OCR A Level Chemistry A: Polyesters and polyamides (H432)

This study guide teaches 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 section sits in Module 6, Organic chemistry and analysis, and each of its lettered learning outcomes, 6.2.3(a) to (c), is taught below through worked examples. H432 is the full A level, and its specification places Module 6 in two examined components: Synthesis and analytical techniques (02) and Unified chemistry (03). No PAG is listed against this section.

Every H432 topic is linked from the OCR chemistry hub; tick outcomes off on the printable checklist. The free diagnostics show where else in H432 you need work.

Outcomes in 6.2.3

Specification reference	Skill	Additional guidance (from the specification)
6.2.3(a)(i)	Describe condensation polymerisation to form polyesters	Formed from carboxylic acids or dicarboxylic acids (or their acyl chlorides) and alcohols or diols
6.2.3(a)(ii)	Describe condensation polymerisation to form polyamides	Formed from carboxylic acids or dicarboxylic acids (or their acyl chlorides) and amines or diamines. You are not expected to recall the structures of synthetic polyesters and polyamides or their monomers
6.2.3(b)(i)	Describe acid and base hydrolysis of the ester groups in polyesters	none
6.2.3(b)(ii)	Describe acid and base hydrolysis of the amide groups in polyamides	none
6.2.3(c)(i)	Predict the repeat unit from given monomer(s), for addition and condensation polymers	See also 4.1.3(j)
6.2.3(c)(ii)	Predict the monomer(s) needed for a given section of polymer	See also 4.1.3(j)
6.2.3(c)(iii)	Predict the type of polymerisation	See also 4.1.3(j)

Since recall of named polymers is not required, every example here gives you the monomers. Expect unfamiliar ones in the exam.

Esters, acyl chlorides and amides come from Carboxylic acids and esters; addition polymers from Alkenes. The amide link in a polyamide, –CONH–, is a secondary amide (6.2.2(b)).

What makes a polymer a condensation polymer

In condensation polymerisation, monomers join and a small molecule is lost at each new link. The small molecule is water when a carboxylic acid group reacts, and hydrogen chloride when an acyl chloride group reacts.

Each monomer needs two reactive groups, in one of two patterns.

• Two different monomers, each with two identical groups: a dicarboxylic acid (or diacyl chloride) with a diol, or with a diamine.
• One monomer carrying one group of each kind: a hydroxycarboxylic acid, or an aminocarboxylic acid.

In addition polymerisation (4.1.3(j)), monomers with a C=C bond join as the double bond opens, and the polymer is the only product.

Link formed	Groups that react	Small molecule lost
Ester, –COO–	–COOH + –OH	H₂O
Ester, –COO–	–COCl + –OH	HCl
Amide, –CONH–	–COOH + –NH₂	H₂O
Amide, –CONH–	–COCl + –NH₂	HCl

Acyl chlorides are far more reactive than carboxylic acids, and they esterify phenols, which carboxylic acids do not readily do (6.1.3(f)). So a diol with both –OH groups on a benzene ring needs a diacyl chloride partner.

6.2.3(a)(i): Forming polyesters

Worked example 1: diacid plus diol

Propanedioic acid, HOOC–CH₂–COOH, reacts with hexane-1,6-diol, HO–(CH₂)₆–OH.

1. –OH from the acid and H from the diol leave as H₂O.
2. The C=O carbon bonds to the diol’s O, making an ester link, –CO–O–.
3. Both ends still carry a reactive group, so the chain grows at each end.
4. Repeat unit: –[CO–CH₂–CO–O–(CH₂)₆–O]–

Equation for n of each monomer:

n HOOC–CH₂–COOH + n HO–(CH₂)₆–OH → –[CO–CH₂–CO–O–(CH₂)₆–O]ₙ– + (2n − 1) H₂O

A chain of n repeat units has 2n − 1 links. For a long chain this is close to 2n, so the equation is often written with 2n H₂O.

When drawing it, display the ester link (C=O and C–O), with a bond through each bracket and n after the closing one.

Worked example 2: diacyl chloride plus an aromatic diol

Benzene-1,4-diol, HO–C₆H₄–OH, has both –OH groups on the ring, so it behaves as a phenol. It is paired with heptanedioyl dichloride, ClOC–(CH₂)₅–COCl.

1. Each new ester link releases HCl, not water.
2. Repeat unit: –[CO–(CH₂)₅–CO–O–C₆H₄–O]–
3. Equation: n ClOC–(CH₂)₅–COCl + n HO–C₆H₄–OH → –[CO–(CH₂)₅–CO–O–C₆H₄–O]ₙ– + (2n − 1) HCl

Worked example 3: a single hydroxycarboxylic acid

5-Hydroxypentanoic acid is HO–(CH₂)₄–COOH. Each molecule’s –OH end condenses with the –COOH end of a neighbour.

• Repeat unit: –[O–(CH₂)₄–CO]–, made from one monomer.
• Equation: n HO–(CH₂)₄–COOH → –[O–(CH₂)₄–CO]ₙ– + (n − 1) H₂O
• Check: monomer C₅H₁₀O₃ (Mr 118.0) minus H₂O (18.0) gives repeat unit C₅H₈O₂ (Mr 100.0).

6.2.3(a)(ii): Forming polyamides

Worked example 4: diacyl chloride plus diamine

Butanedioyl dichloride, ClOC–(CH₂)₂–COCl, reacts with heptane-1,7-diamine, H₂N–(CH₂)₇–NH₂.

1. The N of –NH₂ bonds to the C of –COCl; H and Cl leave as HCl.
2. The new link is the amide link, –CO–NH–.
3. Repeat unit: –[CO–(CH₂)₂–CO–NH–(CH₂)₇–NH]–
4. Equation: n ClOC–(CH₂)₂–COCl + n H₂N–(CH₂)₇–NH₂ → –[CO–(CH₂)₂–CO–NH–(CH₂)₇–NH]ₙ– + (2n − 1) HCl

Butanedioic acid would give the same polymer, losing water instead.

The repeat unit is C₁₁H₂₀N₂O₂, Mr 212.0. A chain of Mr 42 400 therefore holds 42 400 ÷ 212.0 = 200 repeat units.

Worked example 5: a single aminocarboxylic acid

7-Aminoheptanoic acid, H₂N–(CH₂)₆–COOH, has an amine group at one end and a carboxylic acid group at the other.

• Repeat unit: –[NH–(CH₂)₆–CO]–
• One H₂O is lost per link: monomer C₇H₁₅NO₂ (Mr 145.0) becomes repeat unit C₇H₁₃NO (Mr 127.0).

Display the amide link with C=O, C–N and N–H shown.

6.2.3(b): Hydrolysis of polyesters and polyamides

Hydrolysis reverses condensation: water breaks each ester or amide link, returning the monomers or their ions. Both acid and base hydrolysis are done by heating under reflux. An acid protonates basic groups and an alkali deprotonates acidic ones, so the products depend on the conditions.

6.2.3(b)(i): Polyesters

Conditions	Products from the ester link
Hot aqueous acid, e.g. dilute HCl	Carboxylic acid, –COOH, and alcohol, –OH
Hot aqueous alkali, e.g. NaOH(aq)	Carboxylate salt, –COO⁻ Na⁺, and alcohol, –OH

For the polyester in worked example 1:

• Acid: –[CO–CH₂–CO–O–(CH₂)₆–O]ₙ– + 2n H₂O → n HOOC–CH₂–COOH + n HO–(CH₂)₆–OH
• Alkali: –[CO–CH₂–CO–O–(CH₂)₆–O]ₙ– + 2n NaOH → n NaOOC–CH₂–COONa + n HO–(CH₂)₆–OH

6.2.3(b)(ii): Polyamides

Conditions	Products from the amide link
Hot aqueous acid, e.g. dilute HCl	Carboxylic acid, –COOH, and ammonium salt, –NH₃⁺ Cl⁻
Hot aqueous alkali, e.g. NaOH(aq)	Carboxylate salt, –COO⁻ Na⁺, and amine, –NH₂

In acid, the freed amine group accepts a proton, as in 6.2.1(a). For the polyamide in worked example 4, acid hydrolysis gives butanedioic acid and the salt Cl⁻ ⁺H₃N–(CH₂)₇–NH₃⁺ Cl⁻. Alkaline hydrolysis gives sodium butanedioate, NaOOC–(CH₂)₂–COONa, and heptane-1,7-diamine.

Worked example 6: how much alkali hydrolyses a polyester?

A 9.30 g sample of the polyester from worked example 1 is fully hydrolysed by heating with 2.00 mol dm⁻³ NaOH(aq). Take H 1.0, Na 23.0, C 12.0 and O 16.0.

1. Repeat unit C₉H₁₄O₄: Mr = 9(12.0) + 14(1.0) + 4(16.0) = 186.0
2. Amount of repeat units = 9.30 ÷ 186.0 = 0.0500 mol
3. Each repeat unit holds two ester links, so NaOH needed = 2 × 0.0500 = 0.100 mol
4. Volume = 0.100 ÷ 2.00 = 0.0500 dm³ = 50.0 cm³
5. Products: 0.0500 mol sodium propanedioate (Mr 148.0) = 7.40 g; 0.0500 mol hexane-1,6-diol (Mr 118.0) = 5.90 g

Check: 9.30 g + 4.00 g of NaOH = 13.30 g, and 7.40 g + 5.90 g = 13.30 g.

6.2.3(c): Predicting repeat units, monomers and polymer type

6.2.3(c)(iii): Which type of polymerisation?

Look at the monomers or the chain.

• Monomer has a C=C and no pair of groups that can condense: addition. The chain backbone is all carbon.
• Monomers carry –COOH or –COCl together with –OH or –NH₂, two reactive groups per monomer: condensation.
• Chain contains –COO– or –CONH– in the backbone: condensation.

A side group can mislead. Take the chain –CH₂–CH(COOCH₂CH₂CH₃)–CH₂–CH(COOCH₂CH₂CH₃)–. It has ester groups, but they hang off an all-carbon backbone, so it is an addition polymer of propyl propenoate, CH₂=CH–COOCH₂CH₂CH₃.

6.2.3(c)(i): Repeat unit from monomers

• Addition: draw the C=C with its four groups, turn it into C–C and add a bond out of each end. 3-Methylbut-1-ene, CH₂=CH–CH(CH₃)₂, gives –[CH₂–CH(CH(CH₃)₂)]–.
• Condensation, two monomers: join one of each, then remove –OH or –Cl from the free acid end and H from the free –OH or –NH₂ end.
• Condensation, one monomer: remove H from one end and –OH from the other.

6.2.3(c)(ii): Monomers from a section of polymer

Worked example 7. A polymer section reads:

–O–CH(CH₃)–CH₂–CO–O–CH(CH₃)–CH₂–CO–

1. Ester links are in the backbone, so this is a condensation polyester.
2. The pattern repeats every four carbons: –O–CH(CH₃)–CH₂–CO–. Only one kind of unit appears, so there is one monomer.
3. Add –OH to the C=O end and H to the O end.
4. Monomer: HO–CH(CH₃)–CH₂–COOH, 3-hydroxybutanoic acid.

Worked example 8. A section reads:

–CO–C₆H₄–CO–O–(CH₂)₈–O–CO–C₆H₄–CO–O–(CH₂)₈–O–

1. Cut through each ester link between C=O and O.
2. The pieces alternate: –CO–C₆H₄–CO– and –O–(CH₂)₈–O–.
3. Add –OH to each acyl end and H to each oxygen end.
4. Monomers: HOOC–C₆H₄–COOH (a benzenedicarboxylic acid; the diacyl chloride ClOC–C₆H₄–COCl also works) and HO–(CH₂)₈–OH, octane-1,8-diol.

For a polyamide, cut between C=O and N–H in the same way and give back –OH (or –Cl) and H.

Common errors

• Writing H₂O as the by-product when an acyl chloride is used. It is HCl.
• Leaving an –OH or –H at the ends of a repeat unit.
• Including two of one monomer in a repeat unit for a diacid–diol polymer. The repeat unit has one residue of each.
• Cutting an ester link between the O and the carbon chain. Split between the C=O carbon and the next O or N.
• Giving a free amine as the product of acid hydrolysis of a polyamide. In acid it is the ammonium salt.
• Giving a carboxylic acid as the product of alkaline hydrolysis. In alkali it is the carboxylate salt.
• Calling a chain with ester side groups a polyester.
• Using one mole of NaOH per repeat unit when the unit contains two ester or amide links.

Next steps

Condense this into recall sheets with the Polyesters and polyamides revision notes, then test yourself on the Polyesters and polyamides practice questions. The amine reactions behind polyamide hydrolysis are in Amines, and the techniques for planning routes between functional groups are in Organic synthesis.

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.