Worked examples: Polymers

Work through each example by covering the answer, attempting it yourself, then checking your method against the solution. Every example flags the slip that most often costs marks. These are original questions written in SPM style, not past-year papers, and they are built directly on the four content standards of the Form 5 Polymers chapter.

Example 1, Finding the monomer of an addition polymer

Question. A synthetic polymer has the repeating unit –CH₂–CHCl– written inside brackets with subscript n. Name and give the formula of its monomer.

Solution. Take one repeating unit and restore the carbon–carbon double bond between the two carbon atoms. The monomer is CH₂=CHCl, chloroethene (vinyl chloride). The polymer itself is poly(chloroethene), or PVC.

Common slip: writing the monomer with a single bond (CH₂–CHCl) or leaving a bond hanging. The monomer of an addition polymer must be a complete, neutral molecule with the C=C double bond put back.

Example 2, From monomer to repeating unit

Question. Ethene, CH₂=CH₂, undergoes addition polymerisation. Draw the repeating unit of the polymer formed and name it.

Solution. The double bond opens, and the two carbons each form a bond to the next unit. The repeating unit is –CH₂–CH₂– inside brackets with subscript n, and the polymer is polythene (poly(ethene)).

Common slip: keeping the double bond in the repeating unit. In an addition polymer the double bond has opened, so the chain contains only single bonds.

Example 3, Classifying the type of polymerisation

Question. State, with a reason, whether the formation of nylon from a diamine and a dicarboxylic acid is addition or condensation polymerisation.

Solution. It is condensation polymerisation. Each monomer has two functional groups (–NH₂ on the diamine and –COOH on the acid), and every time a link forms a small molecule (water) is eliminated. That elimination of a small molecule is the signature of condensation.

Common slip: calling any reaction that makes a long chain “addition”. The test is not chain length but the mechanism, a released small molecule means condensation.

Example 4, The quick addition-versus-condensation test

Question. Propene (CH₂=CHCH₃) and a mixture of ethane-1,2-diol with benzene-1,4-dicarboxylic acid can both form polymers. For each, state the type of polymerisation and whether a by-product is released.

Solution. Propene has a C=C double bond and one functional feature only, so it forms polypropene by addition polymerisation with no by-product released. The diol-plus-diacid mixture has two reactive groups on each monomer (–OH and –COOH), so it forms terylene (a polyester) by condensation polymerisation, releasing water each time an ester link forms.

Common slip: forgetting to say whether a molecule is given off. The presence or absence of a by-product is often the mark that separates the two types.

Example 5, Vulcanisation as structure to property

Question. Explain why vulcanised rubber is harder and more elastic than unvulcanised natural rubber.

Solution. In unvulcanised rubber the long poly(isoprene) chains are held together only by weak forces, so they slide past one another easily and the rubber is soft and loses its springiness. Vulcanisation heats the rubber with sulfur, and sulfur atoms form cross-links (sulfur bridges) between the chains at the double bonds. These cross-links tie the chains together so they can no longer slide far apart and are pulled back into shape after stretching. The result is rubber that is harder, stronger, more elastic, with a higher elasticity limit, and more resistant to heat and oxidation.

Common slip: answering only that it is “better” or “stronger” without naming the cross-links. The mark is for the structural reason, sulfur cross-links restricting chain movement.

Example 6, Coagulation and preservation of latex

Question. Explain why adding a little ammonia solution to latex prevents it coagulating, while adding acid causes it to coagulate.

Solution. In latex the rubber particles are each wrapped in a membrane carrying a negative charge, so the particles repel one another and stay dispersed. Adding an acid neutralises the charge; the particles collide, join and the latex coagulates. Adding a little alkali (ammonia solution) keeps the negative charge, so the particles keep repelling and the latex stays liquid, which is how it is preserved for transport.

Common slip: saying the acid “dissolves” the rubber. The acid neutralises the surface charge so the particles can clump, it does not dissolve anything.

Example 7, A balanced environmental answer

Question. Discuss one advantage and one environmental problem of using synthetic polymers, and suggest one way to reduce the problem.

Solution. An advantage is that synthetic polymers are light, cheap, strong and waterproof, so they replace heavier materials in packaging and pipes. An environmental problem is that most are non-biodegradable: microorganisms cannot break them down, so waste accumulates on land and in the sea for a very long time. A workable solution is to reduce, reuse and recycle plastics, and to develop biodegradable or photodegradable plastics, rather than disposing of them by open burning.

Common slip: giving an answer that is all advantages or all problems. The marking scheme rewards the balance, a genuine benefit set against a genuine cost, plus a solution.

Example 8, Natural versus synthetic

Question. Classify each of the following as a natural or a synthetic polymer and name its monomer where you can: (a) natural rubber, (b) PVC, (c) protein, (d) polythene.

Solution. (a) Natural rubber is a natural polymer; its monomer is isoprene. (b) PVC is synthetic; its monomer is chloroethene. (c) Protein is a natural polymer; its monomers are amino acids. (d) Polythene is synthetic; its monomer is ethene.

Common slip: calling natural rubber synthetic because it can be processed in a factory. Its origin is the latex of a living tree, so it is a natural polymer even after processing.

Using these examples

Notice that every answer either restores a double bond, names the released molecule, or links a structural change to a property, the habits this chapter rewards. Once the method feels automatic, move on to the practice questions for this chapter and mark yourself the same way. A one-to-one teacher can check that your monomer structures and your vulcanisation reasoning are complete, which is exactly where marks are gained or lost in the SPM Chemistry written papers. Rehearse the eight examples until the reasoning is second nature, and the Polymers questions in Paper 1 and Paper 2 become some of the most reliable marks in the whole Form 5 course.