Molecular Weight of Copolymer Formed From Ozonolysis Products

To determine the molecular weight of the copolymer $\mathbf{U}$, we first deduce the chemical structures of compounds $\mathbf{P}$, $\mathbf{Q}$, $\mathbf{R}$, $\mathbf{S}$, and $\mathbf{T}$ from the given experimental reactions.

1. Identification of Compound $\mathbf{P}$

• Molecular Formula: $\text{C}_9\text{H}_{18}\text{O}_2$
• Degree of Unsaturation (DU): $\text{DU} = C + 1 - \frac{H}{2} = 9 + 1 - \frac{18}{2} = 1$
• Reactions:
  • $\mathbf{P}$ decolorizes bromine water $\implies$ contains one carbon-carbon double bond ($\text{C=C}$). Since $\text{DU} = 1$, all unsaturation is accounted for by this double bond, meaning there are no carbonyl groups or rings present in $\mathbf{P}$. The two oxygen atoms must be present as hydroxyl groups ($-\text{OH}$).
  • $\mathbf{P}$ gives a positive iodoform test $\implies$ contains a $\text{CH}_3-\text{CH(OH)}-$ group.

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2. Identification of $\mathbf{Q}$ and $\mathbf{R}$

Oxidative ozonolysis ($\text{O}_3/\text{H}_2\text{O}_2$) of $\mathbf{P}$ cleaves the alkene double bond ($\text{-CH=CH-}$) to form two carboxylic acids, $\mathbf{Q}$ and $\mathbf{R}$:

• $\mathbf{Q}$ gives a positive iodoform test, while $\mathbf{R}$ does not.
• Oxidation of $\mathbf{Q}$ and $\mathbf{R}$ with pyridinium chlorochromate (PCC) followed by heating yields $\mathbf{S}$ and $\mathbf{T}$, respectively, both of which show positive iodoform tests.
  • PCC oxidizes secondary alcohol groups ($-\text{CH(OH)}-$) into ketones ($-\text{C=O}$).
  • Heating causes $\beta$-keto acids to undergo decarboxylation (loss of $\text{CO}_2$) to form ketones.
  • For $\mathbf{S}$ and $\mathbf{T}$ to yield positive iodoform tests, they must contain a methyl ketone ($\text{CH}_3\text{CO}-$) group.

From these observations:

• Compound $\mathbf{Q}$: $\text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{COOH}$ (4-carbon $\beta$-hydroxy acid)

  • Molar mass of $\mathbf{Q}$ ($\text{C}_4\text{H}_8\text{O}_3$): $M_{\mathbf{Q}} = (4 \times 12) + (8 \times 1) + (3 \times 16) = 104 \text{ g/mol}$
  • $\mathbf{Q} \xrightarrow{\text{PCC}} \text{CH}_3-\text{CO}-\text{CH}_2-\text{COOH} \xrightarrow{\Delta} \text{CH}_3-\text{CO}-\text{CH}_3 \, (\mathbf{S}, \text{acetone}) + \text{CO}_2$

• Compound $\mathbf{R}$: $\text{CH}_3-\text{CH}_2-\text{CH(OH)}-\text{CH}_2-\text{COOH}$ (5-carbon $\beta$-hydroxy acid)

  • Molar mass of $\mathbf{R}$ ($\text{C}_5\text{H}_{10}\text{O}_3$): $M_{\mathbf{R}} = (5 \times 12) + (10 \times 1) + (3 \times 16) = 118 \text{ g/mol}$
  • $\mathbf{R} \xrightarrow{\text{PCC}} \text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{COOH} \xrightarrow{\Delta} \text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_3 \, (\mathbf{T}, \text{butan-2-one}) + \text{CO}_2$

Recombining $\mathbf{Q}$ and $\mathbf{R}$ at the cleaved double bond confirms $\mathbf{P}$: $\mathbf{P} = \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH(OH)}-\text{CH}_2-\text{CH}_3 \quad (\text{C}_9\text{H}_{18}\text{O}_2)$

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3. Calculation of Molecular Weight of Copolymer $\mathbf{U}$

Both $\mathbf{Q}$ and $\mathbf{R}$ are hydroxy acids that undergo condensation copolymerization to form an acyclic polyester $\mathbf{U}$.

• Monomer units: $500 \text{ moles of } \mathbf{Q} + 500 \text{ moles of } \mathbf{R} = 1000 \text{ total monomer units}$.
• For 1 mole of an acyclic (linear) polymer made of $N = 1000$ monomer units, there are $(N - 1) = 999$ ester linkage formations.
• Each ester bond formation eliminates 1 molecule of $\text{H}_2\text{O}$ ($M_{\text{H}_2\text{O}} = 18 \text{ g/mol}$).

The molecular weight of copolymer $\mathbf{U}$ is calculated as: $M_{\mathbf{U}} = 500 \cdot M_{\mathbf{Q}} + 500 \cdot M_{\mathbf{R}} - 999 \cdot M_{\text{H}_2\text{O}}$

Substitute the values: $M_{\mathbf{U}} = 500(104) + 500(118) - 999(18)$ $M_{\mathbf{U}} = 500(104 + 118) - 999(18)$ $M_{\mathbf{U}} = 500(222) - 17982$ $M_{\mathbf{U}} = 111000 - 17982 = 93018 \text{ g/mol}$

Final Answer: The molecular weight of $\mathbf{U}$ is 93018.