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A Level · Cambridge (CIE) · 9701

Form a polyester from a hydroxycarboxylic acid

A hydroxycarboxylic acid links to itself through ester groups, eliminating water to form a polyester.

Polymerisation · Cambridge International AS & A Level Chemistry 9701 · 35.1

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 groups allow a hydroxycarboxylic acid to self-condense?

What you will learn

  • Recognise both functional groups on one monomer
  • Describe self-condensation into an ester-linked chain
  • Deduce the residue retained in the repeat

One monomer carries both ends

A hydroxycarboxylic acid contains an alcohol –OH group and a carboxylic acid –COOH group. The –OH of one molecule can react with the –COOH of another to form an ester link.

Because each molecule has one group of each type, the reaction can continue in both directions through repeated condensation. Water is eliminated as ester links form, and the carbon skeleton between –OH and –COOH remains in the chain.

Keep the monomer residue intact

For a monomer HO–R–COOH, the polyester repeat can be represented as [–O–R–CO–]ₙ. The oxygen from the hydroxyl group and the carbonyl from the carboxyl group appear in the ester linkage.

A hydroxycarboxylic acid is one monomer type with two different functional groups; it is not a mixture of a separate diol and diacid.

Pause and explain

What linkage forms when a hydroxycarboxylic acid self-condenses?

Put the idea to work

Worked example

Deduce the repeat from 2-hydroxyethanoic acid, HOCH₂COOH.

Show the worked solution
  1. Identify –OH and –COOH in the same molecule.
  2. Join the hydroxyl oxygen to a neighbouring carbonyl carbon.
  3. Retain the CH₂ residue and show ester connectivity.

Answer The repeat is [–OCH₂CO–]ₙ; water is eliminated as the ester links form.

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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.

1 · Guided

Write a repeat unit for HOCH₂CH₂COOH polymerising by condensation.

Give me a hint

Keep the two-carbon residue between hydroxyl oxygen and carbonyl carbon.

Compare my reasoning
  1. The monomer has both –OH and –COOH.
  2. Join molecules through –O–CO–.
  3. Bracket the repeated –OCH₂CH₂CO– residue.

[–OCH₂CH₂CO–]ₙ, with water eliminated as ester links form.

Look for these in your work

  • Correct monomer residue
  • Ester linkage
  • Water by-product
2 · Independent

Explain why one type of hydroxycarboxylic-acid monomer can form a chain without a separate diol.

Give me a hint

Count the different functional groups on each molecule.

Compare my reasoning
  1. Each molecule has one hydroxyl group.
  2. Each molecule also has one carboxyl group.
  3. Those two ends link successive molecules in both directions.

Each monomer is bifunctional: its –OH and –COOH groups can form ester links to neighbouring monomers, so one monomer type can make a chain.

Look for these in your work

  • Both groups named
  • Bifunctionality explained
  • Chain growth explained
3 · Transfer

A proposed repeat for HOCH₂COOH is [–CH₂CH₂CO–]ₙ. Identify the error.

Give me a hint

Trace the monomer's atoms and the oxygen in the ester link.

Compare my reasoning
  1. The monomer has one CH₂ group, not two.
  2. Its hydroxyl oxygen remains in the ester linkage.
  3. The repeat should be –OCH₂CO–.

The correct repeat is [–OCH₂CO–]ₙ: the proposed structure adds a carbon and loses the hydroxyl oxygen.

Look for these in your work

  • Carbon count checked
  • Oxygen tracked
  • Repeat corrected

Common mistakes

  • Treating the molecule as two separate monomers
  • Removing the CH₂ carbon skeleton
  • Drawing an amide link
  • Forgetting that condensation eliminates water
Recall without your notes

What can you explain now?

Which two groups allow a hydroxycarboxylic acid to self-condense?

Compare with the explanation

–OH and –COOH

One molecule carries both groups needed to form an ester link.

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Make it stick

How can one hydroxycarboxylic-acid monomer form a polyester chain?

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.

Keep learning Chemistry

Course scope and assessment

Four hundred and forty-two original lessons map the published learning outcomes across AS Chemistry Topics 1–22 and A Level Topics 23–37, with guided, independent and transfer practice and explained answers. Every numbered subject-content topic and listed outcome tag now has a mapped lesson sequence. Topic 32 maps all 12 hydroxy-compound outcome tags in 11 lessons; Topic 33 maps all 15 carboxylic-acid and derivative outcome tags in 16 lessons; Topic 34 maps all 22 nitrogen-compound outcome tags in 20 lessons; Topic 35 maps all 12 polymerisation outcome tags in 16 lessons; Topic 36 maps all four organic-synthesis outcome tags in 14 lessons; Topic 37 maps all 22 chromatography and NMR outcome tags in 18 lessons. Cambridge's 2028–2030 Version 1 syllabus states that there are no significant changes which affect teaching from the preceding edition; the routes are aligned to that edition. This is 37 of the 37 numbered subject-content topics. Full cumulative exam assessment, Paper 3 laboratory competence, Paper 5 investigation practice and human subject-teacher review remain outstanding. Written practice is self-checked and does not certify exam readiness.

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