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Notes/Chemistry/Paper 1/Polymerisation
CAIEAS Level9701§20

Polymerisation

Alkenes joining into long chains: draw the repeat unit from a monomer, find the monomers from a section of chain, do n × Mr sums, and explain why poly(alkene)s are hard to get rid of.

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The Nitrogen Compounds note finished the AS functional groups. This note goes back to the alkenes of the AS Hydrocarbons note, where you met addition polymerisation briefly: many alkene molecules join into one long chain, and nothing else is made.

Here you build that into a full set of exam skills. First the reaction itself, using poly(ethene) and PVC. Then you draw the repeat unit of a polymer from its monomer, and work backwards from a section of chain to the monomer or monomers that made it. After that come unusual monomers (dienes and rings), the chiral carbons in poly(propene), and simple sums with nn and MrM_r. The note ends with why poly(alkene)s are hard to get rid of.

Before you start you should be able to

  • Alkenes and addition polymerisation (AS Hydrocarbons note — “Addition polymerisation”)

  • σ and π bonds, sp² and sp³ carbon (AS Chemical Bonding note — “σ and π bonds, and hybridisation”)

  • Naming alkenes, and chiral centres (AS Introduction to Organic Chemistry note — “Nomenclature (up to C6)” and “Stereoisomerism: cis/trans and optical”)

  • Relative molecular mass and empirical formula (AS Atoms, Molecules and Stoichiometry note)

  • Making an alkene by elimination of HBr (AS Halogen Compounds note — “Elimination: the door back to the alkene”)

  • Esters and carboxylate salts, which appear as side groups here (AS Carboxylic Acids note)

By the end of this page you can

  • Describe addition polymerisation as exemplified by poly(ethene) and poly(chloroethene), PVC: the π bond breaks, new σ bonds form, the carbons change from sp² to sp³, and no atoms are lost

  • Deduce the repeat unit of an addition polymer from a given monomer: the two C=C carbons form the backbone, a bond sticks out through each bracket, other groups stay as side groups, n goes outside

  • Identify the monomer(s) present in a given section of an addition polymer, including both monomers of a copolymer, and name them

  • Handle unusual monomers: dienes that leave a C=C inside the repeat unit, and cycloalkenes whose rings sit in the chain; explain why poly(propene) shows optical isomerism

  • Use Mr(polymer molecule) = n × Mr(repeat unit), and know the empirical formula of an addition polymer equals that of its monomer

  • Recognise the difficulty of disposing of poly(alkene)s: they are non-biodegradable, and burning them gives harmful products (CO, and HCl from PVC)

01

What addition polymerisation is

Syllabus requirement · §20

“

describe addition polymerisation as exemplified by poly(ethene) and poly(chloroethene), PVC

”

Alkenes that add to each other

In the Hydrocarbons note, a small molecule such as HBr\text{HBr} or Br2\text{Br}_2 added across an alkene's C=C\text{C}{=}\text{C}. In addition polymerisation the alkene adds to other molecules of the same alkene. Each molecule's C=C\text{C}{=}\text{C} opens and bonds to the molecules on either side, and this repeats thousands of times to make one very long chain. The small molecules are the monomers; the long chain is the polymer. The monomers only add together and nothing is given off, so the polymer is the only product.

The syllabus names two examples:

  • ethene, CH2=CH2\text{CH}_2{=}\text{CH}_2 → poly(ethene) (plastic bags);
  • chloroethene, CH2=CHCl\text{CH}_2{=}\text{CHCl} → poly(chloroethene), usually called PVC (window frames, pipes).

If a question asks what type of polymerisation turns an alkene into its polymer, the answer is addition. (Polymers made by joining monomers and giving off a small molecule such as water are condensation polymers; you meet them in the A Level Polymerisation note.)

What happens to the bonds

Each carbon of a C=C\text{C}{=}\text{C} is sp²: it has three σ bonds, plus the π bond it shares with the other carbon. When a monomer joins the chain, three things happen to it:

  1. its π bond breaks;
  2. two new σ bonds form, one to the unit on each side;
  3. its two carbons change from sp² to sp³ (trigonal planar → tetrahedral).

So the polymer chain is saturated: a backbone of C–C single bonds, with the other atoms or groups (H, Cl, CH3\text{CH}_3 …) still attached to the same carbons as in the monomer. With no π bond left, a poly(alkene) is unreactive, like an alkane: it does not decolourise bromine water, for example.

Nothing is lost

Addition polymerisation only joins monomers together; no atoms are removed. So a polymer molecule made from nn monomers contains exactly nn monomers' worth of atoms. Three results follow, and all three are examined:

  • the molecular formula of the polymer molecule is n×n \times the monomer's formula;
  • Mr(polymer molecule)=n×Mr(monomer)\text{M}_r(\text{polymer molecule}) = n \times \text{M}_r(\text{monomer});
  • the empirical formula of the polymer is the same as the monomer's.

The “Working with n and Mr” section uses the second result for calculations.

addition polymerisation — every monomer unit undergoes the same three changesMONOMER · propeneHHCCHCH₃gold line = the π bond: exposed, it does the joiningboth C=C carbons sp², trigonal planar, flat1 · the π bond breaks2 · two new σ bonds form — one to each neighbouring unit3 · both carbons sp² → sp³ : trigonal planar → tetrahedralPRODUCT · poly(propene), units joined head-to-tailone repeat unitCH₂CHCH₃CH₂CHCH₃CH₂CHCH₃written asCH₂CHCH₃nno atoms lost: Mr(repeat unit) = Mr(monomer), and the empirical formula is unchangedn counts the repeat units in one molecule: Mr(polymer) = n × Mr(repeat unit)

Propene's C=C (π bond in gold) above, the three bond changes on the arrow, and below the product: a poly(propene) chain section with one repeat unit highlighted, and the same unit written in brackets with a bond through each bracket and n outside.

Ethene and propene, step by step

(i) Show how nn molecules of ethene become poly(ethene), and explain the bracket notation.

(ii) Do the same for propene, CH2=CH(CH3)\text{CH}_2{=}\text{CH(CH}_3\text{)}, and state where the CH3\text{CH}_3 group ends up.

Show full working
  1. 1

    (i) Start with the monomers. Write n CH2=CH2n\,\text{CH}_2{=}\text{CH}_2. Here nn is a large number (usually thousands): it is the number of monomers that join.

    n is part of the notation: it goes in front of the monomer and again after the bracket of the polymer.

  2. 2

    Open every double bond. Each π bond breaks and each carbon forms a new σ bond to the next unit, so the units join into a chain: –CH2–CH2–CH2–CH2–\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–CH}_2\text{–} … Every carbon now has four single bonds.

    This is the three-change picture above: π bond breaks, new σ bonds form, sp² becomes sp³.

  3. 3

    Bracket one repeat unit. The repeat unit is the smallest piece that builds the whole chain when repeated. Here it is –CH2–CH2–\text{–CH}_2\text{–CH}_2\text{–}. Put it in brackets, draw a bond sticking out through each bracket, and write nn after the closing bracket:

    n CH2=CH2  →  –[ CH2–CH2–]n–n\,\text{CH}_2{=}\text{CH}_2 \;\rightarrow\; \text{–[\,CH}_2\text{–CH}_2\text{–]}_n\text{–}

    The bonds through the brackets (often called dangling bonds) join this unit to the units on either side.

    The two commonest errors are forgetting the bonds through the brackets and leaving a C=C inside the brackets. The π bond has gone, so the bond inside is single.

  4. 4

    (ii) Propene. Only the two carbons of the C=C\text{C}{=}\text{C} go into the chain: the CH2\text{CH}_2 and the CH. The CH3\text{CH}_3 was not part of the double bond, so it stays on its carbon as a side group:

    n CH2=CH(CH3)  →  –[ CH2–CH(CH3)–]n–n\,\text{CH}_2{=}\text{CH(CH}_3\text{)} \;\rightarrow\; \text{–[\,CH}_2\text{–CH(CH}_3\text{)–]}_n\text{–}

    This is poly(propene).

    Propene has three carbons, but only two were in the C=C. The third carbon is the CH₃ side group, never part of the chain.

Answer

(i) n CH₂=CH₂ → –[CH₂–CH₂–]ₙ–, with a bond through each bracket (ii) –[CH₂–CH(CH₃)–]ₙ–, poly(propene): the CH₃ is a side group on every second chain carbon

To go backwards from a repeat unit to its monomer, put the C=C back between the two bracketed carbons and keep every side group where it is. The “Finding the monomers” section does this in full.

Four statements about PVC

9701/13 M/J 2024 Q371 mark

Which statement about PVC is correct?

A   Combustion products of PVC are very alkaline and harmful to breathe in.
B   The empirical formula of PVC is the same as the empirical formula of the monomer.
C   Molecules of PVC are unsaturated.
D   The repeat unit of PVC is (CH2CCl2)(\text{CH}_2\text{CCl}_2).

Show full working
  1. 1

    B is true. No atoms are lost, so one PVC repeat unit, –CH2–CHCl–\text{–CH}_2\text{–CHCl–}, contains exactly the atoms of one chloroethene molecule, CH2=CHCl\text{CH}_2{=}\text{CHCl}. Same atoms in the same ratio, so the same empirical formula, C2H3Cl\text{C}_2\text{H}_3\text{Cl}.

    “Empirical formula of an addition polymer = empirical formula of the monomer” is always true. It comes straight from “nothing is lost”.

  2. 2

    A is false. Burning PVC can give hydrogen chloride, HCl\text{HCl}, which dissolves in water to give hydrochloric acid. The products are acidic, not alkaline.

    Remember “PVC burns to give acidic HCl”; it is one of the disposal problems in the last section.

  3. 3

    C is false. Every π bond was used up when the units joined, so the PVC chain has only single bonds: it is saturated.

    Addition polymers of simple alkenes are always saturated.

  4. 4

    D is false. A repeat unit –CH2–CCl2–\text{–CH}_2\text{–CCl}_2\text{–} would come from CH2=CCl2\text{CH}_2{=}\text{CCl}_2, with both Cl atoms on the same carbon. PVC comes from CH2=CHCl\text{CH}_2{=}\text{CHCl}, so its repeat unit has one Cl: –CH2–CHCl–\text{–CH}_2\text{–CHCl–}. Answer B.

    Where the side groups sit is part of the repeat unit. Check D by turning it back into a monomer: it gives the wrong alkene.

Answer

B

Your turn

  1. 19701/21 O/N 2025 Q4(b)(i)1 mark

    Under suitable conditions, propene polymerises to form poly(propene).

    Poly(propene) exhibits stereoisomerism.

    Identify the type of polymerisation that forms poly(propene) from propene.

    Show solution
    1. 1

      Propene is an alkene. Its molecules join through their C=C\text{C}{=}\text{C} bonds and nothing else is formed, so this is addition polymerisation.

      Alkene → polymer is always addition. Don't answer “condensation”: no small molecule such as water is given off.

    Answer

    addition

  2. 22 marks

    A molecule of poly(propene) contains exactly 1500 repeat units.

    (i) How many carbon atoms does this molecule contain?

    (ii) How many C=C double bonds does its chain contain?

    Stuck? Show hint

    Count per repeat unit first, then multiply by n. For (ii), ask which bonds were used up when the units joined.

    Show solution
    1. 1

      (i) Carbons in one repeat unit. The unit –CH2–CH(CH3)–\text{–CH}_2\text{–CH(CH}_3\text{)–} has 3 carbons: two in the chain and one in the CH3\text{CH}_3 side group.

      Side groups count: they are still in the molecule.

    2. 2

      (i) Multiply by the number of units.

      1500×3=4500 carbon atoms1500 \times 3 = 4500 \text{ carbon atoms}

      Count one unit, then multiply by n. The molecular formula is (C₃H₆)₁₅₀₀, i.e. C₄₅₀₀H₉₀₀₀.

    3. 3

      (ii) None. Every π bond was used up when the units joined, so the chain has only single bonds.

      “The polymer is unsaturated” is a common wrong option. A poly(alkene) made from a monoalkene is saturated.

    Answer

    (i) 4500 (ii) none — the chain is saturated

  3. 32 marks

    But-1-ene, C4H8\text{C}_4\text{H}_8, and butane, C4H10\text{C}_4\text{H}_{10}, are both colourless gases.

    Only one of them can undergo addition polymerisation. Identify which, and explain why the other cannot.

    Stuck? Show hint

    What does a molecule need in order to add to another molecule?

    Show solution
    1. 1

      But-1-ene can. It has a C=C\text{C}{=}\text{C} double bond. Its π bond can break so that each molecule bonds to the next. Its polymer is –[ CH2–CH(C2H5)–]n–\text{–[\,CH}_2\text{–CH(C}_2\text{H}_5\text{)–]}_n\text{–}, with an ethyl side group.

      Addition polymerisation needs a C=C: no C=C, no addition polymer.

    2. 2

      Butane cannot. It is an alkane: it has only σ bonds and no π bond, so there is nothing to add across.

      Look at the functional group, not the size. CₙH₂ₙ can be an alkene; CₙH₂ₙ₊₂ is always an alkane.

    Answer

    but-1-ene, because it has a C=C (π bond) to open; butane is saturated, with no double bond

The rest of this note

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Can you do all of these?

  • Only the two C=C carbons go into the backbone; every other group stays on its own carbon as a side group

  • Repeat unit: single bond inside the brackets, a bond through each bracket, n outside

  • Monomers from a chain: split into pairs of carbons, put the C=C back in each pair, name each alkene (longest chain through the C=C, lowest number)

  • Two different pairs = a copolymer made from two monomers

  • –CH(CH₃)–CH(CH₂CH₃)– comes from pent-2-ene, not pent-1-ene

  • A diene can leave a C=C inside the repeat unit; a cycloalkene keeps its ring, joined at the two neighbouring carbons of the old C=C

  • Poly(propene)'s CH(CH₃) carbons have four different groups → chiral centres → optical isomerism

  • Mr(polymer molecule) = n × Mr(repeat unit); empirical formula unchanged

  • Disposal: non-biodegradable (no functional group for enzymes) and harmful combustion products (CO; CO₂; HCl from PVC, which is acidic)

Paper 1 · MCQ
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Paper 2 · Theory
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