Question

In the following reaction sequence, major products X and Y are acyclic monomers.
\(\text{CH}_3\text{I} \xrightarrow{\text{1. KCN}}{\text{2. H}_3\text{O}^+, \Delta}{\text{3. Red P, Br}_2}{\text{4. NH}_3\text{ (excess) \text{X} \)
\(\text{Caprolactam} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{Y} \)
500 mol of X completely reacts with 500 mol of Y to give 1 mol of a single biodegradable acyclic copolymer Z as the only product. The amount of Z formed in grams is _______.
Given: Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, Br : 80}

Hint:

In condensation polymerization, always subtract the mass of small molecules (usually water) eliminated during bond formation from the total mass of monomers.

Answer 1

Correct Answer: 85018

Step 1: Understanding the Question:
We need to identify the monomers \(X\) and \(Y\) formed in the given reaction sequence and then calculate the mass of the biodegradable acyclic copolymer \(Z\) formed from: \[ 500\ \text{mol of }X \] and \[ 500\ \text{mol of }Y \]

Step 2: Key Formula or Approach:

• Determine products stepwise using organic reaction mechanisms.
• Polymerization occurs by condensation with elimination of water molecules.
• Total polymer mass: \[ \text{Mass of monomers} - \text{Mass of eliminated water} \]

Step 3: Detailed Explanation:
(i) Formation of X \[ CH_3I \xrightarrow{KCN} CH_3CN \] (Cyanide substitution gives acetonitrile) Hydrolysis: \[ CH_3CN \xrightarrow{H_3O^+,\Delta} CH_3COOH \] Hell–Volhard–Zelinsky reaction: \[ CH_3COOH \xrightarrow{Red\ P,Br_2} BrCH_2COOH \] Treatment with excess ammonia: \[ BrCH_2COOH \xrightarrow{NH_3} NH_2CH_2COOH \] Thus, \[ X=\text{Glycine} \] Molar mass of glycine: \[ =2(12)+5(1)+14+2(16) \] \[ =24+5+14+32 \] \[ =75\ g\ mol^{-1} \] (ii) Formation of Y Caprolactam on hydrolysis gives: \[ Y=H_2N(CH_2)_5COOH \] (6-aminohexanoic acid) Molar mass: \[ =6(12)+13(1)+14+2(16) \] \[ =72+13+14+32 \] \[ =131\ g\ mol^{-1} \] (iii) Formation of Copolymer Z Total monomer mass: \[ 500(75)+500(131) \] \[ =37500+65500 \] \[ =103000\ g \] Since 1000 monomer molecules combine into one polymer molecule, total peptide/amide bonds formed: \[ 1000-1=999 \] Each bond formation eliminates one water molecule: \[ H_2O=18\ g\ mol^{-1} \] Mass lost: \[ 999\times18 \] \[ =17982\ g \] Thus, \[ \text{Mass of polymer }Z \] \[ =103000-17982 \] \[ =85018\ g \]

Step 4: Final Answer:
The amount of copolymer \(Z\) formed is: \[ \boxed{85018\ g} \]