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Notes/Chemistry/Paper 4/Polymerisation
CAIEA2 Level9701§35.1–35.3

Polymerisation

Polyesters, polyamides and the rest — every mark routes through one question: which groups meet, and what leaves?

240 min read 5 sub-topics
43
question parts
2021–2025 · 23 papers
3 marks
per paper
≈ 3% of the paper
1.9/3
avg difficulty
moderate
#14
most examined
of 15 topics by marks

Polymerisation is a Paper 4 fixture rather than a heavyweight: across 2021–2025 it filled 43 examined parts worth 73 marks over 23 papers, and those marks arrive as one structured question — most often 2–4 marks, never more than 6 — in each paper that tests the topic. All of them are decided by a single idea worth stating before any chemistry — a condensation polymer is built when functional groups MEET and a small molecule LEAVES, so every drawing mark asks which groups met, every bookkeeping mark asks what left, and every degradability mark asks whether water can simply be added back the same way. Addition polymerisation (the AS Polymerisation note) consumed a C=C and lost nothing; condensation consumes two complementary groups — COOH meeting OH or NH₂, COCl meeting OH or NH₂ — and pays H₂O or HCl for the join. The route through: §01 the six recipes that build polyesters and polyamides, §02 drawing the repeat unit with the linkage displayed, §03 reading a printed chain backwards to its monomers — including the unfamiliar polyanhydride — §04 predicting which polymerisation a monomer or a chain section implies, C=C-plus-COOH traps included, and §05 why poly(alkene)s persist while polyesters and polyamides biodegrade.

Before you start you should be able to

  • AS Polymerisation: draw addition repeat units with dangling bonds piercing the brackets, read a drawn chain back to its alkene, and recall why poly(alkene)s are hard to dispose of

  • A2 Carboxylic Acids: acyl chlorides forming esters and amides cold with HCl paid; ester hydrolysis in hot acid or alkali; acyl chloride + carboxylate salt giving an anhydride

  • A2 Nitrogen Compounds: the peptide bond drawn fully displayed, and amide hydrolysis products

  • AS esters: carboxylic acid + alcohol ⇌ ester + water, and the naming of hydroxy- and amino-carboxylic acids

By the end of this page you can

  • Describe the formation of polyesters: diol + dicarboxylic acid, diol + dioyl chloride, and hydroxycarboxylic acids alone — writing the equation with the repeat unit bracketed and the co-product named

  • Describe the formation of polyamides: diamine + dicarboxylic acid, diamine + dioyl chloride, aminocarboxylic acids alone, and amino acids joining through peptide links

  • Deduce the repeat unit from any given monomer or pair: groups joined, small molecule expelled, linkage DRAWN DISPLAYED, brackets round the smallest tiling span, continuation bonds through both brackets, and no expelled atoms left inside

  • Identify the monomers present in a given section of a condensation polymer — ester, amide and anhydride cuts, H₂O added back across each, acyl-chloride alternatives credited where the mark scheme allows OR

  • Predict the type of polymerisation for a given monomer or pair: C=C used → addition with nothing lost; complementary groups → condensation with a small molecule out — including monomers carrying BOTH kinds of handle, decided by which bonds the product shows

  • Deduce the type of polymerisation from a printed polymer section: an ester/amide/anhydride link in the chain means condensation; a saturated C–C backbone with side groups means addition

  • Recognise the three degradability clauses: poly(alkene)s are chemically inert and difficult to biodegrade (non-polar C–C backbone, cannot hydrolyse); some polymers are degraded by the action of light; polyesters and polyamides are biodegradable by acidic and alkaline hydrolysis

01

The second way to build a chain

Syllabus requirement · §35.1

“

describe the formation of polyesters: (a) the reaction between a diol and a dicarboxylic acid or dioyl chloride (b) the reaction of a hydroxycarboxylic acid; describe the formation of polyamides: (a) the reaction between a diamine and a dicarboxylic acid or dioyl chloride (b) the reaction of an aminocarboxylic acid (c) the reaction between amino acids.

”

The second family

The AS Polymerisation note built poly(alkene)s the only way alkenes know how: open a C=C\text{C}{=}\text{C}, chain up, lose nothing. But nylon, the PET of a drinks bottle, PLA surgical thread, a protein, Kevlar — none of these is a poly(alkene). They belong to the second family: condensation polymers, built when molecules carrying the right complementary functional groups meet and join by expelling a small molecule — usually water, sometimes hydrogen chloride. The syllabus names the two product families you must be able to build: polyesters and polyamides. Everything in this note runs off one question, so ask it of every printed structure you meet this topic: WHICH groups are meeting, and WHAT is leaving?

Two joins you already own

Condensation polymerisation invents no new chemistry — it reuses two reactions from earlier notes, one per linkage. For polyesters, the join is ester formation: a carboxylic acid and an alcohol at reflux sit in the familiar equilibrium (AS Carboxylic Acids), paying water; the sharper version, an acyl chloride plus an alcohol at room temperature, runs to completion and pays HCl (the A2 Carboxylic Acids note's addition–elimination). For polyamides, the join is amide formation: an acyl chloride plus ammonia or a primary amine in the cold, paying HCl (the Nitrogen Compounds note), or a carboxylic acid plus an amine, paying water. Either way, one event makes ONE ester or amide link. The polymer twist is purely structural:

Give each monomer TWO groups and the joins chain up

Hand one molecule two alcohol ends and another two acid ends. The first join spends one end of each — and look at what remains: the diol still owns a free −OH-\text{OH}, the diacid still owns a free −COOH-\text{COOH}. Those free ends can only do the one thing they know how to do — react again — so instead of capping a small ester, they extend a growing chain. Because every monomer offers exactly two handles, every fully-reacted molecule sits INSIDE the chain, bonded to two neighbours. Each new link costs one small molecule: condensation = join + expel. That phrase is the entire topic; everything else is bookkeeping.

monomer(s)

class

linkage

small molecule out

diol HO–R–OH + dicarboxylic acid HOOC–R′–COOH

polyester

ester –C(=O)–O–

H₂O

diol HO–R–OH + dioyl chloride ClCO–R′–COCl

polyester

ester –C(=O)–O–

HCl

diamine H₂N–R–NH₂ + dicarboxylic acid HOOC–R′–COOH

polyamide

amide –C(=O)–N(H)–

H₂O

diamine H₂N–R–NH₂ + dioyl chloride ClCO–R′–COCl

polyamide

amide –C(=O)–N(H)–

HCl

hydroxycarboxylic acid HO–R–COOH, ALONE

polyester

ester –C(=O)–O–

H₂O

aminocarboxylic acid H₂N–R–COOH, or amino acids, ALONE

polyamide

peptide (= amide) –C(=O)–N(H)–

H₂O

The six recipes. Rows 5 and 6 need no partner: one molecule carrying BOTH groups supplies its own complement. Peptide link = the amide link inside a protein.

The arithmetic of n + n

Writing the full equation needs a counting convention. Feed nn molecules of each monomer into the pot: n+nn + n monomers give one long chain, and every join expels one small molecule — so the equation ends + 2n H2O\,2n\,\text{H}_2\text{O} (or HCl). Strict honesty: a finite chain built from 2n2n monomers holds 2n−12n - 1 links, so 2n−12n - 1 small molecules truly leave. The published mark schemes accept the tidy 2n2n, and hedging looks worse than the rounding — so write 2n2n, and keep 2n−12n - 1 filed away for the rare bookkeeping exercise that deliberately probes the difference.

§35.1 · the six recipes — every condensation polymer is one of thesePOLYESTERS — every route builds the ester link –C(=O)–O–HO–R–OH + HOOC–R′–COOH→ ester link + ≈2n H₂OHO–R–OH + ClCO–R′–COCl→ ester link + ≈2n HClHO–R–COOH (both groups on ONE molecule)→ ester link + ≈n H₂O — no partner neededH₂O routes: the acid's –OH and the alcohol's –H leave together (alone-rows: ≈n H₂O).HCl routes: the acyl chloride's Cl simply replaces that –OH.POLYAMIDES — every route builds the amide link –C(=O)–N(H)–H₂N–R–NH₂ + HOOC–R′–COOH→ amide link + ≈2n H₂OH₂N–R–NH₂ + ClCO–R′–COCl→ amide link + ≈2n HClH₂N–R–COOH or H₂N–CHR–CO₂H (amino acids)→ amide (peptide) link + ≈n H₂O — no partner neededSame two shapes as the ester column — swap the alcohol's Ofor an amine's NH. The protein peptide bond IS this amide link.read every condensation question with ONE question: WHICH groups meet — and WHAT leaves?WHICH groups meet decides the linkage (acid + alcohol → ester · acid + amine → amide) · WHAT leaves pays the co-productbookkeeping: n + n monomers → one chain + ≈2n small molecules — CAIE accepts 2n, though a finite chain truly makes 2n−1 links

The six recipes as a map: two families, three routes each. Acid partners pay H₂O; acyl-chloride partners pay HCl; a molecule carrying BOTH groups needs no partner.

Building a polyester step by step (invented)

Hexane-1,6-diol, HO(CH2)6OH\text{HO(CH}_2\text{)}_6\text{OH}, reacts with hexane-1,6-dioic acid, HOOC(CH2)4COOH\text{HOOC(CH}_2\text{)}_4\text{COOH}.

(i) Name the linkage formed between the monomers and the small molecule expelled.

(ii) Write the equation for the polymerisation, showing one repeat unit and the amount of co-product formed from nn molecules of each monomer.

Show full working
  1. 1

    (i) Inventory the functional groups. The diol carries two −OH-\text{OH} groups; the diacid carries two −COOH-\text{COOH} groups. Alcohol meeting acid can only make an ester — so this is a polyester, and the co-product is water.

    Classify before you calculate: spotting OH + COOH settles the polymer class instantly. The co-product follows the ACID partner — water for a carboxylic acid, HCl if it had been the acyl chloride.

  2. 2

    Run one join. One −OH-\text{OH} of the diol attacks one −COOH-\text{COOH} of the diacid; an ester link forms and one H₂O leaves — the acid contributing the −OH-\text{OH} half of its carboxyl group, the alcohol contributing an −H-\text{H}.

    Know WHERE the water comes from: acid's –OH plus alcohol's –H. This detail is exactly what §02's bookkeeping check audits — whatever left may not appear in the chain.

  3. 3

    (ii) Why a chain forms. After that first join, BOTH fragments still hold a spare group at their far ends — an −OH-\text{OH} on the diol fragment, a −COOH-\text{COOH} on the diacid fragment. Each free end reacts with fresh monomer the same way, so every unit ends up bonded left AND right: the joins chain up.

    This is the 'two handles' idea doing the work: two groups per monomer is precisely what converts a one-off esterification into a polymerisation.

  4. 4

    Extract the repeat unit. The smallest section that tiles the chain is one acid residue joined to one alcohol residue:

    n HO(CH2)6OH+n HOOC(CH2)4COOH→–[ C(=O)(CH2)4C(=O)–O–(CH2)6–O–]n–+2n H2On\,\text{HO(CH}_2\text{)}_6\text{OH} + n\,\text{HOOC(CH}_2\text{)}_4\text{COOH} \rightarrow \text{–[\,C(=O)(CH}_2\text{)}_4\text{C(=O)–O–(CH}_2\text{)}_6\text{–O–]}_n\text{–} + 2n\,\text{H}_2\text{O}

    with continuation bonds passing through the brackets. Bookkeeping check: the atoms that left in the water — the acid's −OH-\text{OH} and an alcohol −H-\text{H} — appear NOWHERE in the bracketed unit.

    The carbonyl carbons come from the acid and the bridging O from the alcohol. If your drawn unit still shows an –OH or a stray H on the oxygen, the expelled water never actually left.

  5. 5

    Count the co-product. n+n=2nn + n = 2n monomer molecules, one small molecule expelled per link: write 2n H2O2n\,\text{H}_2\text{O}. (A finite chain truly makes 2n−12n - 1 links — the accepted convention is 2n2n; quote it without agonising.)

    '2n H₂O' scores; 'H₂O' with no coefficient or 'n H₂O' reads as unsure bookkeeping. The coefficient travels with whichever small molecule left.

Answer

(i) ester links; H₂O expelled (ii) n HO(CH₂)₆OH + n HOOC(CH₂)₄COOH → –[C(=O)(CH₂)₄C(=O)–O–(CH₂)₆–O–]ₙ– + 2n H₂O

Whatever the monomers, the drill is identical: inventory the groups → name the join and the co-product → extract the tiling section → count 2n. §02 turns this drill into the full five-step drawing method the examiner marks against.

Your turn

Classification and bookkeeping before any drawing — these check you can read the six recipes off a monomer list.

  1. 13 marks

    For each monomer set below, state the class of polymer formed, the linkage formed, and the small molecule expelled.

    (a) benzene-1,4-diamine + hexane-1,6-dioyl chloride

    (b) propane-1,3-diol + octane-1,8-dioic acid

    (c) 4-hydroxybenzoic acid as the ONLY monomer

    Stuck? Show hint

    Which two groups meet in each case — and in (c), does one molecule carry both of them?

    Show solution
    1. 1

      (a) Amine meets acyl chloride: −NH2-\text{NH}_2 groups onto −COCl-\text{COCl} groups. Class: polyamide; linkage: amide; co-product: HCl — the chloride leaves from the acyl chloride and an −H-\text{H} from the nitrogen.

      Acid-chloride partners always pay HCl regardless of whether the other group is OH or NH₂ — the halogen belongs to the acyl chloride, so the co-product decision rides on THAT partner.

    2. 2

      (b) Alcohol meets carboxylic acid: polyester, ester links, H₂O — the acid's −OH-\text{OH} and the alcohol's −H-\text{H} leave together.

      Both monomers carry two groups each, so both belong inside the chain as alternating residues — the standard diol + diacid pattern.

    3. 3

      (c) One molecule wearing BOTH a phenol −OH-\text{OH} and a −COOH-\text{COOH}: it provides its own partner. Polyester, ester links, H₂O — no second monomer required.

      The 'no partner needed' rows of the six-recipe table are examined as identification questions: count the functional groups on the single molecule before hunting for a mate.

    Answer

    (a) polyamide, amide links, HCl (b) polyester, ester links, H₂O (c) polyester (alone), ester links, H₂O

  2. 22 marks

    Three sealed flasks each contain one pure compound and nothing else.

    A   propane-1,3-diol, HOCH2CH2CH2OH\text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH}

    B   4-aminobutanoic acid, H2N(CH2)3COOH\text{H}_2\text{N(CH}_2\text{)}_3\text{COOH}

    C   hexane-1,6-dioic acid, HOOC(CH2)4COOH\text{HOOC(CH}_2\text{)}_4\text{COOH}

    Which flask can produce a polymer on its own? Explain what happens in it, and why the other two flasks do not polymerise at all.

    Stuck? Show hint

    Count the functional groups per molecule — how many handles does a self-polymerising monomer need?

    Show solution
    1. 1

      Flask B polymerises. ✓ Its single molecule carries BOTH reactive groups: an −NH2-\text{NH}_2 at one end and a −COOH-\text{COOH} at the other. Molecules couple head-to-tail — the amine of one attacking the acid of the next — giving a polyamide with amide (peptide-type) links and H₂O expelled.

      This is recipe 6: aminocarboxylic acids are nature's own monomers — proteins are exactly such chains of amino acids, met properly in §05's degradation story.

    2. 2

      A and C cannot. ✗ Propane-1,3-diol offers only −OH-\text{OH} groups; hexane-1,6-dioic acid offers only −COOH-\text{COOH} groups. Neither flask contains the COMPLEMENTARY partner its groups need, so no ester or amide can form — the contents sit unchanged however long you heat them.

      Polymerisation needs complementary groups to MEET. Two same-ended monomers in one pot is a stand-off, however reactive each looks on paper.

    Answer

    flask B — its molecules carry both NH₂ and COOH and self-condense to a polyamide (+H₂O); A (only OH) and C (only COOH) lack a complementary partner

  3. 32 marks

    An industrial reactor is charged with exactly 250 molecules of ethane-1,2-diol and 250 molecules of benzene-1,4-dicarboxylic acid, and the PET chain they form is allowed to grow as long as physically possible.

    (i) How many ester links does that longest chain contain, and how many H₂O molecules leave?

    (ii) State the coefficient conventionally written in the polymerisation equation for nn molecules of each monomer.

    Stuck? Show hint

    Think of beads: a chain of k beads has how many strings connecting them? Each connection squeezes out one water.

    Show solution
    1. 1

      (i) Count links, not molecules. The pot holds 250+250=500250 + 250 = 500 monomer molecules, so the longest chain has 500 residues. A finite chain of kk residues holds k−1k - 1 links between them — the last residue's free end has no further partner. So: 500−1=500 - 1 = 499 ester links, and 499 H₂O leave.

      This is the honest bookkeeping the intro warned about: every link costs one water, and a finite chain always has one link fewer than it has monomers.

    2. 2

      (ii) The accepted convention. With n=250n = 250: write the equation ending in 2n H2O2n\,\text{H}_2\text{O}, i.e. 500. CAIE mark schemes credit 2n2n even though the true physical count is 2n−12n - 1 — never substitute the honest number back into an exam equation.

      Two numbers exist because two jobs exist: the equation's job (showing the stoichiometric pattern) uses 2n; a probing calculation's job (counting actual molecules) uses 2n − 1. Know which context wants which.

    Answer

    (i) 500 − 1 = 499 ester links, hence 499 H₂O (ii) 2n H₂O — i.e. 500 by the accepted convention

Practise reading monomer pairs off as polyester or polyamide, with the right co-productReal past-paper questions · Condensation polymerisation: polyesters and polyamides

The rest of this note

Checking your access…

Can you do all of these?

  • I can state all six condensation recipes — which monomer pairs (or lone bifunctional molecules) build polyesters and polyamides, and whether H₂O or HCl leaves

  • I can write the n-equation for a condensation polymerisation ending in 2n small molecules, knowing a finite chain truly makes 2n−1 links

  • I can deduce a repeat unit from any monomer pair: groups meet, small molecule expelled, linkage drawn DISPLAYED (C=O out), brackets round ONE tiling section, continuation bonds through both brackets, n outside

  • I can spot bystander features — a C=C, a ring, a side chain — that survive untouched INSIDE a repeat unit

  • I can follow the command words: 'one repeat unit' vs 'draw a section containing k residues' vs 'label the repeat unit on your diagram'

  • I can slice a printed chain section back into its monomers — cut the linkage's single bond, add H₂O across it — including unfamiliar links such as the anhydride link

  • I can apply the OR convention: a polymer made via a dioyl chloride credits the diacid or the dioyl chloride as the monomer

  • I can classify a monomer, a pair or a printed section as addition vs condensation instantly — resolving C=C-plus-COOH traps by asking which bonds the product actually shows

  • I can explain why poly(alkene)s resist biodegradation (inert, non-polar C–C backbone, cannot hydrolyse) and why polyesters and polyamides biodegrade (their links hydrolyse in acid or alkali)

  • I can state how light degrades some polymers — UV breaking bonds near carbonyl groups incorporated in the backbone

Now do the questions
43 real Paper 4 parts from 2021–2025, sorted by difficulty, with mark schemes