OCR A Level Chemistry A: Polyesters and polyamides (H432) – Revision Notes

These revision notes condense section 6.2.3, Polyesters and polyamides, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover every learning outcome, 6.2.3(a) to (c), within Module 6, Organic chemistry and analysis, and H432 is the full A level. Longer explanations and worked examples are in the Polyesters and polyamides study guide.

Once these points come back without prompting, try the Polyesters and polyamides practice questions. Every H432 topic is linked from our OCR Chemistry A hub, and the printable checklist lets you tick off each outcome. To spot weak areas elsewhere in H432, take one of the free diagnostics.

Key definitions

• Condensation polymerisation: monomers join with the loss of a small molecule, H₂O or HCl, at every link.
• Addition polymerisation: monomers with C=C join as the double bond opens; the polymer is the only product (4.1.3(j)).
• Polyester: a condensation polymer whose monomer units are joined by ester links, –COO–.
• Polyamide: a condensation polymer whose monomer units are joined by amide links, –CONH–. Each link is a secondary amide.
• Repeat unit: the smallest section of chain that, repeated, gives the whole polymer. Bonds pass out through the brackets at each end.
• Hydrolysis: the reverse of condensation, in which water splits each link; here it is carried out in hot aqueous acid or alkali.

The specification states you will not be expected to recall the structures of synthetic polyesters and polyamides or their monomers. Questions give you the monomers or a section of chain, so the skill being tested is applying the rules below to structures you have not seen before. Practise on unfamiliar examples rather than learning named polymers.

6.2.3(a): Which monomers make which link

Monomer pair or single monomer	Link	Small molecule lost
Dicarboxylic acid + diol	ester	H₂O
Diacyl chloride + diol	ester	HCl
Hydroxycarboxylic acid (one monomer)	ester	H₂O
Dicarboxylic acid + diamine	amide	H₂O
Diacyl chloride + diamine	amide	HCl
Aminocarboxylic acid (one monomer)	amide	H₂O

Two reminders from 6.1.3(f):

• Acyl chlorides react much more readily than carboxylic acids.
• Phenols are not readily esterified by carboxylic acids. A diol with –OH groups on a benzene ring needs a diacyl chloride.

Method in steps: repeat unit from monomers

1. Decide the link from the groups present: –COOH or –COCl with –OH gives ester; with –NH₂ gives amide.
2. For two monomers, write one of each side by side, acid-type monomer first.
3. Remove –OH (or –Cl) from each acyl group and H from each –OH or –NH₂.
4. Join the pieces through the new link and leave an open bond at each end.
5. Put square brackets round the unit, with a bond through each bracket, and add n.
6. Count the small molecules: for n of each of two monomers, 2n − 1, often written 2n. For one monomer, n − 1.

Reminder: ethanedioic acid, HOOC–COOH, with pentane-1,5-diamine, H₂N–(CH₂)₅–NH₂, gives –[CO–CO–NH–(CH₂)₅–NH]–, losing water.

6.2.3(b): Hydrolysis products

Both kinds of hydrolysis are carried out by heating under reflux.

Link	Hot aqueous acid (e.g. HCl(aq))	Hot aqueous alkali (e.g. NaOH(aq))
Ester, –COO–	–COOH + –OH	–COO⁻ Na⁺ + –OH
Amide, –CONH–	–COOH + –NH₃⁺ Cl⁻	–COO⁻ Na⁺ + –NH₂

Logic to remember rather than memorise:

• Acid protonates the basic –NH₂, giving an ammonium salt. The –COOH stays as the acid.
• Alkali deprotonates the –COOH, giving a carboxylate salt. The –NH₂ stays as the amine; the –OH of an alcohol is not affected.
• The –OH from a diol is the same in both conditions.
• A phenol –OH freed by hydrolysis is weakly acidic, so excess alkali turns it into –O⁻ Na⁺ (6.1.1(h)).

Method in steps: hydrolysis products

1. Find every ester or amide link in the backbone.
2. Split each link between the C=O carbon and the O or N.
3. Give –OH to every acyl end; give H to every O or N end.
4. Adjust for conditions: in acid, –NH₂ becomes –NH₃⁺ (with Cl⁻ if HCl is used); in alkali, –COOH becomes –COO⁻ Na⁺.
5. Amount of NaOH: one mole per ester or amide link, so two per repeat unit of a diacid-based polymer.

Worked reminder

A polyester made from 3-hydroxypropanoic acid has repeat unit –[O–CH₂–CH₂–CO]–, C₃H₄O₂, Mr 72.0 (C 12.0, H 1.0, O 16.0).

• A chain of Mr 36 000 holds 36 000 ÷ 72.0 = 500 repeat units.
• Hot NaOH(aq) gives HO–CH₂–CH₂–COO⁻ Na⁺, sodium 3-hydroxypropanoate.
• Hot HCl(aq) gives HO–CH₂–CH₂–COOH, the monomer itself.

6.2.3(c): Predicting from monomers and chains

6.2.3(c)(iii): Spot the type

Clue	Type
Monomer has C=C; chain backbone is all carbon	addition
Each monomer has two groups from –COOH, –COCl, –OH, –NH₂	condensation
–COO– or –CONH– sits in the backbone	condensation
–COO– or –CONH– only in side groups	still addition

Method in steps: monomers from a section of chain

1. Check the backbone. All carbon means addition; go to step 5.
2. Cut every ester or amide link between C=O and O or N.
3. Collect the pieces. One kind of piece means one monomer; two alternating kinds means two monomers.
4. Add –OH (or –Cl for an acyl chloride) to each acyl end and H to each O or N end.
5. For addition: take two backbone carbons as the repeat unit, put C=C between them and remove the open bonds.

Must-know distinctions

• Addition vs condensation: addition gives one product only; condensation also gives H₂O or HCl.
• H₂O vs HCl: water from –COOH; hydrogen chloride from –COCl.
• Acid vs alkaline hydrolysis of a polyamide: acid gives –COOH and –NH₃⁺; alkali gives –COO⁻ and –NH₂.
• Acid vs alkaline hydrolysis of a polyester: acid gives –COOH; alkali gives –COO⁻. The alcohol is the same either way.
• Repeat unit vs monomer: the repeat unit has lost the atoms of the small molecule; the monomer has not.
• Two monomers vs one: a diacid–diol or diacid–diamine polymer has one residue of each in its repeat unit; a single-monomer polymer has one residue.

Quick self-test

1. ClOC–(CH₂)₃–COCl polymerises with a diol. Which small molecule forms?
2. Give the type of polymerisation for CH₂=C(CH₃)–CH₂OH, and its repeat unit.
3. Give the repeat unit formed from ethanedioic acid and pentane-1,5-diamine, and its Mr (C 12.0, H 1.0, O 16.0, N 14.0).
4. How many moles of water form, approximately, when 0.500 mol of the repeat unit in question 3 is made from these monomers? What mass is this (Mr H₂O 18.0)?
5. Give the hydrolysis products of the polymer in question 3 with hot dilute HCl(aq).
6. Give the products with hot NaOH(aq) for the same polymer.
7. A section reads –NH–CH₂–C₆H₄–CH₂–NH–CO–(CH₂)₂–CO–. Give the two monomers.
8. Why must benzene-1,3-diol be polymerised with a diacyl chloride rather than a dicarboxylic acid?
9. Is –CH₂–CH(CONH₂)–CH₂–CH(CONH₂)– a polyamide? Explain.
10. How many moles of NaOH hydrolyse 0.500 mol of the repeat unit in question 3?

Answers

1. HCl.
2. Addition (it has C=C and only one other reactive group); repeat unit –[CH₂–C(CH₃)(CH₂OH)]–.
3. –[CO–CO–NH–(CH₂)₅–NH]–, C₇H₁₂N₂O₂, Mr 156.0.
4. Two water molecules per repeat unit: 1.00 mol, 18.0 g.
5. HOOC–COOH and Cl⁻ ⁺H₃N–(CH₂)₅–NH₃⁺ Cl⁻.
6. NaOOC–COONa (sodium ethanedioate) and H₂N–(CH₂)₅–NH₂.
7. H₂N–CH₂–C₆H₄–CH₂–NH₂ and HOOC–(CH₂)₂–COOH, butanedioic acid (or its diacyl chloride).
8. Both –OH groups are on the ring, so they are phenolic. A carboxylic acid will not readily esterify a phenol; an acyl chloride will.
9. No. The backbone is all carbon and the amide groups are side groups. It is an addition polymer of CH₂=CH–CONH₂.
10. Two amide links per repeat unit: 1.00 mol.

Where marks are usually lost

• Writing H₂O as the by-product when the monomer is a diacyl chloride.
• Drawing a repeat unit with –H or –OH still on its ends, or with no bonds through the brackets.
• Putting two residues of the same monomer in the repeat unit of a two-monomer polymer.
• Splitting an ester link between the O and the carbon chain, which gives wrong monomers.
• Giving the free diamine, not its ammonium salt, as a product of acid hydrolysis.
• Giving the carboxylic acid, not its salt, as a product of alkaline hydrolysis.
• Classifying a polymer by side groups instead of by the backbone.
• Showing an amide link without the H on the nitrogen.
• Using one mole of NaOH per repeat unit when there are two links in it.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.3 Polyesters and polyamides, learning outcomes (a) to (c). Published by OCR.