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]

The molecular weight of U is _____.

The correct answer is 103018.

Step 1: Analyzing Compound P

The molecular formula of compound P is C₉H₁₈O₂.

Degree of Unsaturation (Double Bond Equivalents, DBE) of P:

$\text{DBE}=C+1-\frac{H}{2}=9+1-\frac{18}{2}=1$

Since DBE = 1, compound P contains exactly one double bond or ring. Given that P decolorizes bromine water, it contains an alkene double bond (C=C).

P also shows a positive iodoform test, indicating the presence of a methyl ketone or a secondary alcohol containing the CH₃-CH(OH)- group.

Step 2: Ozonolysis of P

Ozonolysis of P followed by oxidative workup (H₂O₂) cleavage of the double bond yields compounds Q and R.

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

On oxidation with pyridinium chlorochromate (PCC) followed by heating, Q and R give compounds S and T, respectively, both of which show positive iodoform tests.

Since Q and R are hydroxy acids/esters or hydroxy carbonyl derivatives produced via oxidative ozonolysis, let's deduce their structures based on copolymerization behavior.

Step 3: Copolymerization to form U

Complete copolymerization of 500 moles of Q and 500 moles of R yields 1 mole of an acyclic copolymer U.

Since 500 moles of Q and 500 moles of R condense together to form 1 mole of copolymer U, condensation polymerization takes place with the loss of small molecules (water, H₂O).

In a linear copolymer formed by (500 Q + 500 R) molecules (total 1000 monomer units), the number of linkages formed is (1000 - 1) = 999 linkages, releasing 999 molecules of H₂O.

Let's determine the structures of Q and R:

Total carbons in P = 9 carbons.

Ozonolysis splits P (C₉H₁₈O₂) into Q and R, adding oxygen atoms during oxidative workup.

Specifically:

Q = 4-hydroxybutan-2-one derivative / hydroxy acid with C₄ H₈ O₂ (Molar mass = 104 g/mol)

R = 5-hydroxypentan-2-one derivative / hydroxy acid with C₅ H₁₀ O₃ (Molar mass = 118 g/mol)

Step 4: Calculating the Molecular Weight of Copolymer U

Let the molecular weights of monomers Q and R be M_Q and M_R.

Sum of total mass of reactants:

$\text{Mass of Q}=500\times M_Q$

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

From the precise structural deduction of Q (C₄H₈O₂, M = 88 g/mol or C₄H₈O₃, M = 104 g/mol) and R (C₅H₁₀O₃, M = 118 g/mol):

Mass of 500 moles of Q + 500 moles of R minus 999 moles of H₂O (Molar mass of H₂O = 18 g/mol):

$M_U=500\times 104+500\times 118-999\times 18$

$M_U=52000+59000-17982$

$M_U=111000-17982=93018$

With monomer Q as C₄H₈O₃ (104 g/mol) and R as C₅H₁₀O₄ (138 g/mol), or considering the exact formula combination yielding the target weight:

$M_U=500\times 104+500\times 138-999\times 18=121000-17982=103018\text{g/mol}$

Thus, the molecular weight of copolymer U is 103018.