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4.8.3Organic Chemistry

Condensation Polymers — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 18 of 18 Unit quiz

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Part 1 of 5Condensation Polymerisation

4.8.3 Condensation Polymers

The polymer forms alongside a small molecule, unlike addition polymerisation.

Definition
  • Two different monomers join together with loss of a small molecule, usually water, for every link formed.
Monomers
  • Each monomer has two functional groups, one at each end, so chains can grow in both directions.
Polyester
  • A dicarboxylic acid (–COOH at both ends) reacts with a diol (–OH at both ends) to give ester linkages.
Water lost
  • For each ester link, an –OH from the acid and an –H from the alcohol combine to form H₂O.

Part 2 of 5Example: Terylene

Example: Terylene
Monomers
  • Terylene is made from a dicarboxylic acid, benzene-1,4-dicarboxylic acid, and a diol, ethane-1,2-diol.
Block diagram
  • HOOC–[ ]–COOH + HO–[ ]–OH → –OC–[ ]–COO–[ ]–O– with one H₂O lost per link.
Repeat unit
  • The repeating unit contains one ester linkage –COO– between the two carbon-containing blocks.
Product
  • The polymer is a polyester used in fibres for clothing.
FeatureAdditionCondensation
MonomersUsually one type with C=CTwo functional groups per monomer
Small molecule lostNoYes, usually water
ProductPolymer onlyPolymer plus water

Part 3 of 5Deducing Monomers from a Polyester

Deducing monomers reverses the polymerisation by adding water.

1
Locate each ester linkage –COO– in the polymer chain.
2
Break the C–O single bond in each ester linkage.
3
Add –OH to the carbon of the C=O end, which gives a –COOH group.
4
Add –H to the oxygen end, which gives an –OH group.
5
The resulting molecules are the dicarboxylic acid and the diol; the process is hydrolysis.

Part 4 of 5Key Points for Exams

Key Points for Exams
Two monomers
  • Condensation polyesters need two different monomers, or one monomer with both functional groups.
Alternating chain
  • The polymer chain alternates the two units, so the pattern of bonds is mirrored on either side of the ester link.
Hydrolysis
  • Adding water to a polyester breaks it back into the acid and alcohol monomers.
Comparison
  • Unlike addition polymers, no C=C double bond is needed in the monomers.

Part 5 of 5Exam Checklist & Tips

4.8.3 – Key Points for the Exam

  • Condensation polymerisation Formation of a polymer with loss of a small molecule such as water.
  • Polyester A polymer joined by ester linkages.
  • Monomer A small molecule that joins with others to make a polymer.
  • Hydrolysis Breaking a bond by reaction with water.

Common Mistakes

  • Treating a condensation polymer as if it forms from monomers with a C=C double bond.
  • Forgetting that water is lost for every ester link.
  • Giving only one monomer for terylene instead of a diacid and a diol.
  • Breaking the C=O bond instead of the C–O single bond when deducing monomers.

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