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6th Year HL · Chemistry
Organic chemistry · Polymers and society

Addition polymerisation

A carrier bag from the shop, a PVC pipe under the sink and a non-stick frying pan all have something in common at molecular level: they are made of very long chains that formed without losing a single atom along the way. If a monomer has a C=C double bond, thousands of its molecules can link up one after another. Understanding that chain, and doing calculations with it, is the aim of this page.

From monomer to polymer

When 100 kg of ethene gas is fed into a high-pressure reactor with a trace of initiator, exactly 100 kg of solid poly(ethene) is obtained. Not a gram of gas or by-product escapes. How do thousands of separate molecules join up without losing any mass?

The double bond of ethene opens to become the repeating unit of poly(ethene). monomer CH₂=CH₂ n × repeating unit (in the chain) –CH₂–CH₂– the double bond opens: a single bond is left on each side

Other addition polymers on your course follow the same pattern. Poly(phenylethene) (polystyrene) comes from phenylethene, $\mathrm{CH_2=CH-C_6H_5}$, and has the repeating unit $[-\mathrm{CH_2-CH(C_6H_5)}-]_n$. Poly(propene) (polypropylene) comes from propene, $\mathrm{CH_2=CH-CH_3}$, and has the repeating unit $[-\mathrm{CH_2-CH(CH_3)}-]_n$: the $-\mathrm{CH_3}$ is a side group, so $M_{\text{unit}} = 42.08\ \text{g/mol}$, the same as propene. A general alkene $\mathrm{CH_2=CHX}$ gives $[-\mathrm{CH_2-CHX}-]_n$, where X is H, Cl, $\mathrm{CH_3}$, $\mathrm{C_6H_5}$ and so on. The same pattern works with other substituents: tetrafluoroethene, $\mathrm{CF_2=CF_2}$, gives poly(tetrafluoroethene) (PTFE), $[-\mathrm{CF_2-CF_2}-]_n$. The C=C of the monomer is used up, so the chain contains only single bonds. The strong C–F bonds make PTFE very unreactive. Chemical stability is not the same as mechanical hardness: PTFE is very resistant to chemicals, yet it is a soft, waxy material.

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Ethene ($\mathrm{CH_2=CH_2}$) is the fundamental raw material for the industrial production of polyethylene by radical or catalytic addition polymerisation. a) Write the…

Ethene ($\mathrm{CH_2=CH_2}$) is the fundamental raw material for the industrial production of polyethylene by radical or catalytic addition polymerisation.

  1. a) Write the general chemical equation of the polymerisation process, using bracket notation for the repeating unit and stating the meaning of the subscript $n$.
  2. b) Justify why the molecular mass of the repeating unit coincides exactly with that of the starting monomer, explaining whether this type of polymerisation produces any by-product such as water or hydrogen chloride.
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Chloroethene (commonly called vinyl chloride) polymerises to give poly(vinyl chloride) (PVC), a material of great technical importance in pipes and profiles. a) Write,…

Chloroethene (commonly called vinyl chloride) polymerises to give poly(vinyl chloride) (PVC), a material of great technical importance in pipes and profiles.

  1. a) Write, using semi-structural formulae, the chloroethene molecule and the repeating unit of poly(vinyl chloride).
  2. b) Explain whether the presence of the chlorine substituent on the carbon chain affects the molecular mass of the unit relative to the monomer, and explain why it should not be generalised that all vinyl polymers have identical flexibility or mechanical rigidity at room temperature.
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Polytetrafluoroethene (PTFE) is an addition polymer known for its extraordinary chemical inertness and non-stick properties. a) Write the formula of the monomer…

Polytetrafluoroethene (PTFE) is an addition polymer known for its extraordinary chemical inertness and non-stick properties.

  1. a) Write the formula of the monomer tetrafluoroethene and of the constitutional repeating unit of polytetrafluoroethene.
  2. b) State, with justification, whether the bond present in the monomer that allows polymerisation is retained in the resulting polymer, and justify why the fluorine atoms confer great chemical stability against acids and strong oxidising agents without this necessarily implying high mechanical hardness.
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Degree of polymerisation and mean molar mass

Why? Each chain is just $n$ identical units stuck together, so its mass is $n$ times the mass of one unit.

Dividing the mean molar mass by the molar mass of the repeating unit gives the degree of polymerisation. mean molar mass of the polymer ÷ M of unit degree of polymerisation n = units per chain
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In the structural characterisation of linear synthetic polymers, it is essential to correlate the sequence of atoms in the chain with the monomer from which it derives.…

In the structural characterisation of linear synthetic polymers, it is essential to correlate the sequence of atoms in the chain with the monomer from which it derives. A representative fragment of a linear polyethylene chain free of branches is illustrated below.

CH₂ CH₂ CH₂ CH₂ CH₂ CH₂ CH₂ CH₂ Repeating unit (?)
  1. a) From the fragment shown, identify and draw the semi-structural formula of the repeating unit corresponding to the incorporation of one ethene molecule, and that of the corresponding precursor monomer.
  2. b) If an industrial batch of this polymer has a mean molar mass of $M = 84\,150\text{ g}\times\text{mol}^{-1}$, calculate the mean degree of polymerisation $n$, justifying the approximation of neglecting the mass of the end groups.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$.

You can do it on paper and send a photo with your answer.

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Polypropylene is obtained industrially by the successive addition of propene using stereospecific Ziegler-Natta catalysts. a) Write the polymerisation reaction of…

Polypropylene is obtained industrially by the successive addition of propene using stereospecific Ziegler-Natta catalysts.

  1. a) Write the polymerisation reaction of propene, showing the semi-structural formula of the monomer and that of the repeating unit of polypropylene.
  2. b) A commercial isotactic polypropylene has a mean degree of polymerisation $n = 3\,500$. Determine the mean molar mass of the polymer, neglecting the end groups. If, in a mass balance model, the two ends of a chain with the same number of units were bonded phenyl groups ($\mathrm{-C_6H_5}$, molar mass $77.10\text{ g}\times\text{mol}^{-1}$ each), calculate the percentage error that would be made by omitting these end groups in the calculated molar mass.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$.

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Percentage composition of a homopolymer

Elemental analysis of pure PVC always gives about 56.8% chlorine by mass, exactly the same as in a single chloroethene molecule. Why is the percentage composition of a giant molecule the same as that of its monomer?

Mass distribution in the repeating unit of PVC: 56.8% chlorine and 43.2% carbon and hydrogen. repeating unit –CH₂–CHCl– of pure PVC C, H43.2% Cl56.8% share of the mass of the unit
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A sample of polychloroethene (PVC) is analysed in a quality control laboratory. a) If the mean molar mass determined experimentally for the sample is…

A sample of polychloroethene (PVC) is analysed in a quality control laboratory.

  1. a) If the mean molar mass determined experimentally for the sample is $175\,000\text{ g}\times\text{mol}^{-1}$, calculate the mean degree of polymerisation $n$, neglecting the chain ends.
  2. b) Determine the theoretical percentage by mass of chlorine in the repeating unit of pure PVC. If the elemental analysis of the batch reveals an actual chlorine percentage of $54.50\,\%$, explain whether this single piece of data makes it possible to distinguish between degradation with loss of hydrogen chloride and the presence of ethene-derived units in the chain, and propose an additional test to tell them apart.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{Cl}) = 35.45\text{ g}\times\text{mol}^{-1}$.

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Copolymers: two monomers, one chain

Adding a second monomer during synthesis changes the properties of the material, for example making it more flexible or more resistant to chemicals. How do you find the exact proportions in which two monomers have combined in one chain?

A copolymer chain built from two different monomers, A and B, arranged along the backbone. copolymer chain A–B (random arrangement) A B B A A B each A and each B adds its own molar mass to the chain
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The ETFE copolymer is synthesised by joint addition polymerisation of ethene ($\mathrm{CH_2=CH_2}$) and tetrafluoroethene ($\mathrm{CF_2=CF_2}$) with no loss of matter.…

The ETFE copolymer is synthesised by joint addition polymerisation of ethene ($\mathrm{CH_2=CH_2}$) and tetrafluoroethene ($\mathrm{CF_2=CF_2}$) with no loss of matter. A pure sample of this material has a mean molar mass of $180\,000\text{ g}\times\text{mol}^{-1}$ and a fluorine content of $59.34\,\%$ by mass.

  1. a) Calculate the mole fraction of tetrafluoroethene and of ethene in the copolymer.
  2. b) Calculate the total mean number of monomer residues forming each macromolecule and state the mass of ethene theoretically consumed to produce $250\text{ kg}$ of this copolymer.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{F}) = 19.00\text{ g}\times\text{mol}^{-1}$.

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A special copolymer for elastomers is synthesised by joint addition of propene ($\mathrm{CH_2=CH-CH_3}$) and tetrafluoroethene ($\mathrm{CF_2=CF_2}$). A dried sample of…

A special copolymer for elastomers is synthesised by joint addition of propene ($\mathrm{CH_2=CH-CH_3}$) and tetrafluoroethene ($\mathrm{CF_2=CF_2}$). A dried sample of $2,500\text{ g}$ of this copolymer is subjected to combustion and complete mineralisation in an oxidising atmosphere, with all the fluorine collected in aqueous solution as fluoride ions, $\mathrm{F^-}$. Titration of these fluorides requires $44.30\text{ mL}$ of a $\mathrm{La(NO_3)_3}$ $0.200\text{ M}$ solution, quantitatively forming the precipitate $\mathrm{LaF_3}$.

  1. a) Determine the percentage by mass of tetrafluoroethene and of propene in the copolymer.
  2. b) Calculate the propene : tetrafluoroethene molar ratio in the polymer chains and state the mean total degree of polymerisation if the mean molar mass of the polymer is $214\,000\text{ g}\times\text{mol}^{-1}$.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{F}) = 19.00\text{ g}\times\text{mol}^{-1}$.

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In the coatings industry, a copolymer is synthesised by addition using ethene ($\mathrm{CH_2=CH_2}$) and chloroethene ($\mathrm{CH_2=CHCl}$). The polymerisation is…

In the coatings industry, a copolymer is synthesised by addition using ethene ($\mathrm{CH_2=CH_2}$) and chloroethene ($\mathrm{CH_2=CHCl}$). The polymerisation is carried out with the radical initiator azobis(isobutyronitrile) (AIBN). The chains terminate exclusively by bimolecular coupling (combination), so that each polymer molecule contains exactly two initiator fragments of formula $\mathrm{-C(CH_3)_2CN}$ as end groups.

  1. a) Deduce the expression for the molar mass of the polymer in terms of the number of ethene units ($x$) and chloroethene units ($y$) per chain, and state the general formula of this macromolecule.
  2. b) A chemical analysis determines that the copolymer has a $25.00\,\%$ by mass of chlorine and a mean molar mass of $141\,900\text{ g}\times\text{mol}^{-1}$. Calculate the mean number of chloroethene units ($y$) and ethene units ($x$) per macromolecule, and check whether neglecting the mass of the ends would introduce an error greater than $0.2\,\%$ in the calculation of the molar mass.

Data: Atomic molar masses: $M(\mathrm{C}) = 12.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{H}) = 1,008\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{N}) = 14.01\text{ g}\times\text{mol}^{-1}$; $M(\mathrm{Cl}) = 35.45\text{ g}\times\text{mol}^{-1}$.

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Summary

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