Polyesters and PET: ester links and return to monomers
Deduce a diacid/diol polyester, recognise PET's specified backbone and distinguish chemical return to monomers from merely melting the plastic.
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Identify the ester linkage and acid/alcohol condensation with water loss.
- Distinguish physical melting from a chemical change in chain connectivity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which functional-group pair makes the specified polyester?
What you will learn
- Deduce polyester repeat units and monomers from a dicarboxylic acid and diol.
- Describe and draw PET as a polyester and state that it can return to monomers and be re-polymerised.
Use two alcohol ends and two acid ends
A diol HO–R₂–OH and dicarboxylic acid HOOC–R₁–COOH can form a polyester by condensation. The pattern [–O–R₂–O–C(=O)–R₁–C(=O)–]ₙ contains ester links C(=O)–O and releases water at formed acid–alcohol links. The monomers' two ends allow the chain to continue. To deduce them back, restore OH to the acid carbonyl end and H to the alcohol oxygen, giving COOH and OH respectively.
PET is the specified polyester example. Its diol-derived segment is –CH₂–CH₂– and its acid-derived segment is the C₆H₄ aromatic backbone shown by the box, with opposite ring attachment positions. The drawing uses [–O–CH₂–CH₂–O–C(=O)–R–C(=O)–]ₙ and explicitly defines that R as PET's particular acid backbone, rather than any arbitrary atom. The full expanded name of PET is not required by this syllabus objective.
Recovering monomers requires changing the bonds
PET can be converted back into monomers and re-polymerised. In the hydrolysis model, water is used when ester links are broken to restore the acid and alcohol functional groups. The recovered diol is HOCH₂CH₂OH and the acid is HOOC–C₆H₄–COOH. They can form PET again through the stated condensation pattern.
Simply heating to melt PET changes its physical state and does not by itself demonstrate recovery of separate acid and diol monomers. Actual chemical recycling also requires suitable recovery and processing; the stated possibility does not guarantee every discarded item is collected, has pure recovered monomers or returns with no losses. As with polyamides, a finite linear four-monomer model has three links and three water losses, rather than an automatically doubled water count from repeat notation.
Pause and explain
Which statement correctly describes PET monomer recovery?
Worked example
Identify the PET link, recover its two monomer structures and distinguish the chemical recycling claim from melting.
Show the worked solution
- PET contains ester links C(=O)–O rather than amide C(=O)–NH.
- Restoring the acid/alcohol groups gives HOOC–C₆H₄–COOH and HOCH₂CH₂OH.
- Breaking and restoring functional groups is chemical conversion; melting alone leaves a polymer rather than proving monomer recovery.
Answer Ester links; monomers HOOC–C₆H₄–COOH and HOCH₂CH₂OH. PET can return to monomers and be re-polymerised; melting alone does not establish that conversion.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw the PET repeat with its two diol carbons, all their H atoms, carbonyls, ester oxygens and defined acid-backbone box.
Give me a hint
Use oxygen at the linking positions rather than nylon's NH.
Compare my reasoning
- Draw O–CH₂–CH₂–O for the diol-derived part, with four individually shown H.
- Join C(=O)–R–C(=O), defining R as PET's stated C₆H₄ acid backbone.
- Extend bonds across the brackets, label n, and identify the C(=O)–O ester links.
[–O–CH₂–CH₂–O–C(=O)–R–C(=O)–]ₙ, all four diol H shown and R explicitly PET's C₆H₄ backbone.
Look for these in your work
- The ester links and explicit atoms are correct.
- The box and continuing repeat boundaries are defined.
Deduce the acid and diol monomers from the PET repeat and describe why they can form a continuing chain.
Give me a hint
Restore COOH and OH at the appropriate cut ester links.
Compare my reasoning
- The acid is HOOC–C₆H₄–COOH.
- The diol is HOCH₂CH₂OH.
- Both monomers have two relevant functional-group ends, so repeated condensation can create a polyester chain with water loss.
HOOC–C₆H₄–COOH and HOCH₂CH₂OH; both are bifunctional and can re-form PET through ester-link condensation.
Look for these in your work
- Both recovered monomer structures are correct.
- The two ends and chain-forming link type are explained.
A supplied linear model joins two C₈H₆O₄ PET diacid molecules and two C₂H₆O₂ diol molecules with three ester links. Deduce the product formula after water loss and assess whether melting alone would reverse those links.
Give me a hint
Add the complete monomers and subtract H₂O once per stated new link.
Compare my reasoning
- The four starting molecules total C₂₀H₂₄O₁₂.
- Three links eliminate H₆O₃, leaving C₂₀H₁₈O₉, including the model's remaining end groups.
- Melting alone does not restore the monomers; the links must undergo an appropriate chemical conversion.
C₂₀H₁₈O₉ after three water losses; melting alone does not establish chemical recovery of the monomers.
Look for these in your work
- The finite-model atom and link counts agree.
- Physical melting and chemical reversal remain distinct.
Common mistakes
- Drawing amide nitrogen into the specified PET polyester.
- Treating the acid-backbone box as an arbitrary single atom.
- Using a one-OH alcohol as the stated continuing diol monomer.
- Claiming a molten plastic must already be separate recovered monomers.
What can you explain now?
What can happen to PET after conversion back to its monomers?
Compare with the explanation
The recovered monomers can be re-polymerised.
The possibility is chemical conversion and reconstruction, not proof that every item has already been recycled.
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Make it stick
Compare PET and nylon link atoms, recover PET's diacid and diol, then explain monomer recovery and re-polymerisation without guaranteeing real collection or yield.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.