OCR A Level Chemistry A: Polyesters and polyamides (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Here you practise 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. The questions test every learning outcome, 6.2.3(a) to (c), in Module 6, Organic chemistry and analysis, of the full A level. The last two questions, 10 and 11, are longer problems that link several outcomes with earlier topics.

Revise first with the Polyesters and polyamides study guide and the revision notes. Every other H432 topic is linked from our OCR Chemistry A hub, and the printable checklist lists every outcome. For a wider check, try a free diagnostic.

In calculations take C 12.0, H 1.0, O 16.0, N 14.0, Na 23.0. In every structure, –C₆H₄– is a benzene ring with two substituents.

Questions

1. (a) State what is meant by condensation polymerisation. [1] (b) Name the small molecule formed when a diamine reacts with (i) a dicarboxylic acid and (ii) a diacyl chloride. [2]

2. For each polymer section, state the type of polymerisation and give the structure of the monomer.

(a) –CH₂–CH(OCOCH₂CH₃)–CH₂–CH(OCOCH₂CH₃)– [2] (b) –O–CH₂–CH₂–CH₂–CO–O–CH₂–CH₂–CH₂–CO– [2]

3. Decane-1,10-diol, HO–(CH₂)₁₀–OH, reacts with benzene-1,3-dicarbonyl dichloride, ClOC–C₆H₄–COCl, to form a polyester.

(a) Give the repeat unit of the polyester. [2] (b) Name the other product. [1]

4. 3-Aminobutanoic acid, CH₃–CH(NH₂)–CH₂–COOH, forms a polymer on its own. Give the repeat unit and state the type of link formed. [2]

5. A section of a polymer is shown. The ring is 1,3-substituted.

–NH–C₆H₄–NH–CO–(CH₂)₃–CO–NH–C₆H₄–NH–CO–(CH₂)₃–CO–

(a) Name the functional group that links the monomers. [1] (b) Give the structures of two monomers that could form this polymer with loss of water. [2]

6. A polyester contains this repeat unit:

–[O–CH₂–C(CH₃)₂–CH₂–O–CO–(CH₂)₂–CO]–

(a) A sample is refluxed with dilute HCl(aq). Draw or write both organic products. [2] (b) A second sample is refluxed with NaOH(aq) instead. Draw or write both organic products. [2]

7. The polymer from question 4 is hydrolysed.

(a) Give the structure of the organic product with excess hot HCl(aq). [2] (b) Give the structure of the organic product with excess hot NaOH(aq). [2]

8. This question is about the polymer in question 5.

(a) Calculate the relative formula mass of its repeat unit. [2] (b) One chain has Mr 30 600. Calculate the number of repeat units in it. [1] (c) The polymer is made from the dicarboxylic acid and the diamine. Assuming two molecules of water form per repeat unit, calculate the mass of water produced when 1.00 kg of polymer forms. Give your answer to 3 significant figures. [3]

9. 4-Hydroxybut-2-enoic acid is HO–CH₂–CH=CH–COOH.

(a) Explain why this compound can form both an addition polymer and a condensation polymer. [2] (b) Give the repeat unit of the addition polymer. [1] (c) Give the repeat unit of the condensation polymer and name the other product. [2]

10. A polyester is made from benzene-1,3-dicarboxylic acid, HOOC–C₆H₄–COOH, and a diol, HO–(CH₂)ₙ–OH. Its repeat unit has Mr = 164 + 14n.

A 2.75 g sample was heated under reflux with 40.0 cm³ of 1.00 mol dm⁻³ NaOH(aq), an excess. After cooling, the unreacted NaOH was titrated with 0.500 mol dm⁻³ HCl(aq), using phenolphthalein. The titre was 30.0 cm³.

(a) Explain why each repeat unit reacts with two NaOH. [1] (b) Calculate the amount, in mol, of NaOH that reacted with the polyester. [3] (c) Calculate the Mr of the repeat unit. [2] (d) Find n and name the diol. [2]

11. A section of a polymer is shown. The ring is 1,4-substituted.

–NH–C₆H₄–O–CO–CH₂–CO–NH–C₆H₄–O–CO–CH₂–CO–

(a) Name the two types of link in the chain. [2] (b) Give the structures of the two monomers, using a diacyl chloride as one of them. [2] (c) Explain why a diacyl chloride is used rather than propanedioic acid. [2] (d) Give the structures of the organic products when the polymer is heated with excess NaOH(aq). Explain the form of the aromatic product. [3] (e) Give the structures of the organic products when the polymer is heated with excess HCl(aq). [2]

Answers

1. (a) Monomers join together with the loss of a small molecule, such as water or hydrogen chloride, at each link [1] (b) (i) water, H₂O [1]; (ii) hydrogen chloride, HCl [1] Examiner insight: A definition that says only “monomers join” describes addition polymerisation too; the loss of a small molecule is the point that earns the mark.

2. (a) Addition: the backbone is all carbon and the ester groups are side groups [1]; monomer CH₂=CH–OCOCH₂CH₃, ethenyl propanoate [1] (b) Condensation: ester links are in the backbone [1]; monomer HO–CH₂–CH₂–CH₂–COOH, 4-hydroxybutanoic acid [1] Examiner insight: In (a), seeing an ester group and answering “condensation” is the trap; justify the type from the backbone, not the side groups.

3. (a) Ester link correctly formed between the acyl carbon and the diol oxygen [1]; repeat unit –[CO–C₆H₄–CO–O–(CH₂)₁₀–O]– with open bonds at both ends and one residue of each monomer [1] (b) Hydrogen chloride [1] Examiner insight: Water in (b) is a common slip; the by-product comes from the acyl chloride group, so it must contain chlorine.

4. Repeat unit –[NH–CH(CH₃)–CH₂–CO]– [1]; amide (secondary amide) link [1] Examiner insight: Keep the methyl group as a side branch on the carbon next to N; moving it to the backbone or dropping it gives a different polymer.

5. (a) Amide link [1] (b) H₂N–C₆H₄–NH₂ (benzene-1,3-diamine) [1]; HOOC–(CH₂)₃–COOH (pentanedioic acid) [1] Examiner insight: The question says water is lost, so the acid monomer must have –COOH groups; a diacyl chloride is a valid monomer for this polymer but does not fit the wording here.

6. (a) HO–CH₂–C(CH₃)₂–CH₂–OH [1]; HOOC–(CH₂)₂–COOH [1] (b) HO–CH₂–C(CH₃)₂–CH₂–OH [1]; NaOOC–(CH₂)₂–COONa (or ⁻OOC–(CH₂)₂–COO⁻) [1] Examiner insight: The diol is identical in both conditions; only the acid part changes, so writing the free acid in (b) loses that mark.

7. (a) –COOH group retained as the acid [1]; amine present as its ammonium salt: Cl⁻ ⁺H₃N–CH(CH₃)–CH₂–COOH [1] (b) Amine group as free –NH₂ [1]; carboxylate salt: H₂N–CH(CH₃)–CH₂–COO⁻ Na⁺ [1] Examiner insight: Each part has two groups to get right; check both ends of the product against the conditions before you move on.

8. (a) Repeat unit is C₁₁H₁₂N₂O₂ [1]; Mr = 11(12.0) + 12(1.0) + 2(14.0) + 2(16.0) = 204.0 [1] (b) 30 600 ÷ 204.0 = 150 [1] (c) Amount of repeat units = 1000 ÷ 204.0 = 4.90 mol [1]; water = 2 × 4.90 = 9.80 mol [1]; mass = 9.80 × 18.0 = 176 g [1] Examiner insight: Show the formula of the repeat unit before the Mr; if the formula is wrong, the working is what lets an examiner see which later steps are still sound.

9. (a) It has a C=C bond, which can open in addition polymerisation [1]; it also has both an –OH and a –COOH group, which can react with each other between molecules to form ester links [1] (b) –[CH(CH₂OH)–CH(COOH)]– [1] (c) –[O–CH₂–CH=CH–CO]– [1]; water [1] Examiner insight: In (b) only the two former C=C carbons belong in the backbone; the –CH₂OH and –COOH groups become side groups.

10. (a) Each repeat unit contains two ester links, and each is hydrolysed by one NaOH [1] (b) NaOH added = 0.0400 × 1.00 = 0.0400 mol [1]; unreacted NaOH = HCl = 0.0300 × 0.500 = 0.0150 mol [1]; reacted = 0.0400 − 0.0150 = 0.0250 mol [1] (c) Repeat units = 0.0250 ÷ 2 = 0.0125 mol [1]; Mr = 2.75 ÷ 0.0125 = 220 [1] (d) 164 + 14n = 220, so n = 4 [1]; butane-1,4-diol [1] Examiner insight: Forgetting to halve in (c) gives Mr 110 and a non-integer n; an impossible n is a signal to recheck the mole ratio.

11. (a) Amide [1]; ester [1] (b) H₂N–C₆H₄–OH (4-aminophenol) [1]; ClOC–CH₂–COCl (propanedioyl dichloride) [1] (c) The –OH is a phenol group, and phenols are not readily esterified by carboxylic acids [1]; an acyl chloride reacts readily with both the phenol –OH and the –NH₂ [1] (d) NaOOC–CH₂–COONa [1]; H₂N–C₆H₄–O⁻ Na⁺ [1]; phenol is weakly acidic and reacts with excess NaOH, while the amine stays as –NH₂ [1] (e) HOOC–CH₂–COOH [1]; HO–C₆H₄–NH₃⁺ Cl⁻ [1] Examiner insight: In (d), the phenol formed by breaking the ester link is itself deprotonated by excess alkali (6.1.1(h)); giving neutral 4-aminophenol misses that reasoning.

Where marks are usually lost

• Calling a polymer with ester or amide side groups a condensation polymer, as in question 2.
• Giving water as the by-product when a diacyl chloride is used.
• Repeat units with leftover –H or –OH at the ends, or two residues of one monomer.
• Acid hydrolysis of a polyamide written with a free –NH₂ rather than –NH₃⁺.
• Alkaline hydrolysis written with –COOH rather than –COO⁻.
• Forgetting that a phenol product is deprotonated by excess NaOH.
• Using a 1 : 1 ratio of NaOH to repeat units when each unit has two links.
• Dropping or moving side branches when drawing repeat units from single monomers.

Next steps

• Recap the tables and method steps: Polyesters and polyamides revision notes
• Full explanations and worked examples: Polyesters and polyamides study guide
• Find gaps anywhere in H432: all free 10-minute diagnostics
• Ester and acyl chloride reactions: Carboxylic acids and esters practice
• Addition polymers: Alkenes practice
• Every H432 topic in one place: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

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.