Topic 15 · page 20 of 28

Polyesters

A polyester forms when a molecule with two –COOH groups reacts with a molecule with two –OH groups. A water molecule is lost at each ester link. Terylene is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

12 min read15.16

Read first: Esterification

Condensation polymerisation 15.16

Addition polymers, such as poly(ethene), form when alkene molecules add together and nothing is lost. Condensation polymerisation is different: monomers join and a small molecule is lost each time a link forms. For polyesters the small molecule is water.

To make a long chain, each monomer must be able to join at both ends. That means each monomer has two functional groups:

An acid group on one monomer and an alcohol group on another form an ester link, exactly as in esterification, and release water. Each new link leaves a free –COOH or –OH at the end of the chain, which can react again.

Terylene 15.16

Terylene (the fibre polyethylene terephthalate) is made from two monomers:

Making Terylene. The monomers benzene-1,4-dicarboxylic acid, with a carboxyl group at each end of a benzene ring, and ethane-1,2-diol, HOCH2CH2OH, join by losing water at each ester link. The repeat unit, in square brackets with n, is C(=O), the benzene ring, C(=O), O, CH2, CH2, O, with the ester link in orange.
A water molecule is lost at each ester link. The repeat unit is in square brackets.
n HOOC−CX6HX4−COOH+n HOCHX2CHX2OH⟶[−OC−CX6HX4−COOCHX2CHX2OX−]Xn+(2n−1) HX2O n\,\ce{HOOC-C6H4-COOH} + n\,\ce{HOCH2CH2OH} \longrightarrow \ce{[-OC-C6H4-COOCH2CH2O-]_n} + (2n-1)\,\ce{H2O}

The part in square brackets is the repeat unit. It contains two ester links. Many mark schemes write 2n H₂O. For a polymer with a large n the difference between 2n and (2n − 1) is negligible.

When you draw a repeat unit, the open bonds at each end of the bracket pass through the brackets, and the n goes at the bottom right.

MethodFinding the monomers from a polyester

  1. Find the ester links, –COO–, in the chain.
  2. Cut each ester link between the C=O carbon and the single-bonded O.
  3. Add –OH to the carbon of the C=O end, which gives a carboxylic acid group.
  4. Add –H to the O end, which gives an alcohol group.

Worked exampleA different polyester

Hexanedioic acid, HOOC(CH₂)₄COOH, and ethane-1,2-diol form a polyester. Write the repeat unit and say how many ester links it contains.

  1. Join an acid end to an alcohol end with loss of water.
  2. The repeat unit is –OC(CH₂)₄COOCH₂CH₂O–.
  3. It contains two ester links, one at each –COO–.

Check yourself

  1. Explain what is meant by condensation polymerisation, and why the monomers each need two functional groups.
  2. Name the two monomers used to make Terylene.
  3. Write an equation for the formation of a polyester from hexanedioic acid and ethane-1,2-diol.
  4. Identify the ester link in the repeat unit –OC–C₆H₄–COOCH₂CH₂O–.
  5. How does condensation polymerisation differ from addition polymerisation in terms of products?

Answers cover these until you have tried the questions

  1. A reaction in which monomers join to form a polymer and a small molecule (water) is lost at each link. Each monomer needs two functional groups so that it can join to a neighbour at each end, which lets the chain keep growing in both directions.
  2. Benzene-1,4-dicarboxylic acid and ethane-1,2-diol.
  3. n HOOC(CHX2)X4COOH+n HOCHX2CHX2OH⟶[−OC(CHX2)X4COOCHX2CHX2OX−]Xn+(2n−1) HX2On\,\ce{HOOC(CH2)4COOH} + n\,\ce{HOCH2CH2OH} \longrightarrow \ce{[-OC(CH2)4COOCH2CH2O-]_n} + (2n-1)\,\ce{H2O} (or 2n H₂O).
  4. The ester link is the –COO– group, with the carbon of the benzene ring side as C=O and the single-bonded oxygen bonded to CH₂: –C(=O)–O–CH₂–.
  5. In addition polymerisation the monomers add together and the polymer is the only product. In condensation polymerisation a small molecule, here water, is also produced.