The shape of the lesson
By the end of the lesson, learners can
- From chain to monomers, and back: Work through each example: find the repeating pattern first, then cut at the links.
- Addition or condensation?: Toggle between the addition and condensation examples in the simulator.
- C–C backbone only: addition; restore the C=C.
- –COO– or –CONH– in the chain: condensation.
- Cut: OH back on C=O, H back on O or N.
- Polyamides hydrogen bond chain to chain.
One picture every idea returns to
To deduce monomers from a condensation polymer, find each ester (–COO–) or amide (–CONH–) link, break the C–O or C–N bond, and add back OH to the carbonyl carbon and H to the O or N (or Cl to the carbonyl carbon if a diacyl dichloride is asked for). An addition polymer has a backbone of carbon atoms only; its monomer is found by taking a two-carbon repeat unit and putting back the C=C. A monomer with a C=C bond undergoes addition; monomers with two reactive groups (OH, NH2, COOH, COCl) undergo condensation. Properties follow from the intermolecular forces between chains: the N–H and C=O groups of polyamides form hydrogen bonds between chains, which makes nylon and Kevlar strong and high-melting.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Read the chain, find the monomers.. Given a section of polymer, can you say how it was made and what it was made from? Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: From chain to monomers, and back | Work through each example: find the repeating pattern first, then cut at the links. Project tab 3 of the lab (Condensation polymers lab, monomer finder) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Addition or condensation? | Toggle between the addition and condensation examples in the simulator. Project tab 4 of the lab (Condensation polymers lab, addition or condensation?) and let a student drive it while the class predicts. | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Addition or condensation?, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| When cutting an amide link, the C=O stays with the acid monomer and the N–H with the amine: do not split the C=O away from its carbon. | A polymer with a C–C backbone and side groups such as –CN or –COOCH 3 is still an addition polymer, even if the side group is an ester. Polyesters cannot hydrogen bond chain to chain, because they have no H on O or N. |
Polymer deductions: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
A section of a polymer is –NH(CH2)4NHCO(CH2)3CO–NH(CH2)4NHCO(CH2)3CO–. (a) Identify the type of polymer. (b) Give the structures of the two monomers.
Draw the repeat unit of the polymer formed from 3-hydroxypropanoic acid, HOCH2CH2COOH, and name the type of polymer.
For each monomer, state whether it forms an addition or a condensation polymer: (a) CH2=CHC6H5, (b) H2NCH(CH3)COOH, (c) CH2=CHCOOH.
Explain why nylon-6,6 has a higher melting point than poly(ethene) of similar chain length.
A polymer section is –CH2CCl2CH2CCl2–. Give the monomer and explain how you know the type of polymerisation.
Suggest why poly(lactic acid) chains cannot hydrogen bond to each other, whereas protein chains can.
Answer key and marking notes
- (a) A polyamide (condensation polymer).
- (b) Butane-1,4-diamine, H2N(CH2)4NH2.
- Pentanedioic acid, HOOC(CH2)3COOH (or its diacyl dichloride ClOC(CH2)3COCl).
- –OCH2CH2CO–.
- A polyester, formed by condensation with loss of water.
- (a) Addition: it has a C=C.
- (b) Condensation: it has NH2 and COOH groups (gives a polyamide).
- (c) Addition through the C=C under normal conditions; the single COOH cannot by itself give a condensation chain.
- Nylon chains contain N–H and C=O groups.
- The H on N of one chain forms hydrogen bonds with a lone pair on the O of a C=O on a neighbouring chain.
- Poly(ethene) has only id-id forces; hydrogen bonds need more energy to overcome, so nylon melts higher.
- Monomer: 1,1-dichloroethene, CH2=CCl2.
- The backbone is carbon only, with no ester or amide links, so it is an addition polymer.
- Poly(lactic acid) is a polyester: its chain has C=O and C–O but no H attached to O or N.
- Protein chains have N–H groups that act as hydrogen bond donors to C=O groups on other chains.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The diagram shows a section of polymer Q. Which monomer was used to make Q?
- A4-hydroxybutanoic acid, HOCH2CH2CH2COOH
- Bpropane-1,3-diol, HOCH2CH2CH2OH, and carbon dioxide
- Cbut-3-enoic acid, CH2=CHCH2COOH
- Dbutanedioic acid, HOOCCH2CH2COOH
Which polymer is an addition polymer?
- A–[CH2CH(CN)]n–
- B–[NH(CH2)5CO]n–
- C–[OCH2CH2OOCC6H4CO]n–
- D–[OCH(CH3)CO]n–
A short section of a protein is –NHCH2CO–NHCH(CH3)CO–NHCH(CH2OH)CO–. How many different amino acids are present in this section?
- A3
- B1
- C2
- D6
Why does nylon-6,6 have a higher melting point and greater strength than poly(ethene) of similar chain length?
- AN–H groups on one chain form hydrogen bonds with C=O groups on neighbouring chains
- Bnylon chains are joined to each other by covalent cross-links
- Cnylon contains ionic bonds between its chains
- Dpoly(ethene) chains have permanent dipoles that repel each other
Which pair of monomers would react to form a polyamide?
- AH2N(CH2)4NH2 and ClOC(CH2)8COCl
- BHO(CH2)4OH and HOOC(CH2)8COOH
- CCH2=CHCONH2 and CH2=CH2
- DH2N(CH2)4NH2 and HO(CH2)4OH
Polymer S has the repeat unit –CH2C(CH3)(COOCH3)–. Which statement about S is correct?
- AS is an addition polymer made from CH2=C(CH3)COOCH3; the ester group is a side chain
- BS is a polyester made by condensation, because it contains an ester group
- CS is made from ethene and methyl ethanoate with loss of water
- DS is a polyamide made from an amino acid
Hexane-1,6-diamine reacts with decanedioyl dichloride, ClOC(CH2)8COCl. Which is the repeat unit of the polymer (nylon-6,10)?
- A–NH(CH2)6NHCO(CH2)8CO–
- B–NH(CH2)6NHCO(CH2)10CO–
- C–NH2(CH2)6NH2CO(CH2)8COCl–
- D–O(CH2)6OCO(CH2)8CO–
Kevlar and nylon-6,6 are both polyamides. Why is Kevlar much stronger?
- Aits flat benzene rings make the chains rigid and straight, so they pack in parallel and hydrogen bond along their whole length
- BKevlar chains are joined by covalent cross-links through the benzene rings
- CKevlar has ester links, which are stronger than amide links
- DKevlar has no hydrogen bonding, so its chains slide less
Quiz key
Q is a polyester with only one kind of residue, –OCH2CH2CH2CO–. Adding H to the O end and OH to the C=O end gives HOCH2CH2CH2COOH, a hydroxycarboxylic acid that polymerises with itself. A diol alone has no carbonyl group, and CO2 is not a monomer here. But-3-enoic acid would give an addition polymer with a C–C backbone. Butanedioic acid has no OH group, so it cannot form ester links on its own.
–[CH2CH(CN)]n– has an all-carbon backbone; it is poly(propenenitrile), made by opening the C=C of CH2=CHCN. The nitrile is just a side group. The NH...CO chain is nylon-6, a polyamide. The O...CO chains are polyesters (PET and poly(lactic acid)), with ester links in the backbone, so all three are condensation polymers.
Cutting at the peptide links gives three residues with side groups H (glycine), CH3 (alanine) and CH2OH (serine): three different amino acids. 1 would need identical side groups. 2 overlooks that CH2OH is a different side group from CH3. 6 counts every NH and CO group separately, but each amino acid supplies both an NH and a CO.
The amide link contains N–H (an H on an electronegative atom) and C=O (an O with lone pairs), so hydrogen bonds form between chains. These need far more energy to overcome than the id-id forces between non-polar poly(ethene) chains. Nylon is a thermoplastic with no covalent cross-links, and it is molecular, not ionic. Poly(ethene) has essentially no permanent dipoles; its chains attract each other only through id-id forces.
A diamine with a diacyl dichloride gives amide links, –CONH–, with loss of HCl: a polyamide. A diol with a diacid gives a polyester. The alkenes would undergo addition polymerisation; the amide in propenamide is only a side group. A diamine and a diol have no carbonyl group, so neither can form ester or amide links with the other.
The backbone is two carbons with no O or N in it, so S (poly(methyl 2-methylpropenoate), Perspex) is formed by addition across a C=C; the COOCH3 group hangs off the chain. A polyester has the ester link in the chain itself. No water is lost in addition polymerisation, and S has the same atoms as its single monomer. There is no nitrogen, so it cannot be a polyamide.
Each NH2 loses one H and each COCl loses its Cl (as HCl), giving –NH(CH2)6NHCO(CH2)8CO–: 6 carbons from the diamine and 10 from the acid unit (8 CH2 + 2 C=O). Writing (CH2)10 double-counts the carbonyl carbons. Keeping NH2 and COCl ignores the HCl eliminated. The O version is a polyester from a diol, not a diamine.
The 1,4-substituted benzene rings and amide links make Kevlar chains planar and rod-like, so they align in ordered sheets with N–H...O=C hydrogen bonds between neighbouring chains all along their length. Flexible (CH2) chains in nylon pack less regularly. Kevlar is not cross-linked covalently. It has amide links, not ester links. Hydrogen bonding is the main reason for its strength, so saying it has none is wrong.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.