In the following reaction sequence, major products X and Y are acyclic monomers.

${\mathrm{CH}}_{3}I\stackrel{1.\mathrm{KCN}}{\to }\stackrel{2.H^{+},\Delta }{\to }\stackrel{3.\mathrm{Red}P,{\mathrm{Br}}_2,{\mathrm{NH}}_3\left(\mathrm{excess}\right)}{\to }X$

$\mathrm{Caprolactam}\stackrel{H_3{O}^{+},\Delta }{\to }Y$

$500\mathrm{mol}\mathrm{of}X\mathrm{completely}\mathrm{reacts}\mathrm{with}500\mathrm{mol}\mathrm{of}Y\mathrm{to}\mathrm{give}1\mathrm{mol}\mathrm{of}a\mathrm{single}\mathrm{biodegradable}\mathrm{acyclic}\mathrm{copolymer}Z\mathrm{as}\mathrm{the}\mathrm{only}\mathrm{product.}$

$\text{The amount of}Z\text{formed in grams is}\underset{\hspace{1em}}{‾}.$

$\text{Given: Atomic mass (in amu): H = 1, C = 12, N = 14, O = 16, Br = 80}$

The correct answer is 85018.

Let us determine the chemical structure of products X and Y step-by-step from the given reaction sequence.

Step 1: Formation of Product X
Starting material is methyl iodide (${\mathrm{CH}}_{3}I$).
1. Nucleophilic substitution with KCN yields acetonitrile:
${\mathrm{CH}}_{3}I+\mathrm{KCN}\to {\mathrm{CH}}_3\mathrm{CN}+\mathrm{KI}$
2. Acidic hydrolysis ($H^{+},\Delta$) of acetonitrile converts the nitrile group to a carboxylic acid, forming acetic acid:
${\mathrm{CH}}_3\mathrm{CN}\underset{\Delta }{\overset{H^{+}}{\to }}{\mathrm{CH}}_3\mathrm{COOH}$
3. Reaction with Red P, ${\mathrm{Br}}_2$ followed by excess ${\mathrm{NH}}_3$ is the Hell-Volhard-Zelinsky (HVZ) reaction followed by amination. This converts acetic acid to 2-bromoacetic acid and then to glycine (aminoacetic acid, X):
${\mathrm{CH}}_3\mathrm{COOH}\underset{2.{\mathrm{NH}}_3\left(\mathrm{excess}\right)}{\overset{1.\mathrm{Red}P,{\mathrm{Br}}_2}{\to }}{H}_{2}N-{\mathrm{CH}}_2-\mathrm{COOH}$
Thus, X is Glycine ($H_{2}N-{\mathrm{CH}}_2-\mathrm{COOH}$).

Step 2: Formation of Product Y
Acidic hydrolysis ($H_3{O}^{+},\Delta$) of Caprolactam opens the seven-membered cyclic amide ring to yield 6-aminohexanoic acid (Y, aminocaproic acid):
$\text{Caprolactam}\underset{\Delta }{\overset{H_3{O}^{+}}{\to }}{H}_{2}N-{\left({\mathrm{CH}}_2\right)}_5-\mathrm{COOH}$
Thus, Y is 6-aminohexanoic acid ($H_{2}N-{\left({\mathrm{CH}}_2\right)}_5-\mathrm{COOH}$).

Step 3: Formation of Copolymer Z
Glycine (X) and 6-aminohexanoic acid (Y) undergo condensation copolymerization to form Nylon-2-nylon-6, a biodegradable acyclic copolymer.
Reaction of 500 mol of X with 500 mol of Y forms 1 mol of copolymer Z:
$500X+500Y\to 1Z+(500+500-1)H_{2}O$
Since a single acyclic chain polymer Z is formed from 1000 monomer units (500 units of X + 500 units of Y), the total number of peptide bond formations involves the elimination of $(500+500-1)=999$ molecules of water ($H_{2}O$).

Step 4: Molar Mass Calculations
Let us calculate the molar masses of X, Y, and $H_{2}O$ using the given atomic masses (H = 1, C = 12, N = 14, O = 16):
1. Molar mass of Glycine (X, $C_2{H}_5{\mathrm{NO}}_2$):
$M_X=(2\times 12)+(5\times 1)+14+(2\times 16)=24+5+14+32=75\mathrm{g}/\mathrm{mol}$
2. Molar mass of 6-aminohexanoic acid (Y, $C_6{H}_{13}{\mathrm{NO}}_2$):
$M_Y=(6\times 12)+(13\times 1)+14+(2\times 16)=72+13+14+32=131\mathrm{g}/\mathrm{mol}$
3. Molar mass of Water ($H_{2}O$):
$M_{H_{2}O}=(2\times 1)+16=18\mathrm{g}/\mathrm{mol}$

Step 5: Mass Balance Calculation for Z
By conservation of mass: