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K324 / 2027
Organic Chemistry

Lesson 8 of 9

Condensation polymers and depolymerisation

Recognise the linkage that is made or broken.

O-Level 6092 (2026) / SEC G3 K324 (2027)

Pure only
  • I can describe polymers and monomers.Syllabus

    Syllabus K324, 11.4(a). Describe polymers and monomers

  • I can deduce addition-polymer structures.Syllabus

    Syllabus K324, 11.4(d). Deduce addition-polymer structures

  • I can compare recycling routes.Syllabus

    Syllabus K324, 11.4(h). Compare recycling routes

  • I can recognise condensation polymers.Syllabus

    Syllabus K324, 11.4(e). Recognise condensation polymers

  • I can identify man-made fibre uses.Syllabus

    Syllabus K324, 11.4(f). Identify man-made fibre uses

  • I can explain depolymerisation.Syllabus

    Syllabus K324, 11.4(i). Explain depolymerisation

Condensation polymers and depolymerisation

Recognise the linkage that is made or broken.

Condensation polymerisation joins monomers with suitable functional groups, eliminating small molecules such as water. Nylon is a polyamide, with -C(=O)-NH- links. Terylene is a polyester, with -C(=O)-O- links. Different polymers may differ in their monomer units, their linkages, or both.

Read partial chain structures
PolymerA simplified repeating patternLinkage
Nylon[-NH-A-NH-C(=O)-B-C(=O)-]nAmide: -C(=O)-NH- between neighbouring units
Terylene[-O-A-O-C(=O)-B-C(=O)-]nEster: -C(=O)-O- between neighbouring units

A and B stand for the intervening carbon-containing sections, not chemical element symbols. Follow the chain through the bracket boundary to find each linkage. These partial structures explain the polymer class without requiring industrial manufacture details or mechanisms.

Nylon: recognise the amide links

A partial chain with two carbonyl groups flanking carbon-containing segment B and NH groups flanking segment A. Open bonds at both edges continue the polymer.

An amide link is -C(=O)-NH-. Boxes A and B stand for the intervening carbon-containing sections, as in a partial structure; they are not element symbols.

Terylene: recognise the ester links

A partial chain with two carbonyl groups flanking carbon-containing segment B and O groups flanking segment A. Open bonds at both edges continue the polymer.

An ester link is -C(=O)-O-. Boxes A and B stand for the intervening carbon-containing sections, as in a partial structure; they are not element symbols.

Worked exampleDeduce a condensation repeat unit

How would HO-A-OH and HOOC-B-COOH join, and how does replacing the alcohol with H2N-A-NH2 change the polymer?

Try it first. Then show the worked stepsHide the worked steps
  1. The diol has two alcohol ends and the diacid two carboxylic-acid ends, so repeated joining can build a long chain.
  2. Each ester link eliminates water and connects -O-A-O- with -C(=O)-B-C(=O)-.
  3. With a diamine instead, -NH-A-NH- replaces the oxygen-containing segment; amide links form.
Answer

The diol/diacid gives a polyester; the diamine/diacid gives a polyamide. To deduce monomers from the chain, cut the linkages and restore -OH at acid ends and H at alcohol/amine ends.

Nylon and Terylene are useful man-made fibres: applications include clothing, curtains, fishing line, parachutes and sleeping bags. Suitability depends on properties such as strength, durability and flexibility; knowing the material name is less useful than connecting a property to the job.

Hydrolyse a polyester
  1. Add water under acidic conditions

    Acid acts as a catalyst; heating can speed the process.

  2. Break ester links

    Water contributes H and OH across the broken linkage.

  3. Recover monomers

    Alcohol and carboxylic-acid groups are restored; extensive cleavage can return the original monomers.

For a chain made from HO-A-OH and HOOC-B-COOH, hydrolysis restores these two monomer types. This is depolymerisation: breaking a polymer into its monomers. It is different from merely melting the plastic, which leaves the long molecules largely intact.

Worked exampleRecover both ends when a polyester is hydrolysed

A polyester repeat contains -O-A-O-C(=O)-B-C(=O)-, with the chain continuing at both ends. Deduce the two monomer types after complete acid-catalysed hydrolysis.

Try it first. Then show the worked stepsHide the worked steps
  1. Mark every -C(=O)-O- ester link, including the one at the repeat boundary.
  2. Break the bond between the carbonyl carbon and the single-bonded oxygen. Restore OH to each carbonyl carbon to form the two carboxylic-acid ends.
  3. Restore H to each alcohol-derived oxygen. Segment A has two alcohol ends; segment B has two acid ends.
Answer

HO-A-OH and HOOC-B-COOH. Water provides the H/OH; the acid is a catalyst. Cutting a C-C bond or returning only one end group would give the wrong monomers.

Check your understanding

A polymer chain contains the link -C(=O)-NH- again and again. Is it a polyester?

  1. No. -CONH- is an amide link, so it is a polyamide such as nylon.
  2. No. -CONH- is an addition polymer link, like the one in poly(ethene).
  3. Yes. Any condensation polymer containing C=O groups is a polyester.
Show the answer

No. -CONH- is an amide link, so it is a polyamide such as nylon.

Look at the atom next to the carbonyl carbon. N gives an amide link (polyamide); O gives an ester link, -C(=O)-O- (polyester).