The polyester in your shirt, the nylon in your toothbrush bristles, the Kevlar in a bulletproof vest, and the proteins in your muscles are all condensation polymers — made by the same type of reaction, with water lost at every link, and the same ester or amide bond repeated millions of times.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
A PET plastic bottle and a polyester shirt are made from the same polymer. A nylon rope and a silk thread have the same type of chemical linkage — both are polyamides. The proteins in your muscles are also polyamides.
Before reading: what do you think the structural difference is between a polyester and a polyamide? What connecting group holds the monomers together in each case? And why do you think water is lost when these polymers form?
Wrong: Addition polymers and condensation polymers both release a small molecule during formation.
Right: Addition polymers form by monomers adding together with no by-product (e.g., polyethylene from ethene). Condensation polymers form with the loss of a small molecule like water or HCl. The presence or absence of a by-product is the defining distinction between the two polymerisation types.
In condensation polymerisation, every bond that forms between two monomers also releases a small molecule — usually water — so the polymer has fewer atoms than the sum of its monomers, unlike addition polymers where every atom is retained.
Requirements: Monomers must be bifunctional — each monomer has two reactive groups (one at each end), allowing the chain to grow from both ends simultaneously. When two groups react to form a linkage, a small molecule is eliminated:
| Feature | Addition polymerisation | Condensation polymerisation |
|---|---|---|
| Monomer requirement | ONE functional group — C=C per monomer | TWO functional groups per monomer (or two bifunctional monomer types) |
| By-product | NONE — all atoms retained in polymer | H₂O (or HCl) — released at every bond formed |
| Polymer formula | Same empirical formula as monomer | Different — atoms lost in by-product |
| Linkage type | C-C (from C=C pi bond opening) | -COO- ester or -CO-NH- amide |
| Monomer type | Alkene (vinyl compound) | Diol + diacid; diamine + diacid; hydroxy acid; amino acid |
| Examples | PE, PP, PVC, PTFE, PS | PET, Nylon 6,6, Dacron, silk, proteins, collagen |
| Hydrolysable? | No — C-C bonds not susceptible to water | Yes — ester and amide bonds cleave with acid, base, or enzymes |
| Environmental persistence | Very high — no hydrolysable linkage; degrades only by UV/fragmentation | High, but ester/amide bonds can hydrolyse slowly; proteins rapidly digested |
Polyesters form when hydroxyl groups (-OH) and carboxyl groups (-COOH) react — the same esterification from Module 7, but repeated n times using bifunctional monomers to produce a polymer chain instead of a small ester molecule.
Diol: ethylene glycol (ethane-1,2-diol) — HOCH₂CH₂OH — two -OH groups
Diacid: terephthalic acid (benzene-1,4-dicarboxylic acid) — HOOC-C₆H₄-COOH — two -COOH groups
High MP (~260°C) — benzene ring rigidity. High tensile strength. Gas barrier properties. Recyclable (RIC code 1).
Uses: fizzy drink bottles, water bottles, polyester fabric (Dacron/Terylene — spun PET fibres), food packaging, photographic film.
Polyamides form by the same condensation mechanism as polyesters, but using amino groups (-NH₂) instead of hydroxyl groups — and the amide linkage that results is the same bond that holds every protein in your body together.
Diamine: hexane-1,6-diamine — H₂N(CH₂)₆NH₂ — 6 carbons, -NH₂ at each end
Diacid: hexanedioic acid (adipic acid) — HOOC(CH₂)₄COOH — 6 carbons total, -COOH at each end
"6,6" = 6 carbons in each monomer
High MP (~265°C). N-H···O=C H-bonds between chains. High tensile strength + flexibility. Absorbs water (hygroscopic).
Uses: toothbrush bristles, clothing, ropes, parachutes, fishing line, gears, bearings, tyre cords.
The two-way conversion skill — polymer from monomers, and monomers from polymer — is the core exam skill of IQ6 and applies identically to polyesters and polyamides.
Because ester and amide bonds form by condensation of functional groups, they can be broken by the reverse — hydrolysis (adding H₂O across the bond). This explains:
Draw the repeat unit in correct square bracket notation for each of the following. State the linkage type and the by-product.
(a) Butane-1,4-diol (HOCH₂CH₂CH₂CH₂OH) + hexanedioic acid (HOOC(CH₂)₄COOH)
(b) Hydroxy acid: lactic acid (HO-CH(CH₃)-COOH) — polymerises with itself
(c) Hexane-1,6-diamine (H₂N(CH₂)₆NH₂) + hexanedioic acid (HOOC(CH₂)₄COOH)
For each repeat unit below, identify the linkage type, name and draw the monomers, and classify the polymer.
(i) [-O-(CH₂)₄-O-CO-(CH₂)₄-CO-]ₙ
(ii) [-NH-(CH₂)₅-CO-]ₙ (Nylon 6 — single amino acid monomer)
(iii) [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ (PET)
| Feature | This polyester | Polystyrene |
|---|---|---|
| Monomers | Butane-1,4-diol + hexanedioic acid (bifunctional, -OH and -COOH) | Styrene — one C=C only |
| By-product | H₂O (2 per repeat unit) | None |
| Linkage | -COO- (ester) | -C-C- |
| Hydrolysable? | Yes — ester bonds cleave under acid/base | No — C-C bonds not hydrolysed |
Q1. Which correctly describes condensation polymerisation?
Q2. A polymer has repeat unit [-CO-(CH₂)₄-CO-NH-(CH₂)₆-NH-]ₙ. What are the monomers and polymer type?
Q3. PET is susceptible to hydrolysis, but polystyrene is not. Which statement best explains this difference?
Q4. A student draws the repeat unit of Nylon 6,6 as [-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ and states it contains one amide linkage per repeat unit. What is wrong?
Q5. Which monomer pairing CANNOT undergo condensation polymerisation to form a polyamide?
Q6. (3 marks) (a) Draw the repeat unit of PET using the correct square bracket notation. (b) Name both monomers by IUPAC name and state the linkage type. (c) State the by-product and how many moles are produced per mole of repeat unit.
Q7. (4 marks) A polymer X has repeat unit [-O-(CH₂)₂-O-CO-C₆H₄-CO-]ₙ. (a) Identify the linkage type and classify the polymer. (b) Identify the two monomers by name and structural formula. (c) Explain why polymer X is hydrolysable while polyethylene is not, using bond chemistry. (d) State one application of polymer X and explain how the benzene ring in one monomer influences the polymer's melting point.
Q8. (5 marks) Compare addition polymerisation (using polypropylene as an example) with condensation polymerisation (using Nylon 6,6 as an example) across the following criteria: (i) monomer functional group requirements; (ii) by-product of polymerisation; (iii) linkage type and how it forms; (iv) hydrolysability; (v) environmental persistence in ocean water. Give structural justification for each point.
At the start you predicted the connecting group in polyesters and polyamides and why water is lost. Now you know: polyesters have an ester linkage (-COO-) formed when -OH reacts with -COOH, losing H₂O. Polyamides have an amide linkage (-CO-NH-) formed when -NH₂ reacts with -COOH, also losing H₂O. Water is lost because the H from -OH or -NH₂ and the OH from -COOH combine and leave — the linkage is what remains between the two carbon chains.