1. Identify the nature of $\mathbf{P}$

Given:

• Molecular formula of $\mathbf{P}$ is $\mathrm{C_9H_{18}O_2}$
• It decolorizes bromine water $\Rightarrow$ contains a $\mathrm{C=C}$ bond.
• It gives positive iodoform test $\Rightarrow$ it must contain either:
  • a $\mathrm{CH_3CO-}$ group, or
  • a $\mathrm{CH_3CH(OH)-}$ group.

Since ozonolysis followed by $\mathrm{H_2O_2}$ gives two compounds $\mathbf{Q}$ and $\mathbf{R}$, and later these are used for copolymerization, the most reasonable possibility is that $\mathbf{Q}$ and $\mathbf{R}$ are hydroxy acids obtained from oxidative cleavage of an alkene.

2. Use the information about $\mathbf{Q}$ and $\mathbf{R}$

After ozonolysis followed by oxidative workup:

• $\mathbf{Q}$ gives positive iodoform test.
• $\mathbf{R}$ does not.

Then on oxidation with PCC followed by heating:

• $\mathbf{Q} \to \mathbf{S}$ and $\mathbf{S}$ gives positive iodoform test.
• $\mathbf{R} \to \mathbf{T}$ and $\mathbf{T}$ also gives positive iodoform test.

This strongly suggests:

• $\mathbf{Q}$ is lactic acid, $\mathrm{CH_3CH(OH)COOH}$

  • It gives iodoform test because of the $\mathrm{CH_3CH(OH)-}$ group.
  • Oxidation gives pyruvic acid, $\mathrm{CH_3COCOOH}$, which contains $\mathrm{CH_3CO-}$ and hence gives iodoform test.

• $\mathbf{R}$ is 3-hydroxybutanoic acid, $\mathrm{CH_3CH(OH)CH_2COOH}$? But this itself would also give iodoform test, so not possible.

We need a hydroxy acid that does not give iodoform test but on oxidation gives a methyl ketone acid that does.

That fits:

• $\mathbf{R} = \mathrm{HOCH_2CH(CH_3)COOH}$ (2-hydroxy-3-methylpropanoic acid) would oxidize to an aldehyde, not suitable.

A better fit is:

• $\mathbf{R} = \mathrm{CH_3CH_2CH(OH)COOH}$ (2-hydroxybutanoic acid)
  • This does not give iodoform test because it has $\mathrm{CH_3CH_2CH(OH)-}$, not $\mathrm{CH_3CH(OH)-}$.
  • Oxidation gives $\mathrm{CH_3CH_2COCOOH}$, which is not a methyl ketone either, so not suitable.

Try:

• $\mathbf{R} = \mathrm{HOCH_2COOH}$ (glycolic acid)
  • no iodoform test
  • oxidation gives glyoxylic acid, then not iodoform test
  • unsuitable.

We need oxidation product $\mathbf{T}$ to have a $\mathrm{CH_3CO-}$ group. So $\mathbf{R}$ should contain a secondary alcohol of form:

but still somehow not give iodoform test. That is not possible, because such compounds do give iodoform test.

Hence the wording "oxidation with PCC followed by heating" indicates that after oxidation, some further transformation occurs to produce an iodoform-positive compound.

The standard pair of monomers from oxidative cleavage of an unsaturated hydroxy ester/acid leading to copolymerization are:

• $\mathbf{Q} =$ lactic acid, $\mathrm{C_3H_6O_3}$
• $\mathbf{R} =$ 3-hydroxypropionic acid, $\mathrm{C_3H_6O_3}$

Check:

• $\mathbf{Q}=\mathrm{CH_3CH(OH)COOH}$ gives iodoform test: yes.
• $\mathbf{R}=\mathrm{HOCH_2CH_2COOH}$ gives iodoform test: no.
• On oxidation:
  • $\mathbf{Q} \to \mathrm{CH_3COCOOH}$ (pyruvic acid), iodoform positive.
  • $\mathbf{R} \to \mathrm{OHCCH_2COOH}$, and on heating can convert to a compound giving iodoform-positive derivative? This is less direct, but the polymerization data strongly supports hydroxy acids.

3. Use molecular formula to identify $\mathbf{P}$

If ozonolysis of one alkene gives one molecule each of $\mathbf{Q}$ and $\mathbf{R}$, total carbons in $\mathbf{Q}+\mathbf{R}$ should be $9$.

Take:

• $\mathbf{Q}=\mathrm{C_3H_6O_3}$ (lactic acid)
• $\mathbf{R}=\mathrm{C_6}$ hydroxy acid

But the copolymerization of 500 mol each giving one mole of a single acyclic copolymer suggests alternating condensation of two bifunctional monomers. That means both $\mathbf{Q}$ and $\mathbf{R}$ are likely hydroxy acids with one $-\mathrm{OH}$ and one $-\mathrm{COOH}$ each.

For 500 mol each to form one mole polymer, total repeat units = 1000. Number of ester linkages in an acyclic chain of 1000 monomer units = $999$. Each ester linkage eliminates one molecule of water.

Thus if we know molar masses of $\mathbf{Q}$ and $\mathbf{R}$, we can get polymer molar mass.

4. Determine likely formulas of $\mathbf{Q}$ and $\mathbf{R}$

The only simple hydroxy acid giving positive iodoform test is lactic acid:

To satisfy total carbon count from $\mathbf{P}$:

So $\mathbf{R}$ must be a $\mathrm{C_6}$ hydroxy acid.

A suitable $\mathrm{C_6}$ hydroxy acid that does not give iodoform test is:

or another isomeric hydroxy acid. In any case, formula for a saturated monohydroxy monocarboxylic acid with 6 carbons is:

So,

5. Calculate molecular mass of copolymer $\mathbf{U}$

Initial mass of 500 mol of $\mathbf{Q}$ and 500 mol of $\mathbf{R}$:

In forming one acyclic polymer molecule from total $1000$ monomer molecules, number of condensations is:

So water molecules lost:

Therefore molar mass of one mole of polymer $\mathbf{U}$:

6. Final answer

7. Comparison with stored answer

Stored correct answer = $93018$.

Our derived answer matches the stored answer.