Lesson 31 of 34
The repeat unit must preserve the correct bonds and functional groups
Addition opens a multiple bond; condensation joins functional groups while losing a small molecule.
A-Level 9476 (2026-2027)
I can recognise polymers as macromolecules built from monomers.Syllabus
Syllabus 9476, 11.10(a). Recognise polymers as macromolecules built from monomers.
I can distinguish addition and condensation polymerisation.Syllabus
Syllabus 9476, 11.10(b). Distinguish addition and condensation polymerisation.
The repeat unit must preserve the correct bonds and functional groups
Addition opens a multiple bond; condensation joins functional groups while losing a small molecule.
Polymers are very large molecules built from monomers. The syllabus counts a molecule as a polymer if its average relative molecular mass is at least 1000, or if it has at least 100 repeat units. A polymer sample contains chains of different lengths, which is why we use an average relative molecular mass.
| Type | What happens | How to recover the monomer |
|---|---|---|
| Addition, exemplified by poly(alkenes) | A C=C pi bond is opened and new C-C sigma links extend the chain; no small molecule is eliminated. | Choose a two-carbon backbone repeat and restore the C=C between those two carbons, retaining substituents. |
| Condensation, exemplified by polyesters | A diol and a dicarboxylic acid, or related acyl derivative, form repeated ester links while losing H2O or HCl. | Break each ester link at C(=O)-O and restore OH/H as appropriate. |
| Condensation, exemplified by polyamides | Amine and carboxylic-acid/acyl-chloride groups form repeated amide links. | Break C(=O)-N links and restore the acid and amine groups. |
A polyester repeat unit keeps both carbonyls
The bracketed repeat unit is O-R-O-C(=O)-R-prime-C(=O), with bonds crossing both brackets. R is the diol spacer and R-prime the dicarboxylic-acid spacer. Each carbonyl carbon has a double bond to oxygen and two single bonds in the backbone.
Worked exampleRecover an addition monomer
A polymer repeat is -CH2-CH(CH3)-. Which alkene forms it?
Try it first. Then show the worked stepsHide the worked steps
- The two backbone carbons arose from one C=C pair.
- The methyl group is a side-chain substituent, not a third backbone atom of that repeat.
- Restore a double bond between the two backbone carbons.
Propene, CH2=CHCH3. The repeat has a saturated backbone even though its monomer was unsaturated.
To make a chain, each monomer needs at least two reactive groups so that linking can continue. A diol and a dicarboxylic acid can both react at each end, so the chain grows. An alcohol and a carboxylic acid with one group each just make a small ester. In an exact equation for a chain of set length, the number of small molecules lost depends on the number of links and end groups, so do not read it straight from the repeat unit.
Check your understanding
A polymer is made from a diol and a dicarboxylic acid. What type of polymer is it, and what else forms?
Show the answer
A polyester, made by condensation, with water given off
Each OH reacts with a COOH to form an ester link and lose H2O. Losing a small molecule at each link is what makes it condensation.