Form a polyester from a diol and a dicarboxylic acid
Join the two functional groups on each monomer to form ester links, with water eliminated at each link.
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 35. Track the functional groups, repeat-unit connectivity and small-molecule by-products. Written practice is self-checked
Before you begin
- Recognise alkenes, carboxylic acids, acyl chlorides, alcohols, amines, amides and ester linkages.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which two functional-group types form the ester links in this polyester route?
What you will learn
- Identify both bifunctional monomers
- Show ester-link formation
- Account for water eliminated during chain formation
Each monomer can link at both ends
A diol has two –OH groups and a dicarboxylic acid has two –COOH groups. A hydroxyl group and a carboxyl group react to form an ester linkage, –COO–, with water eliminated.
Because each monomer is bifunctional, the reaction can continue at both ends to make a long chain. The polymer contains alternating residues from the diol and dicarboxylic acid.
Count links as well as atoms
For one idealised pair of monomers in an extended chain, two ester links form, so two water molecules are eliminated per diol–diacid pair. The repeat unit must preserve each carbon skeleton and show the ester connections.
Do not write the diol and diacid as a simple mixture in the product. The new covalent links join them into one polyester chain.
Pause and explain
What small molecule is eliminated when a diol reacts with a dicarboxylic acid?
Worked example
Deduce the polyester repeat from HOCH₂CH₂OH and HOOCCH₂COOH.
Show the worked solution
- Keep the diol residue –OCH₂CH₂O–.
- Keep the diacid residue –COCH₂CO– between carbonyl carbons and its carbon chain.
- Join residues through –O–CO– ester links and account for water.
Answer An equivalent repeat is [–OCH₂CH₂OCOCH₂CO–]ₙ; water is eliminated as ester links form.
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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.
Use HOCH₂CH₂OH and HOOC–C₆H₄–COOH to sketch the repeating residues in the polyester.
Give me a hint
Alternate the diol and diacid residues through –O–CO– links.
Compare my reasoning
- The diol residue is –OCH₂CH₂O–.
- The diacid residue keeps both carbonyls and the aromatic ring.
- Connect the residues with ester links.
[–OCH₂CH₂OCO–C₆H₄–CO–]ₙ, with water eliminated as the ester links form.
Look for these in your work
- Both monomer residues retained
- Ester linkage shown
- Water identified
State why both monomers need two functional groups to form a long-chain polyester.
Give me a hint
Think about how a growing chain continues at each end.
Compare my reasoning
- Each monomer can form one link at each end.
- A second hydroxyl or carboxyl group allows continued reaction.
- Repeated linking produces an extended chain.
Each bifunctional monomer can connect to neighbours on both sides, allowing the polyester chain to grow beyond a single ester molecule.
Look for these in your work
- Bifunctionality explained
- Two chain ends considered
- Polymer growth linked
A proposed repeat omits the aromatic ring from the dicarboxylic acid. Explain the atom-accounting error.
Give me a hint
Compare the polymer residues with the original diacid skeleton.
Compare my reasoning
- The aromatic ring is part of the dicarboxylic acid residue.
- Only the –OH hydrogen and carboxyl –OH are removed as water at a link.
- The ring must remain in the polyester repeat.
The ring must remain between the two carbonyl groups; condensation removes water but does not remove the diacid's carbon skeleton.
Look for these in your work
- Residue traced
- Eliminated atoms distinguished
- Repeat corrected
Common mistakes
- Using only one –OH or –COOH group and stopping the chain
- Dropping a carbon from either monomer
- Writing an amide link instead of –COO–
- Including water as a repeat-unit atom
What can you explain now?
Which two functional-group types form the ester links in this polyester route?
Compare with the explanation
A hydroxyl group and a carboxyl group
The diol –OH and dicarboxylic acid –COOH form –COO–.
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Make it stick
How do a diol and a dicarboxylic acid make an ester-linked chain, and what is eliminated?
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.