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, H₂O, or hydrogen chloride, 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. -COOH, -OH, -NH₂). 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 H₂O or HCl).
Polyesters
Polyesters contain ester links (-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 (-COO-) in its backbone, formed by condensation reactions between carboxylic acid and alcohol functional groups.
Polyamides
Polyamides contain amide links (-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 (-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 (-COOH), remove the -OH group.
- From amine groups (-NH₂) or alcohol groups (-OH), remove an -H atom.
- Draw a covalent bond joining the remaining carbonyl carbon (C=O) to the heteroatom (-O- or -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 (-CO-O-) or amide (-CO-NH-) link.
- Break the bond between the carbonyl carbon (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 (-COOH).
- Add an -H atom to the oxygen or nitrogen atom to reconstruct the alcohol (-OH) or amine (-NH₂).
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 (-CO-NH-).
Step 2: Cleave the amide bond. Split the polymer at the 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 -OH to the carbonyl groups: HOOC-CH₂-COOH (propanedioic acid).
- Add -H to the nitrogen groups: H₂N-CH₂CH₂-NH₂ (ethane-1,2-diamine).
Answer: The monomers are propanedioic acid and ethane-1,2-diamine.
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