Unlike addition polymerisation, which occurs by opening double bonds in unsaturated monomer molecules, condensation polymerisation involves the reaction of bifunctional monomers with the elimination of a small molecule byproduct.
🔑 Key Principle
In condensation polymerisation, each monomer must have at least two functional groups (bifunctional). Each time a link is formed between two monomer molecules, a small molecule (usually water, \\( \text{H}_2\text{O} \\), or hydrogen chloride, \\( \text{HCl} \\)) is lost.
Addition vs Condensation Polymerisation
It is important to contrast these two types of polymerisation:
- Addition Polymerisation: Monomer contains a C=C double bond. No small molecule is lost. The backbone consists entirely of carbon atoms.
- Condensation Polymerisation: Monomers contain reactive functional groups (e.g. \\( \text{-COOH} \\), \\( \text{-OH} \\), \\( \text{-NH}_2 \\)). A small molecule is lost. The backbone contains heteroatoms (oxygen or nitrogen) in addition to carbon.
A reaction where monomer molecules join together to form a long chain polymer, with the simultaneous release of a small molecule byproduct (such as \\( \text{H}_2\text{O} \\) or \\( \text{HCl} \\)).
Polyesters
Polyesters contain ester links (\\( \text{-COO-} \\)) in their polymer backbones. They are formed by reacting:
- A dicarboxylic acid and a diol
- Or a single monomer that contains both a carboxylic acid group and an alcohol group (a hydroxycarboxylic acid)
Formation of Terylene
Terylene is a common synthetic polyester formed by the reaction of benzene-1,4-dicarboxylic acid and ethane-1,2-diol. Water is lost during the reaction:
\( n\text{HOOC-C}_6\text{H}_4\text{-COOH} + n\text{HO-CH}_2\text{CH}_2\text{-OH} \rightarrow \text{[-CO-C}_6\text{H}_4\text{-CO-O-CH}_2\text{CH}_2\text{-O-]}_n + 2n\text{H}_2\text{O} \)
A polymer containing repeating ester groups (\\( \text{-COO-} \\)) in its backbone, formed by condensation reactions between carboxylic acid and alcohol functional groups.
Polyamides
Polyamides contain amide links (\\( \text{-CONH-} \\)) in their polymer backbones. They are formed by reacting:
- A dicarboxylic acid and a diamine
- Or a single monomer containing both an amine and a carboxylic acid group (an amino acid)
Formation of Nylon-6,6
Nylon-6,6 is a synthetic polyamide made from hexanedioic acid (a 6 carbon dicarboxylic acid) and 1,6-diaminohexane (a 6 carbon diamine). The numbers in Nylon-6,6 indicate the number of carbons in each monomer chain:
\( n\text{HOOC-(CH}_2)_4\text{-COOH} + n\text{H}_2\text{N-(CH}_2)_6\text{-NH}_2 \rightarrow \text{[-CO-(CH}_2)_4\text{-CO-NH-(CH}_2)_6\text{-NH-]}_n + 2n\text{H}_2\text{O} \)
Formation of Kevlar
Kevlar is an aromatic polyamide known for its extreme strength, used in bulletproof vests and car tyres. It is formed from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine:
\( n\text{HOOC-C}_6\text{H}_4\text{-COOH} + n\text{H}_2\text{N-C}_6\text{H}_4\text{-NH}_2 \rightarrow \text{[-CO-C}_6\text{H}_4\text{-CO-NH-C}_6\text{H}_4\text{-NH-]}_n + 2n\text{H}_2\text{O} \)
A polymer containing repeating amide (or peptide) groups (\\( \text{-CONH-} \\)) in its backbone, formed by condensation reactions between carboxylic acids and amines.
How to Draw Repeat Units and Identify Monomers
You must be able to perform two vital skills in A-Level exam questions:
1. Drawing a Repeat Unit from Monomers
When drawing the repeating unit of a condensation polymer:
- Align the functional groups of the monomers next to each other.
- Remove the parts that form the small molecule byproduct:
- From carboxylic acid groups (\\( \text{-COOH} \\)), remove the -OH group.
- From amine groups (\\( \text{-NH}_2 \\)) or alcohol groups (\\( \text{-OH} \\)), remove an -H atom.
- Draw a covalent bond joining the remaining carbonyl carbon (\\( \text{C=O} \\)) to the heteroatom (\\( \text{-O-} \\) or \\( \text{-NH-} \\)).
- Extend bonding lines through square brackets at both ends and write the subscript \\( n \\).
2. Identifying Monomers from a Polymer
To reverse the process and identify the monomers from a polymer repeat unit chain:
- Locate the ester (\\( \text{-CO-O-} \\)) or amide (\\( \text{-CO-NH-} \\)) link.
- Break the bond between the carbonyl carbon (\\( \text{C=O} \\)) and the oxygen or nitrogen atom.
- Add back the components of water:
- Add an -OH group to the carbonyl carbon to reconstruct the carboxylic acid (\\( \text{-COOH} \\)).
- Add an -H atom to the oxygen or nitrogen atom to reconstruct the alcohol (\\( \text{-OH} \\)) or amine (\\( \text{-NH}_2 \\)).
When writing overall equations for polymerisation, pay close attention to the number of water molecules produced. If you start with \\( n \\) molecules of a dicarboxylic acid and \\( n \\) molecules of a diamine, you form \\( 2n-1 \\) links in a linear chain. In examinations, this is simplified and written as producing \\( 2n \\) H₂O molecules because \\( n \\) is a very large number (so \\( 2n-1 \approx 2n \\)).
Step 1: Locate the links. We have a carbon nitrogen single bond between the carbonyl groups and the nitrogen atoms: this is an amide link (\\( \text{-CO-NH-} \\)).
Step 2: Cleave the amide bond. Split the polymer at the \\( \text{C-N} \\) bond: \[ \text{[-CO-CH}_2\text{-CO- } | \text{ -NH-CH}_2\text{CH}_2\text{-NH-]} \]
Step 3: Add back the -OH and -H.
- Add \\( \text{-OH} \\) to the carbonyl groups: \\( \text{HOOC-CH}_2\text{-COOH} \\) (propanedioic acid).
- Add \\( \text{-H} \\) to the nitrogen groups: \\( \text{H}_2\text{N-CH}_2\text{CH}_2\text{-NH}_2 \\) (ethane-1,2-diamine).
Answer: The monomers are propanedioic acid and ethane-1,2-diamine.
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