The polymer chains are held together by hydrogen bonding in a polymer X. Polymer X is formed from monomers Y and Z. What are Y and Z?
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Polymers like Nylon (polyamides) are characterized by strong interchain hydrogen bonding between the $\text{N-H}$ proton of one chain and the $\text{C=O}$ oxygen of an adjacent chain. This strong attraction is the source of their high tensile strength and crystalline nature.
Step 1: Determine the identity of polymer X.
The problem states that the polymer X chains are held together by hydrogen bonding. This is a characteristic feature of polyamides (like Nylon) and polyurethanes. The strongest intermolecular hydrogen bonding, particularly between polymer chains, is characteristic of Nylon (polyamides).
The general structure of a polyamide (Nylon) involves the amide linkage ($\text{-CO-NH-}$) formed by the condensation of a diamine and a dicarboxylic acid. The $\text{H}$ on the $\text{N}$ and the $\text{O}$ on the $\text{C=O}$ group form interchain hydrogen bonds.
Step 2: Identify the monomers Y and Z that form a polyamide.
Polyamides are formed by the condensation polymerization of:
1. A diamine ($\text{H}_2\text{N-R-NH}_2$).
2. A dicarboxylic acid ($\text{HOOC-R'-COOH}$).
Step 3: Match the required monomers to the labels.
Monomer B: $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$. This is Hexamethylenediamine (a diamine). This is a monomer for Nylon-6,6.
Monomer E: $\text{HO}_2\text{C}-(\text{CH}_2)_4\text{CO}_2\text{H}$. This is Adipic acid (a dicarboxylic acid). This is the other monomer for Nylon-6,6.
The polymerization of B and E yields Nylon-6,6, which is known for its strong interchain hydrogen bonding.
\[
n \text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2 + n \text{HOOC}(\text{CH}_2)_4\text{COOH} \to [\text{-NH}-(\text{CH}_2)_6-\text{NH}-\text{CO}-(\text{CH}_2)_4-\text{CO-}]_n + 2n\text{H}_2\text{O}.
\]
Step 4: Conclude the final answer.
The monomers Y and Z are B (Hexamethylenediamine) and E (Adipic acid).
