AQA A-Level Chemistry: Polymers (7405)

This guide teaches section 3.3.12 Polymers of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. It covers both sub-sections: 3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers. The whole section is A-level only, so it is not part of the AS course. The specification places all of section 3.3 on A-level Paper 2, and A-level Paper 3 can test any content.

For quick recall, use the polymers revision notes. To test yourself, try the polymers practice questions. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are. Addition polymers (3.3.4.3) are taught in the alkenes study guide, and hydrogen bonding and dipole–dipole forces in bonding.

What this topic covers

Spec section	What you must be able to do	Status
3.3.12.1	Describe condensation polymers formed from dicarboxylic acids and diols, dicarboxylic acids and diamines, and amino acids	A-level only
3.3.12.1	Recognise the repeating units and linkages in polyesters (eg Terylene) and polyamides (eg nylon 6,6 and Kevlar)	A-level only
3.3.12.1	State typical uses of these polymers	A-level only
3.3.12.1	Draw the repeating unit from monomer structure(s), or from a section of chain	A-level only
3.3.12.1	Draw the monomer(s) from a section of the polymer	A-level only
3.3.12.1	Explain the intermolecular forces between condensation polymer molecules	A-level only
3.3.12.2	State that polyalkenes are chemically inert and non-biodegradable	A-level only
3.3.12.2	State that polyesters and polyamides can be broken down by hydrolysis and are biodegradable	A-level only
3.3.12.2	Explain why polyesters and polyamides can be hydrolysed but polyalkenes cannot	A-level only
3.3.12.2	Weigh up the advantages and disadvantages of disposal methods, including recycling	A-level only

Section 3.3.12 lists no required practical. It suggests making nylon 6,6 (AT k, PS 1.2) and researching the problems of disposing of different polymers.

Condensation polymerisation (3.3.12.1)

In condensation polymerisation each monomer has two reactive functional groups. When two monomers join, a small molecule is lost, usually water. Because every monomer has a group at each end, the chain can keep growing in both directions.

Compare this with addition polymerisation, where a C=C bond opens and nothing else is produced. In a condensation polymer the main chain contains linkages (ester or amide groups), not just C–C bonds.

The specification names three combinations:

Monomers	Linkage formed	Type of polymer
dicarboxylic acid + diol	ester, –COO–	polyester
dicarboxylic acid + diamine	amide, –CONH–	polyamide
amino acids	amide, –CONH–	polyamide (polypeptide)

Polyesters: Terylene

Terylene is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. The –COOH of one monomer reacts with the –OH of the other. The OH from the acid and the H from the alcohol leave as water, and an ester link forms.

n HOOC–C₆H₄–COOH  +  n HO–CH₂CH₂–OH
        →  –[ CO–C₆H₄–CO–O–CH₂CH₂–O ]ₙ–  +  2n H₂O

Here C₆H₄ is a benzene ring joined at positions 1 and 4. The repeating unit contains one residue of each monomer and one complete ester link, plus the bonds that continue to the next units.

Typical uses of Terylene: fibres for clothing (often blended with cotton) and plastic drinks bottles.

Polyamides: nylon 6,6 and Kevlar

Nylon 6,6 is made from hexanedioic acid and hexane-1,6-diamine. The –COOH reacts with –NH₂, water is lost, and an amide link forms. The name “6,6” comes from the six carbon atoms in each monomer.

n HOOC–(CH₂)₄–COOH  +  n H₂N–(CH₂)₆–NH₂
        →  –[ CO–(CH₂)₄–CO–NH–(CH₂)₆–NH ]ₙ–  +  2n H₂O

Kevlar is made from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine. Its repeating unit is –[CO–C₆H₄–CO–NH–C₆H₄–NH]–.

Typical uses: nylon 6,6 for clothing, carpets and ropes; Kevlar for bulletproof vests and body armour, and for ropes and protective clothing where great strength for low weight is needed.

In the laboratory, nylon 6,6 is usually made with hexanedioyl dichloride in place of the acid. The acyl chloride is more reactive, and HCl is lost instead of water. This links to the reactions of acyl chlorides with primary amines in 3.3.9.2.

Polyamides from amino acids

An amino acid has an –NH₂ group and a –COOH group on the same molecule, so one monomer is enough. The –NH₂ of one molecule reacts with the –COOH of the next, losing water.

n H₂N–CH(CH₃)–COOH   →   –[ NH–CH(CH₃)–CO ]ₙ–  +  n H₂O
     2-aminopropanoic acid

When different amino acids join, the polymer is a polypeptide. Proteins and amino acid structure are covered in 3.3.13.

Worked example 1. Draw the repeating unit of the polyester made from butanedioic acid and propane-1,3-diol, and state what else forms.

• Write the acid: HOOC–CH₂CH₂–COOH. Write the diol: HO–CH₂CH₂CH₂–OH.
• Remove OH from one end of the acid and H from one end of the diol; join C to O. Do the same at the other end.
• Repeating unit: –[CO–CH₂CH₂–CO–O–CH₂CH₂CH₂–O]–, with the n outside the brackets.
• The other product is water.

Repeating units from a section of chain

1. Find the linkages (–CO–O– or –CO–NH–) along the chain.
2. Find the pattern that repeats. For a two-monomer polymer it contains two linkages’ worth of atoms: one diacid residue and one diol or diamine residue.
3. Cut through the middle of a linkage (between C and O, or C and N) at each end, so the unit begins with the CO or the O/NH. Draw extension bonds and put n outside.

Monomers from a section of chain

1. Break every ester or amide link between C and O, or between C and N.
2. Add OH to each C=O carbon you cut, giving –COOH.
3. Add H to each O or N you cut, giving –OH or –NH₂.

Worked example 2. A section of a polymer is –NH–(CH₂)₄–NH–CO–(CH₂)₈–CO–NH–(CH₂)₄–NH–CO–(CH₂)₈–CO–. Give the repeating unit and the monomers.

• The links are –CO–NH–, so this is a polyamide.
• Repeating unit: –[NH–(CH₂)₄–NH–CO–(CH₂)₈–CO]–.
• Breaking the C–N bonds and adding H to N and OH to C gives H₂N–(CH₂)₄–NH₂ (butane-1,4-diamine) and HOOC–(CH₂)₈–COOH (decanedioic acid).

Worked example 3. A sample of nylon 6,6 has an average Mr of 4.52 × 10⁴. How many repeating units are in an average chain? (Ar: H = 1.0, C = 12.0, N = 14.0, O = 16.0)

• Repeating unit formula: C₁₂H₂₂N₂O₂. Mr = 12(12.0) + 22(1.0) + 2(14.0) + 2(16.0) = 226.0.
• Check: Mr(hexanedioic acid) + Mr(hexane-1,6-diamine) − 2 × Mr(H₂O) = 146.0 + 116.0 − 36.0 = 226.0.
• Number of units = 4.52 × 10⁴ / 226.0 = 200.

Worked example 4. Calculate the percentage atom economy for making nylon 6,6 from hexanedioic acid and hexane-1,6-diamine, taking two water molecules lost per repeating unit.

• Atom economy = 226.0 / (146.0 + 116.0) × 100 = 86.3%.
• It is below 100% because water is a by-product. Addition polymerisation has an atom economy of 100%.

Intermolecular forces between condensation polymers

All polymer chains have van der Waals forces between them. Condensation polymers also have stronger forces, because their linkages are polar.

• Polyamides (nylon 6,6, Kevlar) form hydrogen bonds between chains. The H on the N–H group of one chain is attracted to a lone pair on the O of a C=O group on a neighbouring chain. N–H is very polar because N is much more electronegative than H.
• Polyesters (Terylene) have permanent dipole–dipole forces between the polar C=O groups of neighbouring chains. There is no H bonded to O or N in the ester link, so there is no hydrogen bonding between the chains.
• Kevlar has flat, rigid chains because of the benzene rings. The chains line up parallel and form many hydrogen bonds with each other. That is why Kevlar fibres are so strong.

Stronger forces between chains mean the polymer needs more energy to melt and is stronger as a fibre. This is why polyamides and polyesters make good fibres, while polyalkenes such as poly(ethene), which have only van der Waals forces, are softer.

Biodegradability and disposal (3.3.12.2)

Why polyesters and polyamides can be hydrolysed

Ester and amide links contain a polar C=O bond, so the carbon is δ+. Nucleophiles such as water or hydroxide ions can attack this carbon, and the link breaks. This is hydrolysis, the reverse of the condensation reaction. It happens faster with an acid or alkali. Because these polymers can be broken down like this, they are biodegradable.

• Acid hydrolysis of a polyester gives the dicarboxylic acid and the diol. Acid hydrolysis of a polyamide gives the dicarboxylic acid and the ammonium salt of the diamine, because the amine groups are protonated.
• Alkaline hydrolysis (eg with NaOH) gives the carboxylate salt, plus the diol from a polyester or the diamine from a polyamide.

Worked example 5. Terylene is boiled with aqueous sodium hydroxide. Name the organic products.

• The ester links are hydrolysed. The diol is released as ethane-1,2-diol.
• The acid groups end up as their sodium salt in alkali: sodium benzene-1,4-dicarboxylate.

Why polyalkenes cannot be hydrolysed

A polyalkene chain contains only C–C and C–H bonds. These bonds are strong and non-polar, so there is no δ+ carbon for a nucleophile to attack. Polyalkenes are therefore chemically inert and non-biodegradable. (PVC has polar C–Cl bonds, but its main chain is still made of C–C bonds.)

Methods of disposal

Method	Advantages	Disadvantages
Landfill	Simple; handles mixed waste without sorting	Uses up land; polyalkenes remain for a very long time; waste of a resource
Incineration	Reduces volume; heat released can generate electricity	Releases CO₂, a greenhouse gas; burning PVC releases toxic gases such as HCl, which must be removed
Mechanical recycling	Saves raw materials (crude oil) and energy; less landfill	Polymers must be collected, sorted by type and cleaned; quality may fall each time
Feedstock recycling	Polymer is broken down into monomers or other chemicals for new products	Needs energy and specialist plant

When you evaluate a method, give an advantage and a disadvantage, each tied to the chemistry.

Common errors

• Drawing a condensation repeating unit with only one monomer residue when there are two monomers.
• Forgetting to show water (or HCl) as the second product in the equation.
• Leaving –OH or –H on the ends of a repeating unit; the unit ends with extension bonds.
• Writing the amide link as –CO–NH₂– or the ester link as –CO–OH–.
• Saying polyesters form hydrogen bonds between chains; the ester link has no N–H or O–H.
• Saying polyalkenes cannot be hydrolysed because “their bonds are weak”; C–C and C–H are strong and non-polar.
• Giving the acid, not the salt, as a product of alkaline hydrolysis.
• Listing “recycling is good” without a reason or a disadvantage.

Where next

Use the polymers revision notes, then the polymers practice questions and the exam preparation guide. For addition polymers and PVC, go back to the alkenes revision notes.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.12 Polymers (3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers), A-level only.