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Organic Chemistry

Topic 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)

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 macromolecules built from monomers. The syllabus uses an average relative molecular mass of at least 1000 or at least 100 repeat units as its recognition convention. A polymer sample usually contains chains of different lengths, so an average relative molecular mass is meaningful.

Two ways to build a chain
TypeWhat happensHow 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 polyestersA 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 polyamidesAmine 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.

The monomers can be HO-R-OH and HOOC-R'-COOH. The bracket-crossing bonds show the chain continues; n is the number of repeats, not a bond or an atom.

Worked example

Recover an addition monomer

A polymer repeat is -CH2-CH(CH3)-. Which alkene forms it?

  1. The two backbone carbons arose from one C=C pair.
  2. The methyl group is a side-chain substituent, not a third backbone atom of that repeat.
  3. Restore a double bond between the two backbone carbons.
Answer

Propene, CH2=CHCH3. The repeat has a saturated backbone even though its monomer was unsaturated.

A chain needs monomers with enough reactive sites to continue linking. A diol plus a dicarboxylic acid can extend at both ends; a monofunctional alcohol plus a monocarboxylic acid mainly gives a small ester. In an exact finite-chain condensation equation, the number of small molecules lost depends on the number of links and end groups, so do not infer it blindly from the repeat-unit bracket.