Polyamides and nylon: monomers, links and water loss
Build the specified dicarboxylic-acid/diamine pattern, recognise amide links and reverse them to recover the starting functional groups.
Section 11.8 · Polymers
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 COOH and NH₂ groups and interpret a backbone box as a segment.
- Count actual links in a specified finite linear chain.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which connection identifies an amide link here?
What you will learn
- Deduce polyamide connectivity from a dicarboxylic acid and diamine and reverse it to the monomers.
- Describe and draw the required nylon polyamide pattern with amide links and the small-molecule loss.
Two functional groups allow the chain to continue
A dicarboxylic acid has two COOH groups, HOOC–R₁–COOH, and a diamine has two NH₂ groups, H₂N–R₂–NH₂. Their condensation forms amide links, –C(=O)–NH–, and water. The nylon polyamide pattern is [–NH–R₂–NH–C(=O)–R₁–C(=O)–]ₙ. Both ends can link onward because the monomers are bifunctional, rather than stopping after one acid/alcohol-style small molecule.
R₁ and R₂ in this diagram are the specified hydrocarbon backbone segments of the two monomers, not single element symbols or necessarily identical groups. Each amide N retains one H; each carbonyl has a C=O double bond. At a repeat boundary, a carbonyl C links to the next unit's N. To reverse an amide link to monomers, restore OH at the acid-derived carbonyl end and H at the nitrogen end, recovering COOH and NH₂.
Count water from actual links, especially at chain ends
Each amide link formed in the stated acid-plus-amine condensation releases one H₂O. A supplied finite linear chain made by joining four alternating monomers has three links and therefore three water molecules eliminated. Do not automatically use the number of monomers as the water count. Repeat notation hides end groups, so a per-repeat formula is not automatically the formula of that complete finite chain.
For a supplied nylon example, hexanedioic acid HOOC(CH₂)₄COOH totals C₆H₁₀O₄, and the diamine H₂N(CH₂)₆NH₂ totals C₆H₁₆N₂. Their repeating contribution after two condensation losses is C₁₂H₂₂N₂O₂. This is an illustrative backbone choice within the nylon polyamide pattern; the general required drawing uses the two backbone boxes and amide linkages, rather than asserting every polyamide has those same carbon counts.
Pause and explain
How do you recover the acid and amine groups from the stated amide link?
Worked example
A specified finite linear chain joins two C₆H₁₀O₄ diacid molecules and two C₆H₁₆N₂ diamine molecules with three amide links. Find its formula after water loss.
Show the worked solution
- The four starting monomers total C₂₄H₅₂N₄O₈.
- Three new amide links eliminate 3H₂O, removing H₆O₃.
- The complete supplied chain therefore has C₂₄H₄₆N₄O₅, including its remaining end groups.
Answer C₂₄H₄₆N₄O₅ after three water losses in the stated four-monomer finite-chain model.
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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.
Draw the generic nylon repeat and identify both monomer-derived backbone boxes.
Give me a hint
Retain the N–H and C=O bonds while joining the groups.
Compare my reasoning
- Use NH–R₂–NH for the diamine-derived part and C(=O)–R₁–C(=O) for the acid-derived part.
- Join them through C(=O)–NH amide links, including the link across a repeat boundary.
- Add brackets and n with continuations, and explain that R₁/R₂ are backbone segments.
[–NH–R₂–NH–C(=O)–R₁–C(=O)–]ₙ, with amide links, N–H bonds, C=O bonds and two labelled backbone segments.
Look for these in your work
- The nylon/polyamide connections are correct.
- Boxes, repeat boundaries and required bonds are explained.
Deduce the two monomers from that generic repeat, showing the functional groups at both ends.
Give me a hint
Cut amide links and restore the groups rather than changing the backbone boxes.
Compare my reasoning
- The acid-derived segment becomes HOOC–R₁–COOH.
- The amine-derived segment becomes H₂N–R₂–NH₂.
- Each monomer is bifunctional; re-forming an acid–amine link releases H₂O.
HOOC–R₁–COOH and H₂N–R₂–NH₂; a dicarboxylic acid and diamine, with water lost per formed link.
Look for these in your work
- Both monomers have the correct two end groups.
- Their backbones and the condensation loss are retained.
In the specified four-monomer model, use C=12, H=1, N=14 and O=16 to check the mass of C₂₄H₄₆N₄O₅ against two diacids plus two diamines minus three waters.
Give me a hint
Use the actual three links, not four monomers as the water multiplier.
Compare my reasoning
- The diacid molar mass is 146 and diamine 116 g/mol, so the supplied four-monomer total is 2×146+2×116=524.
- Three waters contribute 3×18=54 g/mol, leaving 470 g/mol for the model chain.
- C₂₄H₄₆N₄O₅ gives 24×12+46+4×14+5×16=470 g/mol, confirming conservation including end groups.
524−54=470 g/mol, agreeing with C₂₄H₄₆N₄O₅; three links produce three waters in this supplied model.
Look for these in your work
- Masses use the stated atom values.
- The finite-chain link count and end groups are respected.
Common mistakes
- Replacing the amide N with oxygen and calling it the same link.
- Removing the acid carbonyl double bond from the drawing.
- Treating backbone boxes as literal single atoms or necessarily identical segments.
- Confusing number of monomers, repeat units and actual water-releasing links.
What can you explain now?
Which functional-group pair forms the specified nylon polyamide?
Compare with the explanation
A dicarboxylic acid and a diamine.
Their two reactive ends form continuing amide links, with water eliminated at each stated acid–amine link.
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Make it stick
Draw nylon's generic repeat, recover both bifunctional monomers and count water from actual links in a stated finite 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.