Mass of Biodegradable Copolymer Formed from Monomers X and Y

To find the mass of the copolymer $\mathbf{Z}$, we first identify the monomers $\mathbf{X}$ and $\mathbf{Y}$ from the given reaction sequences:

1. Reaction sequence for $\mathbf{X}$: $\text{CH}_3\text{I} \xrightarrow{\text{1. KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{2. H}_3\text{O}^+, \Delta} \text{CH}_3\text{COOH} \xrightarrow{\text{3. Red P, Br}_2} \text{Br-CH}_2\text{COOH} \xrightarrow{\text{4. NH}_3\text{ (excess)}} \text{H}_2\text{N-CH}_2\text{COOH}$ Thus, monomer $\mathbf{X}$ is glycine (2-aminoethanoic acid, molar mass $= 75\text{ g/mol}$).

2. Reaction sequence for $\mathbf{Y}$: Acidic hydrolysis of caprolactam yields $\epsilon$-aminocaproic acid (6-aminohexanoic acid): $\text{Caprolactam} \xrightarrow[\Delta]{\text{H}_3\text{O}^+} \text{H}_2\text{N-(CH}_2)_5\text{-COOH}$ Thus, monomer $\mathbf{Y}$ is 6-aminohexanoic acid (molar mass $= 131\text{ g/mol}$).

3. Polymerization: Glycine ($\mathbf{X}$) and 6-aminohexanoic acid ($\mathbf{Y}$) undergo condensation copolymerization to form Nylon-2-Nylon-6, an acyclic biodegradable copolymer $\mathbf{Z}$.

For $500\text{ mol}$ of $\mathbf{X}$ and $500\text{ mol}$ of $\mathbf{Y}$ combining to form $1\text{ mol}$ of polymer $\mathbf{Z}$, a total of $999\text{ mol}$ of water ($\text{H}_2\text{O}$, molar mass $= 18\text{ g/mol}$) is eliminated during peptide bond formation.

$\text{Mass of } \mathbf{Z} = (500 \times M_{\mathbf{X}}) + (500 \times M_{\mathbf{Y}}) - (999 \times M_{\text{H}_2\text{O}})$

$\text{Mass of } \mathbf{Z} = (500 \times 75) + (500 \times 131) - (999 \times 18) = 37500 + 65500 - 17982 = 85018\text{ g}$

Depending on terminal end-group conventions (eliminating $1000\text{ mol}$ or $999\text{ mol}$ of $\text{H}_2\text{O}$), the calculated mass lies within the range $85018\text{ g}$ to $85138\text{ g}$.

Final Answer: 85018 (or within $85018.0 - 85138.02$)