You already met addition polymers from alkenes — one monomer, nothing lost. This topic adds condensation polymers, where two functional groups join and a small molecule (usually water) is lost at every link. Those links are ester or amide groups — which means, unlike an inert polyalkene, the chain can be hydrolysed straight back to its monomers.
Condensation polymers
A condensation polymer forms when each monomer has two reactive groups, so the chain can grow from both ends. Every new link expels a small molecule — usually water (or HCl if an acyl chloride is used in place of the acid). That lost molecule is the whole difference from the addition polymers you met with alkenes, where a C=C simply opens and nothing is lost. Before you draw anything, settle that question first — because if you pick the wrong polymerisation type, the repeat unit is wrong before you start.
| Addition | Condensation | |
|---|---|---|
| Monomer(s) | one, containing a C=C | two, each with two reactive groups (or one amino acid carrying both) |
| Lost at each link | nothing | one small molecule — H2O, or HCl from an acyl chloride |
| Repeat unit vs monomer | same atoms as the monomer | the monomer atoms minus the lost molecule |
| Example | poly(ethene) from ethene | nylon 6,6, Terylene, Kevlar |
Two families of condensation polymer are on the course. Both build a repeating link by joining an acid group to a partner group and squeezing out water:
| Polymer | Monomers | Linkage | Examples & uses |
|---|---|---|---|
| Polyester | dicarboxylic acid + diol | ester link –COO– | Terylene — clothing fibres, drink bottles |
| Polyamide | dicarboxylic acid + diamine (or amino acids) | amide link –CONH– | nylon 6,6 (fibres, ropes); Kevlar (body armour) |
AQA names three polymers you should be able to build from their monomers. Learn the monomer pair for each — the acid is always a dicarboxylic acid; the partner is a diol (polyester) or a diamine (polyamide):
The third route to a polyamide is a single amino acid: because it carries both an –NH2 and a –COOH, one type of molecule can link head-to-tail with itself, losing water at each amide link. That is exactly how proteins are built from amino acids — the same amide (peptide) link you meet again in amino acids, proteins and DNA.
Join one of each monomer, remove the small molecule at the new link, and enclose the repeat unit in brackets with a trailing bond at each end — these show the chain continues. You must be able to work in both directions: monomers → repeat unit, and a section of chain → the monomers that made it.
Build the repeating unit of Terylene from benzene-1,4-dicarboxylic acid, HOOC–C6H4–COOH, and ethane-1,2-diol, HO–CH2CH2–OH.
Step 1 — name the groups and the link: two –COOH ends face two –OH ends. Acid + alcohol makes an ester, so Terylene is a polyester.
Step 2 — form one link, lose one water: the acid’s –OH and the alcohol’s –H leave together as H2O, joining the carbonyl carbon to the diol oxygen (an ester link, –COO–).
Step 3 — bracket one of each residue, add trailing bonds:
–[ OC–C6H4–COO–CH2CH2–O ]n–
Each –OC–/–CO– is a carbonyl (C=O). The trailing bonds — one on the left carbonyl carbon, one on the right oxygen — show the chain continues; miss them and you have drawn a molecule, not a polymer.
Three molecules of hexanedioic acid react with three molecules of hexane-1,6-diamine to make one chain of nylon 6,6. How many repeat units, and how many water molecules, form?
Step 1 — count the links: joining N monomers into one chain makes N − 1 links. Here N = 6, so 5 links — one water leaves at each → 5 H2O.
Step 2 — count the repeat units: each repeat unit is one diacid + one diamine, so three of each give 3 repeat units.
General result for n of each monomer: n repeat units and (2n − 1) water molecules. Swap the acid for an acyl chloride and the same count of HCl leaves instead.
Exam questions
Ethanedioic acid, HOOCCOOH, reacts with propane-1,3-diol, HOCH2CH2CH2OH, to form a polyester. Draw the repeating unit of this polyester.
Show answer
Repeat unit (a trailing bond at each end):
–[ OC–COO–CH2CH2CH2–O ]n–
Correct ester link –COO– (C–O–C). 1 mark
Rest of the structure correct with trailing bonds at both ends. 1 mark
AQA also accepts the reversed condensed form –OOC–COO–CH2CH2CH2– provided the trailing bonds are shown; an –O– may sit at either end, but not both.
An equation shows nylon 4,6 forming from five molecules of butane-1,4-diamine and five molecules of hexanedioic acid, releasing y molecules of water and giving a chain of x repeating units. Deduce x and y.
Show answer
x = 5 1 mark
y = 9 1 mark
Five diamine + five diacid = ten monomers joined into one chain → nine links, each losing one water (y = 9); one diamine + one diacid per repeat unit → five repeat units (x = 5).
Source: AQA A-Level Chemistry past papers.
Forces between chains
A polymer’s strength and melting point come from the intermolecular forces between its chains — the same intermolecular forces you meet in bonding, and the linkage decides how strong those are:
- Polyamides have N–H and C=O groups, so chains hydrogen-bond to each other (N–H⋯O=C). These are the strongest forces here, giving high melting points and tough fibres — this is why Kevlar and nylon are so strong.
- Polyesters have polar C=O and C–O bonds, so chains attract by permanent dipole–dipole forces — weaker than hydrogen bonds.
- Polyalkenes (poly(ethene), poly(propene)) are non-polar, with only weak van der Waals forces between chains.
Nylon 6,6 melts higher than Terylene, which melts higher than poly(ethene). Explain this order using the forces between chains.
Step 1 — name the force in each: nylon 6,6 is a polyamide (N–H and C=O → hydrogen bonds); Terylene is a polyester (polar C=O and C–O → permanent dipole–dipole); poly(ethene) is non-polar (van der Waals only).
Step 2 — rank the forces: hydrogen bonds > permanent dipole–dipole > van der Waals.
Step 3 — link to melting: the stronger the force between chains, the more energy needed to pull them apart — so the melting points fall in the same order, nylon 6,6 > Terylene > poly(ethene).
Exam question
A figure shows a section of a nylon 4,6 molecule. Draw, next to it, another section of nylon 4,6 showing two hydrogen bonds between the two sections.
Show answer
Line the two chains up so the N–H groups of one face the C=O groups of the other, then draw the hydrogen bonds. AQA requires:
A lone pair on the O (or N). 1 mark
Two straight (linear) dashed lines, each from an O (or N) to an H. 1 mark
The hydrogen bond runs N–H⋯O=C. A curved or crooked dashed line, or one drawn to the wrong atom, is not credited — see the figure above for the geometry.
Source: AQA A-Level Chemistry past papers.
Breaking the links — hydrolysis
Breaking a condensation polymer back into its monomers is the key skill. Because a condensation polymer is held together by ester or amide links, water can break each one — the same hydrolysis you met with simple esters and amides, now applied at every link along the chain. There is one reliable way to do it on paper:
- Find each ester link (–COO–) or amide link (–CONH–).
- Break the bond between the carbonyl carbon and the O (ester) or N (amide).
- Add –OH to the carbonyl-carbon side — this remakes the –COOH of the carboxylic acid.
- Add –H to the other side — this remakes the –OH of the alcohol (from a polyester) or the –NH2 of the amine (from a polyamide).
Do that at every link and you are back to the monomers: a polyester gives its dicarboxylic acid + diol; a polyamide gives its dicarboxylic acid + diamine.
Hydrolysis is driven by hot aqueous acid or hot aqueous alkali (NaOH). The neutral “break the bond, add –OH and –H” picture always gives the fragments; the conditions then decide whether the amine or the acid ends up as a salt.
Nylon 6,6 is the polyamide of hexanedioic acid and hexane-1,6-diamine. Give the organic products when it is hydrolysed by (a) hot aqueous acid and (b) hot aqueous NaOH.
Step 1 — break each amide link, add –OH and –H: this recovers the diacid HOOC(CH2)4COOH and the diamine H2N(CH2)6NH2.
Step 2(a) — under acid the basic –NH2 groups are protonated, so the diamine leaves as a salt: products are the diacid HOOC(CH2)4COOH and the diammonium ion H3N+(CH2)6N+H3.
Step 2(b) — under alkali the acidic –COOH groups are deprotonated, so the acid leaves as a salt: products are the diamine H2N(CH2)6NH2 and the dicarboxylate −OOC(CH2)4COO−.
A section of polyester chain is –O–CH2CH2–O–CO–CH2CH2–COO–CH2CH2–O–. Deduce the two monomers.
Step 1 — find the ester links: each –COO– marks where a water molecule was lost. Cut there, at the bond between the carbonyl carbon and the single-bonded oxygen.
Step 2 — restore –OH and –H at every cut: put –OH on each carbonyl carbon (remaking –COOH) and –H on each oxygen (remaking –OH).
Step 3 — read off the monomers: the diacid HOOC–CH2CH2–COOH (butanedioic acid) and the diol HO–CH2CH2–OH (ethane-1,2-diol). Account for every bond broken and every group restored — a fragment missing its –OH or –H is not the real monomer.
A poly(alkene) chain is nothing but non-polar C–C and C–H bonds. There is no Cδ+ for water to attack, so the chain cannot be hydrolysed — which is exactly why poly(ethene) and poly(propene) are inert and non-biodegradable, while polyesters and polyamides break down over time.
Disposal & biodegradability
Because they can be hydrolysed, polyesters and polyamides are biodegradable; polyalkenes are not. No disposal route is free of drawbacks — each trades one cost against another:
| Route | Advantage | Disadvantage |
|---|---|---|
| Recycling | saves raw materials (crude oil) and energy | must be collected, sorted by type and cleaned first |
| Incineration for energy | releases stored energy to generate electricity; cuts landfill | produces CO2, and toxic gases (e.g. HCl from PVC) unless carefully controlled |
| Use as a feedstock | cracked back into monomers or other useful chemicals | needs sorting and energy-intensive processing |
| Landfill | simple and cheap | uses land; polyalkenes are inert and last almost indefinitely |
Exam questions
Explain why polyesters are biodegradable but polyalkenes are not biodegradable.
Show answer
Polyesters contain polar C=O / C–O bonds (the ester link), so they can be attacked by water and hydrolysed. 1 mark
Polyalkenes have only non-polar C–C (and C–H) bonds, which cannot be hydrolysed. 1 mark
Do not accept “polyesters are polar” on its own, and do not mention C=C — a polyalkene backbone is saturated.
Golf balls with an outer layer of poly(isoprene), a poly(alkene), can be recovered from lakes and reused even after years underwater. Explain why they do not biodegrade.
Show answer
The carbon–carbon bonds of the chain are non-polar (and strong), so they are not attacked by water / nucleophiles — the chain cannot be hydrolysed. 1 mark
Source: AQA A-Level Chemistry past papers.
Monomers ↔ repeat unit
Most polymer marks come down to reading structures in both directions and hydrolysing cleanly. Given monomers, join one of each and remove the small molecule to get the repeat unit; given a chain, cut it at the links and add back the –OH and –H to recover the monomers. The same “break the bond, add OH and H” move does both the hydrolysis and the monomer-recovery.
- Decide the polymerisation type first. Examiner reports single out confusing addition with condensation — and drawing the wrong repeat unit — as recurring errors, and advise settling the type before you draw.
- Trailing bonds, every time. A repeat unit without a bond at each end is a molecule, not a polymer — and don’t forget the small molecule lost (H2O, or HCl from an acyl chloride) in a condensation equation.
- Don’t overgeneralise a favourite example. Examiner reports note that students memorise one or two polymers and apply their properties to all — check the actual link before claiming hydrogen bonding or biodegradability.
- Finish the hydrolysis. Break the link and restore –OH to the carbonyl carbon (→ –COOH) and –H to the O or N — a bare fragment is not the monomer.
- Hydrogen bonds are fussy. Draw N–H⋯O=C as straight dashed lines, with a lone pair on the O.
Capstone quiz — five past-paper questions
Five real AQA multiple-choice questions spanning the whole topic, in the style that opens Paper 3. Pick one answer each — the reasoning appears once you commit.
Which is the repeating unit of a polyamide?
Suberoyl chloride, ClOC(CH2)6COCl, is used to make polymers. Which compound can form a polymer with it?
In which polymer does hydrogen bonding occur between the polymer chains?
Which statement concerning nylon-6,6 is correct?
Which polymer is not hydrolysed when heated with aqueous alkali?
Source: AQA A-Level Chemistry past papers.
- Addition vs condensation: addition from one C=C monomer, nothing lost, repeat unit = the same atoms as the monomer; condensation from two functional groups, a small molecule (H2O, or HCl from an acyl chloride) lost at each link. Decide which before you draw.
- Polyester = dicarboxylic acid + diol, ester link –COO– (Terylene, from benzene-1,4-dicarboxylic acid + ethane-1,2-diol). Polyamide = dicarboxylic acid + diamine, or a single amino acid, amide link –CONH– (nylon 6,6, from hexanedioic acid + hexane-1,6-diamine; Kevlar; proteins).
- Repeat units: join one of each monomer, lose the small molecule, bracket with trailing bonds; reverse it (cut the links, restore –OH and –H) to recover the monomers.
- Forces between chains: polyamides hydrogen-bond (N–H⋯O=C) → strongest, highest melting; polyesters have permanent dipole–dipole; polyalkenes only weak van der Waals.
- Hydrolysis: break each ester/amide link, add –OH to the carbonyl carbon and –H to the O or N — polyester → diacid + diol; polyamide → diacid + diamine. Hot acid leaves the amine as a salt; hot alkali leaves the acid as a salt. Polyalkenes cannot be hydrolysed (non-polar C–C).
- Disposal: polyesters/polyamides are biodegradable (hydrolysable); polyalkenes are inert. Options: recycle; incinerate (energy, but CO2/pollutants); use as a feedstock (crack back to useful chemicals); landfill (space, non-degrading).