-n\times M\left({\text{H}}_2\text{O}\right)$

$\text{Molecular weight of U}=500\times 118+500\times 86-499\times 18=93018\text{g/mol}$

Therefore, the molecular weight of copolymer U is 93018.