UNIT 2 CHEMISTRY · PEARSON AAQ · E1.8.4
Condensation Polymerisation
FREE CHEMISTRY REVISION · TOPIC 66
Condensation Polymerisation
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.8.4
What you will learn
- Explain how two functional groups let each monomer join at both ends.
- Draw polyester repeating units from a dicarboxylic acid and a diol.
- Identify ester links and water formation; distinguish polyester formation from polyamide formation with diamines.
1. Functional groups build a chain
Condensation polymerisation joins monomers through reactions between functional groups, with a small molecule eliminated each time a link forms. For the dicarboxylic-acid/diol route here, that small molecule is water. A dicarboxylic acid has two –COOH groups; a diol has two –OH groups. Each molecule can react at both ends, allowing a long chain rather than just one small ester.
Write the acid as HOOC–R–COOH and the diol as HO–R′–OH. R and R′ represent the remaining carbon-containing parts, not extra functional groups or individual atoms. Using monomers with two reactive groups favours linear chains; monomers with more than two can form branches or networks. A monofunctional alcohol or acid can cap an end and limit chain growth.
2. One ester linkage, one water
A carboxylic acid group reacts with an alcohol group: R–C(=O)–OH + H–O–R′ → R–C(=O)–O–R′ + H₂O. The acid contributes –OH and the alcohol contributes H to the water. The alcohol oxygen stays in the ester linkage. The ester functional group is –C(=O)–O–; simply writing C–O without the carbonyl is incomplete.
Polyesters contain many ester linkages in their backbone. Heating and suitable catalysts can help the reaction; removing water drives esterification towards products and helps obtain long chains. Polymer manufacture uses controlled industrial conditions, not a single universal temperature.
3. From monomers to repeating unit
The generic repeat can be written [–O–R′–O–C(=O)–R–C(=O)–]ₙ. It contains one residue from each monomer. Reversing the order or starting the repeat at another point is valid if the continuous chain has the same connectivity. Count the ester linkage across a repeat boundary as well as the one inside it.
A conventional long-chain representation is nHO–R′–OH + nHOOC–R–COOH → [–O–R′–O–C(=O)–R–C(=O)–]ₙ + 2nH₂O, with end groups omitted. This is an idealised repeat-unit equation, not an exact finite-chain end-group count. For one open linear chain formed from n diol plus n diacid molecules, there are 2n−1 links and 2n−1 waters, with terminal groups retained. The reliable rule is always one water per ester linkage actually formed.
4. Diamines: a different polymer family
The specification heading also names diamines. A diamine, H₂N–R′–NH₂, has two amine groups. A dicarboxylic acid reacting with a diamine forms a polyamide, with –C(=O)–NH– links and water elimination, rather than a polyester. The idealised repeat is [–NH–R′–NH–C(=O)–R–C(=O)–]ₙ. Nylon-6,6 is a familiar example from hexane-1,6-diamine and hexanedioic acid.
Do not draw an ester –C(=O)–O– where an amide –C(=O)–NH– belongs. The detailed polyester requirement uses a diol, not a diamine. Other condensation routes can release other small molecules; water is specific to the acid/alcohol and acid/amine routes taught here.
5. Compare addition and condensation
| Feature | Addition: alkene examples | Condensation: polyester example |
|---|---|---|
| Monomers | C=C-containing monomers | Diol + dicarboxylic acid, each bifunctional |
| What joins | New C–C bonds form as C=C opens | Functional groups react to make ester linkages |
| Small molecule | None eliminated | One H₂O per ester linkage |
| Repeat atoms | Same atoms as the alkene monomer | Monomer residues after water loss |
| Backbone | Carbon backbone in the specified addition polymers | Contains ester links with oxygen in the backbone |
Polyesters can be used in textile fibres and bottles. PET is made from ethane-1,2-diol and benzene-1,4-dicarboxylic acid; the aromatic ring adds rigidity. Having hydrolysable ester bonds does not mean every polyester rapidly biodegrades outdoors. Structure, crystallinity and disposal conditions matter.
Worked and applied examples
Example 1 · Predict a polyester repeat
Ethane-1,2-diol plus hexanedioic acid gives [–O–CH₂–CH₂–O–C(=O)–(CH₂)₄–C(=O)–]ₙ. Keep the four central CH₂ groups of the acid: the two –COOH carbons are already shown as carbonyl carbons.
Example 2 · Check the repeat formula
Hexanedioic acid C₆H₁₀O₄ plus ethane-1,2-diol C₂H₆O₂ minus 2H₂O gives C₈H₁₂O₄ per idealised repeat. C: 6+2=8; H: 10+6−4=12; O: 4+2−2=4. End-group corrections apply to a finite chain.
Example 3 · Recover monomers
For [–O–(CH₂)₄–O–C(=O)–(CH₂)₂–C(=O)–]ₙ, split ester links and restore OH/H. The monomers are butane-1,4-diol, HO–(CH₂)₄–OH, and butanedioic acid, HOOC–(CH₂)₂–COOH.
Example 4 · Count water exactly
If 7 ester links are formed, 7 water molecules are eliminated. If 10 diol and 10 diacid molecules join into one open, unbranched chain, 20 starting molecules require 19 joining links, giving 19 waters. Do not blindly use 20 for this finite-chain question.
Example 5 · Identify a polyamide
A chain containing –C(=O)–NH– comes from an amine-containing monomer in this context. A diamine plus a diacid can make a polyamide; replacing NH with O changes the polymer family.
Example 6 · Why ethanol does not replace a diol
Ethanol has only one –OH. It can form an ester at one end, but cannot connect two diacid residues through two alcohol groups. A bifunctional diol is needed for sustained chain growth in this route.
Common misconceptions and examination tips
- Draw C=O as a double bond; carbonyl carbon has total bond order four.
- Keep both oxygen atoms of each ester linkage: carbonyl O and single-bonded O.
- Do not leave acid –OH groups inside the completed repeat backbone.
- One water forms per link, not automatically per monomer. Identify whether chain ends are included.
- Diamines make polyamides with dicarboxylic acids. Diols make polyesters.
- When evaluating waste, distinguish chemical hydrolysability from rapid biodegradation in a particular environment.
Check your understanding
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