Build polyesters and polyamides from their monomers, retain the correct links and count condensation molecules accurately.
Two functional sites allow a chain to keep growing
Condensation polymerisation joins monomers through reactions between functional groups, releasing a small molecule such as water or HCl per newly formed link in the examples here. A diol with a dicarboxylic acid forms polyester links, –C(=O)–O–. A diamine with a dicarboxylic acid forms polyamide links, –C(=O)–NH–.
Two different monomers are not compulsory. An amino acid contains both amine and acid groups, so one monomer type can form a polyamide; a hydroxycarboxylic acid can similarly form a polyester. Monofunctional molecules terminate a chain because they cannot link onwards at a second site.
Diacyl chlorides can replace dicarboxylic acids in related condensations, releasing HCl rather than water. Do not assume a carboxylic acid and diamine instantly give a polymer in cold water: salt formation and suitable polymerisation conditions must be considered. Addition polymerisation of an alkene opens C=C and releases no small molecule in the ideal propagation step.
Draw the smallest correct repeating section
For a diol–diacid polyester, remove the appropriate H and OH at each linking site and join the alcohol O to the acid carbonyl C. For a diamine–diacid polyamide, join amine N to carbonyl C and leave one H on a nitrogen that started as NH₂. Put the repeat inside brackets with bonds extending through both sides and n outside.
A repeat unit is not a complete finite polymer molecule. Do not cap every repeated fragment with OH or H; those end groups belong only at the ends of a finite chain. When recovering monomers, cut each ester/amide link and restore the OH or H required by the specified starting groups.
| Polymer | Monomers | Repeating backbone written from left to right |
|---|---|---|
| Terylene / PET | Ethane-1,2-diol + benzene-1,4-dicarboxylic acid | –O–CH₂–CH₂–O–C(=O)–C₆H₄–C(=O)– |
| Nylon 6,6 | Hexane-1,6-diamine + hexanedioic acid | –NH–(CH₂)₆–NH–C(=O)–(CH₂)₄–C(=O)– |
| Kevlar | Benzene-1,4-diamine + benzene-1,4-dicarboxylic acid | –NH–C₆H₄–NH–C(=O)–C₆H₄–C(=O)– |
Diagram placeholder
Terylene, nylon 6,6 and Kevlar monomer-to-repeat panels to add
Labels to include:
- Full diol/diamine and diacid groups for each row
- Para attachment points on each aromatic ring
- Ester –C(=O)–O– or amide –C(=O)–NH– link highlighted
- Nylon six-carbon diamine and six-carbon diacid, including both carbonyl C atoms
- Brackets, n and continuing bonds
- No duplicated carbonyl O or missing N–H
Draw each row as displayed or unambiguous skeletal structures. The nylon diacid has four CH₂ groups plus two carbonyl carbons; six CH₂ groups would give the wrong polymer.
Amino acids can polymerise without a second monomer type
For H₂NCH(CH₃)COOH, the repeat is –NH–CH(CH₃)–C(=O)–. The methyl group remains a side chain; it does not enter the main backbone. The link between repeats is the amide/peptide bond. The same connectivity rule applies when several different amino-acid residues occur in a protein.
In the extension example HOCH(CH₃)COOH, condensation gives –O–CH(CH₃)–C(=O)–, a polyester repeat. Some bifunctional molecules can also form rings; whether polymer or cyclic product predominates depends on conditions, so structure prediction alone is not a guarantee of the experimental product.
Count links when end groups matter
A single unbranched chain formed from m separate bifunctional monomer molecules contains m − 1 new links and releases m − 1 small molecules, provided no rings, branches or other reactions form. With n diol molecules and n diacid molecules all in one chain, that means 2n − 1 waters. With n amino-acid molecules in one chain, it means n − 1 waters.
An idealised repeat-unit equation may use 2n waters while suppressing end groups; that is a long-chain shorthand, not the exact finite-chain count. June 2023 Paper 2 Q04.5 explicitly depended on unreacted end groups. Decide whether the question shows an infinite repeat or a finite molecule before choosing the coefficient.
Constructed example: four diacid and four diol molecules form one linear chain. Eight starting molecules make seven links, releasing seven waters. If each monomer pair had a combined mass of 228 and each water 18, the finite-chain relative mass is 4 × 228 − 7 × 18 = 786, including the remaining end groups.
Repeating-unit mass is a useful approximation
PET has repeat formula C₁₀H₈O₄ and repeat-unit Mᵣ = 192.0 using C = 12.0, H = 1.0 and O = 16.0. A chain with 125 repeats has approximate Mᵣ = 125 × 192.0 = 24 000 when end-group masses are neglected. If exact end groups are supplied, include them.
A commercial sample contains a distribution of chain lengths, so a quoted average relative molecular mass need not imply every chain contains exactly the same integer number of repeats. Keep “polyamide/polyester”, the polymer class, separate from “condensation”, the polymerisation type.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Why can an amino acid form a condensation polymer without a second type of monomer?Show answer
It has both an amine and a carboxylic acid functional group, so each molecule can react at both ends with others of the same type.
Q2. Which monomers make nylon 6,6?Show answer
Hexane-1,6-diamine, H₂N(CH₂)₆NH₂, and hexanedioic acid, HOOC(CH₂)₄COOH. Each has six carbon atoms including the diacid carbonyl carbons.
Q3. How many waters form when six amino-acid molecules make one open, linear peptide chain?Show answer
Five: six molecules need five new peptide links. This assumes no cyclic product or branching.
Q4. Three diols and three diacids make one open chain. How many ester links and water molecules form?Show answer
Six monomer molecules require five new links, releasing five waters. Both chain ends remain unreacted.
Q5. Estimate the repeat count of PET with average Mᵣ = 38 400, ignoring end groups.Show answer
38 400/192 = 200 repeats on average. An average does not mean all chains have exactly 200 repeats.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- AQA 7405 organic chemistry specification — 3.3.12 coverage and required skills.
- Chemrevise: Polymers — Coverage checklist; explanations, data exercises and quick checks on this page are original Finesse material.
- AQA June 2023 Paper 2 mark scheme — Q04.5–04.6 pp19–21 and report Q04.5–04.6 p4: finite-chain end groups and correctly drawn polyamide hydrogen bonding.
- AQA June 2023 Paper 2 examiner report — Read alongside the question-specific marking guidance; not a universal wording checklist.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.