Full chapter
Polymers and Organic Chemistry
Read organic structures, predict reactions and explain why a protein or plastic behaves as it does.
A-Level 8873, revised syllabus (2026-2027)
Read and name an organic structure
A formula must show which atoms are joined, not just how many there are.
An organic structure is a map of which atoms are joined. A functional group is the atom or group of atoms that gives a compound its typical reactions. First find the functional group, then the carbon chain. The same molecular formula can describe compounds with different connections, and so different chemistry.
| Representation | What it tells you |
|---|---|
| Empirical: CH2O | The simplest whole-number atom ratio, C:H:O = 1:2:1. |
| Molecular: C3H6O3 | The actual numbers of atoms in one molecule. |
| Structural: CH3CH(OH)CO2H | The middle carbon bears OH; the final carbon belongs to COOH. Brackets keep the connectivity unambiguous. |
| Displayed | Every atom and bond is shown, including each C-H and O-H bond. |
| Skeletal | Carbon atoms are implicit at line ends and corners; enough attached hydrogen atoms are implied to give each carbon four bonds. Heteroatoms and their attached hydrogens, such as OH, are shown. |
To turn a condensed formula into a displayed formula, draw the carbon chain first. Then draw out each bracketed group where it is attached. Write COOH as C(=O)-O-H and CHO as C(=O)-H. Finally draw every C-H bond and check the bonds: four for each carbon, two for each oxygen and one for each hydrogen. In a skeletal formula, use the same check to find the hidden C-H bonds.
A displayed formula keeps every atom visible
Three carbons form a chain. The first has three hydrogens; the second has one hydrogen and an OH group; the third has a double-bonded oxygen and an OH group. There are three carbons, six hydrogens and three oxygens.
The same molecule in skeletal form
The left endpoint is CH3, the middle corner is CH attached to an explicit OH, and the right corner is the carboxyl carbon with C=O and OH. It represents the same lactic-acid connectivity as the displayed formula.
| Class and general structure | Example and naming cue |
|---|---|
| Alkane: CnH2n+2 for an acyclic saturated hydrocarbon | CH3CH3, ethane; suffix -ane. |
| Alkene: CnH2n for an acyclic hydrocarbon with one C=C | CH2=CHCH3, propene; locate the C=C and use -ene. |
| Benzene: C6H6, a six-carbon aromatic ring | Use a hexagon containing a circle to represent the delocalised ring. A fully displayed benzene ring is not required. |
| Halogenoalkane: R-X; CnH2n+1X for a saturated acyclic monohalogen compound | CH3CH2Br, bromoethane; fluoro-, chloro-, bromo- or iodo- gives the substituent. |
| Alcohol: R-OH; CnH2n+2O for a saturated acyclic monohydric alcohol | CH3CH(OH)CH3, propan-2-ol; number the OH position. |
| Aldehyde: R-CHO; CnH2nO for a saturated acyclic monoaldehyde | CH3CHO, ethanal; the carbonyl carbon has at least one H, and its carbon counts in the chain. |
| Ketone: R-CO-R'; CnH2nO for a saturated acyclic monoketone | CH3COCH3, propanone; the carbonyl carbon is bonded to two carbon groups. |
| Carboxylic acid: R-COOH; CnH2nO2 for a saturated acyclic monocarboxylic acid | CH3COOH, ethanoic acid; the COOH carbon is carbon 1. |
| Ester: R-COO-R'; CnH2nO2 for a saturated acyclic monoester | CH3COOCH2CH3, ethyl ethanoate; name the O-attached alkyl group first, then the acid-derived -oate part. |
| Amine: R-NH2 for a primary amine; N can also carry further carbon groups | CH3CH2NH2, ethylamine (ethanamine). For a saturated acyclic monoamine, CnH2n+3N. |
| Amide: R-CONH2; N-substituted forms include R-CONHR' | CH3CONH2, ethanamide; C=O is directly attached to N. A saturated acyclic monoamide has CnH2n+1NO. |
| Amino acid: contains both NH2 and COOH; an alpha-amino acid is H2N-CH(R)-COOH | H2NCH2COOH, aminoethanoic acid; the amino group is on the carbon next to COOH. For a saturated acyclic compound with one of each group: CnH2n+1NO2. |
Here R and R' stand for groups that contain carbon. They do not have to be the same. In methanal and methanoic acid, the group on the carbonyl carbon is H instead. General formulae only work under their stated conditions. Rings, extra double bonds or extra functional groups change them. A formula alone does not tell you the family: propanal and propanone are both C3H6O.
To name a simple compound, choose the parent chain that contains the main functional group. Number it so that the group or the double bond gets a low number. The stems meth-, eth-, prop-, but-, pent- and hex- mean 1 to 6 carbons. Show the branches and their positions. CH3CH(CH3)CH2OH is 2-methylpropan-1-ol: the carbon joined to OH is carbon 1, so the methyl branch is on carbon 2.
| Alcohol class | Carbon groups attached to the OH-bearing carbon | Example |
|---|---|---|
| Primary | One carbon group; methanol is conventionally treated with primary alcohols. | CH3CH2OH |
| Secondary | Two carbon groups. | CH3CH(OH)CH3 |
| Tertiary | Three carbon groups. | (CH3)3COH |
Check your understanding
What is the name of CH3CH2COOCH3?
Show the answer
Methyl propanoate
The CH3 on the oxygen gives methyl. CH3CH2COO- has three carbons including the carbonyl carbon, so it is propanoate.
Change connectivity or fixed spatial arrangement
Systematic drawing prevents repeated structures and missing isomers.
Constitutional (structural) isomers have the same molecular formula, but their atoms are joined differently. You can change the carbon chain, move the functional group to a different position, or use a different functional group that fits the formula. Turning a chain round or drawing it the other way does not make a new isomer.
Worked exampleFind every alcohol with four carbons
Deduce the constitutional alcohol isomers with molecular formula C4H10O. Restrict the task to alcohols.
Try it first. Then show the worked stepsHide the worked steps
- Use the straight four-carbon skeleton. OH can be on carbon 1 or carbon 2: butan-1-ol, CH3CH2CH2CH2OH; and butan-2-ol, CH3CH(OH)CH2CH3. Positions 3 and 4 duplicate 2 and 1 after reversing the chain.
- Use the branched three-carbon skeleton with a methyl branch. OH can be on an end carbon: 2-methylpropan-1-ol, (CH3)2CHCH2OH.
- Or put OH on the central carbon: 2-methylpropan-2-ol, (CH3)3COH.
- Check each carbon has four bonds and each structure has C4H10O. These are the four alcohol constitutional isomers; the restriction matters because other functional-group classes can also share this formula.
Four alcohol isomers: two primary, one secondary and one tertiary.
Worked exampleSearch across functional-group families
A compound has formula C3H6O2 and is known to be a saturated, acyclic carboxylic acid or ester with no other functional group. Deduce all possibilities and decide which could yield methanol on alkaline hydrolysis.
Try it first. Then show the worked stepsHide the worked steps
- For the acid family, one carbon belongs to COOH. The remaining two form an ethyl group: CH3CH2COOH, propanoic acid. With only three carbons there is no different branched acid skeleton.
- For an ester, count the carbonyl carbon as part of the acid-derived fragment. Split the total of three carbons between the acid-derived and alcohol-derived fragments. The alcohol fragment must contain at least one carbon.
- A 2 + 1 split gives CH3COOCH3, methyl ethanoate. A 1 + 2 split gives HCOOCH2CH3, ethyl methanoate. Methanoate has H directly attached to its carbonyl carbon; do not insert an extra CH3.
- All three structures have 3 C, 6 H and 2 O. Reversing a drawing does not create another structure, but swapping the two ester fragments does change connectivity.
- Methyl ethanoate gives methanol and ethanoate on alkaline hydrolysis. Ethyl methanoate gives ethanol and methanoate; propanoic acid undergoes neutralisation instead.
The specified family restriction gives three constitutional isomers. Methanol production selects methyl ethanoate. The general formula alone does not identify the functional group.
Cis-trans isomerism keeps the same connections but changes the arrangement in space around a C=C bond. A double bond is one sigma bond and one pi bond. Rotating it would break the sideways overlap that forms the pi bond, so it cannot rotate freely. Each carbon of the double bond must carry two different groups for cis and trans forms to exist.
But-2-ene has two fixed arrangements
Each carbon in the C=C has H and CH3 attached. In cis-but-2-ene the two CH3 groups lie on the same side; in trans-but-2-ene they lie on opposite sides. Rotation about the C=C is restricted by the pi bond.
Worked exampleSeparate the two kinds of isomer count
Deduce the acyclic alkenes with formula C4H8, including cis-trans forms.
Try it first. Then show the worked stepsHide the worked steps
- A straight skeleton gives but-1-ene, CH2=CHCH2CH3, and but-2-ene, CH3CH=CHCH3. A branched skeleton gives 2-methylpropene, CH2=C(CH3)2.
- But-2-ene has H and CH3 on each double-bonded carbon, so both cis and trans forms exist.
- But-1-ene and 2-methylpropene each contain a CH2 end of the double bond. Its two identical H substituents prevent cis-trans isomerism.
There are three constitutional alkene isomers, or four distinct alkene structures when cis- and trans-but-2-ene are counted separately. Rings are excluded by the question.
Check your understanding
Which of CHCl=CHBr and CH2=CClBr has cis-trans isomers?
Show the answer
CHCl=CHBr only, because each C=C carbon has two different groups
In CHCl=CHBr the two H atoms can be on the same or opposite sides. In CH2=CClBr one carbon carries two H atoms, so swapping them gives the same molecule.
Read shape from sigma and pi bonding
Describe the local geometry around a carbon before the whole molecule.
| Molecule | Carbon bonding and local shape | Angles and freedom of movement |
|---|---|---|
| Ethane, CH3CH3 | Each carbon forms four sigma bonds and has a tetrahedral arrangement. The C-C bond is one sigma bond from head-on overlap. | About 109.5° around each carbon. Rotation about C-C is possible without breaking the sigma bond; the whole molecule is not planar. |
| Ethene, CH2=CH2 | Each carbon has three regions of sigma bonding in a trigonal planar arrangement. Parallel p orbitals overlap sideways above and below the plane, forming one pi bond. | About 120° around each carbon. The molecule is planar; rotation about C=C is restricted. |
| Benzene, C6H6 | Each carbon has three sigma bonds in a planar hexagonal framework. The six parallel p orbitals overlap around the ring, giving delocalised pi electron density above and below it. | About 120° around every carbon. All six C-C bonds are equivalent; benzene is not a ring of alternating independent single and double bonds. |
Benzene: a planar ring with delocalised pi bonding
A hexagon with an internal circle is the preferred representation of benzene. Each corner represents a carbon attached to one hydrogen. The circle represents delocalised pi bonding, not an extra atom or separate small ring.
A multiple bond counts as one direction of electron density when you predict the shape around an atom. In ethene, each carbon has a double bond but only three directions, so it is trigonal planar. You do not need hybridisation labels here. Use the number and arrangement of sigma bonds, and say where sideways pi overlap can happen.
Worked exampleApply the pattern to an unfamiliar chain
Predict the approximate angles around each carbon in propene, CH2=CHCH3.
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- The two double-bonded carbons each have three sigma-bond directions. Their local arrangements are trigonal planar with angles near 120°.
- The CH3 carbon has four sigma-bond directions, so it is tetrahedral with angles near 109.5°.
- The C=C consists of one sigma and one pi bond; the adjacent C-C single bond is a sigma bond. Do not assign 120° to every carbon just because the molecule contains one double bond.
The alkene region is locally planar; the methyl carbon is tetrahedral.
Check your understanding
What shape is expected around the carbonyl carbon in CH3CHO?
Show the answer
Trigonal planar, about 120 degrees, with one sigma and one pi bond in C=O
The carbon has three regions of electron density: bonds to CH3, H and O. These spread out in a plane at about 120 degrees.
An alkane substitutes; an alkene adds
Follow the bond that changes and account for every product.
Make a guess. You are not marked.
Orange bromine water is shaken with hexene, and separately with hexane, in the dark. What do you see?
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Only the hexene decolourises it, by addition
Bromine adds across the C=C bond in hexene. Hexane has only single bonds, and without UV light it does not react.
Alkanes have only strong C-C and C-H sigma bonds and no reactive functional group. So they are unreactive under ordinary conditions. They do burn, and they react with chlorine by substitution in ultraviolet light. Alkenes have a pi bond, which can be replaced by two new sigma bonds in an addition reaction. Addition joins reactants into one product and loses no small molecule. Substitution replaces one atom or group with another.
| Starting compound and conditions | Equation | What changes |
|---|---|---|
| Ethane; oxygen, ignition | 2C2H6 + 7O2 → 4CO2 + 6H2O | Complete combustion oxidises carbon and reduces oxygen. Hydrogen remains at oxidation number +1. Limited oxygen can give CO and/or carbon instead. |
| Ethane; Cl2, ultraviolet light, room temperature | C2H6 + Cl2 → C2H5Cl + HCl | Substitution replaces H by Cl. Further substitutions can occur, so one equation does not promise a single pure product. |
| Ethene; oxygen, ignition | C2H4 + 3O2 → 2CO2 + 2H2O | Complete combustion. |
| Ethene; Br2 in CCl4 | CH2=CH2 + Br2 → CH2BrCH2Br | Addition across C=C gives 1,2-dibromoethane. Bromine colour is discharged; no HBr is a product of this addition. |
| Ethene; H2, Ni catalyst, heat | CH2=CH2 + H2 → CH3CH3 | Hydrogenation is addition and reduction. A pi bond is replaced by two C-H bonds. |
Worked exampleTransfer the reaction to a longer alkene
Predict the product when but-2-ene reacts with Br2 in CCl4.
Try it first. Then show the worked stepsHide the worked steps
- Locate the C=C between carbon 2 and carbon 3: CH3CH=CHCH3.
- Change that C=C to C-C and attach one Br to each of those carbons.
- The product is CH3CHBrCHBrCH3, 2,3-dibromobutane. No carbon is added or removed, and both Br atoms appear in the product.
CH3CH=CHCH3 + Br2 → CH3CHBrCHBrCH3.
Check your understanding
A compound decolourises bromine. Why is that evidence of a reaction, but not enough to identify the compound?
Show the answer
Many alkenes, and some other compounds, decolourise bromine, so more evidence is needed.
Decolourising shows bromine was used up. Combine it with the formula and other tests to identify one compound.
Conditions choose the reaction route
Water favours substitution of bromoethane; ethanolic base promotes elimination.
Make a guess. You are not marked.
Bromoethane is heated with sodium hydroxide dissolved in ethanol, not water. What is the main organic product?
Show the answer
Ethene, by elimination of HBr
In ethanol, hydroxide acts mainly as a base and removes HBr, so elimination gives an alkene.
Elimination removes atoms or groups from two neighbouring carbons and forms a double bond. Substituting Br with OH is different: the carbon chain stays saturated. For bromoethane, the solvent and conditions decide the product, so writing only "NaOH" is not a complete answer.
| Route | Reagents and essential conditions | Balanced change |
|---|---|---|
| Bromoethane → ethanol | Aqueous NaOH; heat, usually under reflux. | CH3CH2Br + NaOH → CH3CH2OH + NaBr |
| Bromoethane → ethene | NaOH in ethanol; heat. | CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O |
| Ethanol → ethene | Concentrated H3PO4 catalyst; heat. | CH3CH2OH → CH2=CH2 + H2O |
| Ethanol → ethanoic acid | Acidified K2Cr2O7 or acidified KMnO4; heat under reflux. | CH3CH2OH + 2[O] → CH3COOH + H2O |
| Ethanol complete combustion | Oxygen; ignition. | C2H5OH + 3O2 → 2CO2 + 3H2O |
A reflux condenser cools the vapour so that it drips back into the flask. This lets you heat for a long time without losing volatile substances. When ethanol is oxidised to the acid, enough oxidant and a long heating time let the reaction go past the aldehyde to the carboxylic acid. Acidified dichromate turns from orange to green. Acidified permanganate loses its purple colour as it is reduced.
In organic chemistry, oxidation usually adds bonds from carbon to oxygen or removes C-H bonds. Reduction does the opposite. [O] and [H] stand for the oxygen or hydrogen supplied in a balanced equation. They are a way of keeping count, not real single atoms in a bottle.
Worked examplePlan a two-step conversion
Convert 1-bromopropane into propanoic acid using analogous reactions from this course.
Try it first. Then show the worked stepsHide the worked steps
- CH3CH2CH2Br → CH3CH2CH2OH using aqueous NaOH and heat: substitution replaces Br by OH.
- Heat the primary alcohol with acidified K2Cr2O7 or acidified KMnO4 under reflux to form CH3CH2COOH.
- Check that the carbon skeleton remains three carbons in both steps. Ethanolic NaOH would instead divert the first step towards propene.
Use aqueous NaOH and heat, then an acidified oxidising agent and heat under reflux.
Check your understanding
Why is ethanol to ethene an elimination, not a substitution?
Show the answer
Water is removed (OH from one carbon, H from the next) and a C=C forms.
In elimination a small molecule is lost and a double bond forms. In substitution, another group would replace OH on the same carbon.
Link oxidation level with acid-base behaviour
Carbonyl chemistry changes C=O; acid-base chemistry transfers protons.
| Compound and change | Reagent and conditions | Equation |
|---|---|---|
| Ethanal → ethanol | LiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat. | CH3CHO + 2[H] → CH3CH2OH |
| Propanone → propan-2-ol | LiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat. | CH3COCH3 + 2[H] → CH3CH(OH)CH3 |
| Ethanal → ethanoic acid | Acidified K2Cr2O7 or acidified KMnO4; heat. | CH3CHO + [O] → CH3COOH |
In carbonyl reduction, one hydrogen adds to the carbonyl carbon and one to the oxygen, so C=O becomes CH-OH. An aldehyde gives a primary alcohol, and a ketone gives a secondary alcohol. LiAlH4 is used in dry conditions, usually dry ether, and water is added afterwards. Do not write it as an aqueous reagent. An aldehyde can be oxidised to a carboxylic acid with the same carbon chain. A ketone is not oxidised by these usual conditions, so do not give it a carboxylic acid product.
Ethanoic acid gives a proton from its COOH group. With an alkali it forms a carboxylate salt and water. With a carbonate it forms a salt, water and carbon dioxide. The group changes from -COOH to -COO-, and the carbon chain stays the same.
| Reaction | What the equation explains |
|---|---|
| CH3COOH + NaOH → CH3COONa + H2O | One mole of ethanoic acid neutralises one mole of hydroxide. |
| 2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2 | Two acidic protons are needed per carbonate ion; effervescence indicates gas evolution. |
| CH3CH2NH2 + HCl → [CH3CH2NH3]+Cl- | Ethylamine accepts a proton at nitrogen, forming ethylammonium chloride. It is not converted into an amide by this acid-base step. |
Worked examplePredict a carbonyl product without guessing the name
Reduce CH3COCH2CH3 using H2/Ni.
Try it first. Then show the worked stepsHide the worked steps
- The starting carbonyl carbon is attached to CH3 and CH2CH3, so it is a ketone.
- Replace C=O by CH-OH while preserving both attached carbon groups: CH3CH(OH)CH2CH3.
- The product is butan-2-ol, a secondary alcohol. Both oxygen and all four carbons remain in the molecule.
Butan-2-ol forms; reduction adds hydrogen without extending or shortening the carbon chain.
Check your understanding
Ethylamine reacts with aqueous acid. Where is the positive charge in the product?
Show the answer
On nitrogen, as its lone pair bonds to H+ to give CH3CH2NH3+.
The lone pair on N accepts the proton. The counter-ion, such as Cl-, balances the charge.
Make and break ester or amide links
The functional group and reaction medium determine the products.
Condensation joins molecules and removes a small molecule, often water. Hydrolysis uses water to break a bond. Heating with aqueous acid or alkali speeds it up. Ester links join a carbonyl group to O, and amide links join it to N. Find that linking bond before you try to name the products.
| Link formed | Reagents and conditions | Organic product |
|---|---|---|
| Ester, -C(=O)-O- | Ethanoic acid + ethanol; concentrated H2SO4 catalyst, heat. | CH3COOH + HOCH2CH3 ⇌ CH3COOCH2CH3 + H2O |
| Amide, -CONH- | Ethanoic acid + ethylamine in the presence of DCC (dicyclohexylcarbodiimide). | CH3CONHCH2CH3, N-ethylethanamide. The acid-derived fragment remains attached to the carbonyl carbon; the ethyl group remains on nitrogen. |
The ester link -C(=O)-O- is neutral. The dash at the end is a covalent bond to the next carbon group. In a carboxylate ion, -COO-, the minus sign is a charge. These are different structures. So the ester links in a polyester do not give the polymer a negative charge.
Keep the two fragments attached to the right atoms
Structural fragments show the carbonyl carbon joined to oxygen in an ester and to nitrogen in an amide. Hydrolysis cleaves the acyl C-O or acyl C-N bond, restoring an acid-derived fragment and an alcohol- or amine-derived fragment. The diagram is a connectivity guide, not a reaction mechanism.
| Conditions | Equation | Why this form appears |
|---|---|---|
| Aqueous acid and heat | CH3COOCH2CH3 + H2O ⇌ CH3COOH + CH3CH2OH | The acidic mixture contains the carboxylic acid. Acid catalyses the reaction. |
| Aqueous NaOH and heat | CH3COOCH2CH3 + NaOH → CH3COONa + CH3CH2OH | Any acid formed is deprotonated by the alkali. The isolated acid-derived product in this medium is the carboxylate salt. |
| Conditions | Net ionic equation | Product check |
|---|---|---|
| Aqueous acid and heat | CH3CONH2 + H2O + H+ → CH3COOH + NH4+ | The nitrogen product is protonated in acid; write ammonium, not free ammonia as the final major form. |
| Aqueous alkali and heat | CH3CONH2 + OH- → CH3COO- + NH3 | The acid fragment is a carboxylate; nitrogen leaves as ammonia. With NaOH, the salt is sodium ethanoate. |
Worked exampleTransfer hydrolysis to an unfamiliar ester
Predict the organic products when methyl propanoate, CH3CH2COOCH3, is heated with aqueous NaOH.
Try it first. Then show the worked stepsHide the worked steps
- Identify the acid-derived fragment CH3CH2CO- and the O-attached methyl fragment.
- Breaking the ester link gives a propanoate ion and methanol in the alkaline mixture.
- Include Na+ as counter-ion: CH3CH2COOCH3 + NaOH → CH3CH2COONa + CH3OH. All four carbons are accounted for.
Sodium propanoate and methanol form. Propanoic acid would require subsequent acidification.
Separating the product is about reasoning, not a memorised list of apparatus. After making an ester, ask what else is left: excess acid, acid catalyst, alcohol and water may all be present. Washing with water removes impurities that dissolve in water. Washing with a carbonate solution neutralises acid. If two liquid layers form, separate the layer that holds the ester, then dry it to remove the last water. Use the solubility and boiling-point data given to choose further steps. Never decide which layer is which only from whether it is on top.
Check your understanding
CH3CONHCH2CH3 is heated with aqueous acid. What are the organic products?
Show the answer
Ethanoic acid and ethylammonium ions, CH3CH2NH3+
Hydrolysis breaks the amide link. The ethyl group stays on nitrogen, and the acid protonates the ethylamine released.
Turn monomers into chains
Identify the repeat unit and the link formed between monomers.
A polymer is a very large molecule built from many monomer units. In this course, a molecule counts as a polymer if its average relative molecular mass is at least 1000 or it has at least 100 repeat units. A polymer sample contains chains of different lengths, so an average is used. The n in the formula does not mean every chain has exactly the same length.
| Feature | Addition polymerisation | Condensation polymerisation |
|---|---|---|
| Starting structures | Typically monomers with C=C, such as ethene. | Each chain-building monomer needs at least two link-forming sites. A diol has two OH groups and a dicarboxylic acid has two COOH groups; an amino acid has one amino group and one carboxyl group. These allow links to form repeatedly. |
| Bond change | The C=C pi bond is replaced by single bonds connecting monomer units. | New links form between functional groups, commonly ester or amide links. |
| Small product | No small molecule is eliminated in forming the addition chain. | A small molecule is eliminated per link in the examples here, usually water. |
| Repeat unit | Contains all the atoms of its contributing alkene monomer. | Accounts for the atoms left after the small molecule is removed; it may contain fragments of two different monomers. |
The substituent stays on its original carbon
The propene double bond becomes a single bond in the polymer backbone. The methyl substituent remains attached to the second carbon. The repeat-unit brackets include two backbone carbons; continuation bonds extend through the bracket boundaries.
For poly(ethene), nCH2=CH2 → [-CH2-CH2-]n. The bonds on both sides of the repeat unit must go through the brackets. To find the monomer from an addition repeat unit, take two neighbouring backbone carbons, put back their double bond and keep their groups attached. Do not break a C-H bond to make the monomer.
An alpha-amino acid has NH2 and COOH on the same carbon: H2N-CH(R)-COOH. In aminoethanoic acid, R = H. In a protein, the carboxyl group of one amino acid joins the amino group of the next, forming a peptide bond, -C(=O)-NH-. A peptide bond is an amide link. Repeating this condensation builds a chain with a repeating -NH-CH(R)-CO- backbone.
Worked exampleBuild a peptide link and keep the side chains
Show one dipeptide formed with aminoethanoic acid, H2NCH2COOH, on the left and 2-aminopropanoic acid, H2NCH(CH3)COOH, on the right.
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- Remove OH from the left molecule's COOH and H from the right molecule's NH2, accounting for H2O overall.
- Join the left carbonyl carbon to the right nitrogen: H2NCH2CONHCH(CH3)COOH.
- The central -CONH- is the peptide link. One amino end and one carboxyl end remain, allowing further chain extension. The CH3 side chain stays attached to its original alpha carbon.
One water molecule is removed per peptide link in this condensation description. A linear chain of N amino acids contains N - 1 peptide links, not N.
Check your understanding
What monomer gives the repeat unit [-CH2-CHCl-]n?
Show the answer
CH2=CHCl, chloroethene
Restore the C=C between the two backbone carbons. Chlorine stays on its carbon. The polymer is PVC.
Change a protein shape without cutting its backbone
Folding, denaturation and hydrolysis are different processes.
Make a guess. You are not marked.
A protein is gently heated until it denatures. What happens to its peptide bonds?
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They stay intact; weaker interactions break
Gentle heating breaks weaker interactions such as hydrogen bonds, so the protein unfolds. Breaking peptide bonds needs hydrolysis.
The order of amino acids gives a chain held together by covalent bonds. But many proteins only work when the chain folds into a particular three-dimensional shape. Attractions between different parts of the chain, or between chains, hold that shape. An enzyme's active site depends on this shape, which links protein structure to the specificity explained in Reaction Kinetics.
| Interaction | Structural origin | How conditions can disturb it |
|---|---|---|
| Hydrogen bonding | Attractions involving suitable N-H or O-H groups and lone pairs on nearby O or N atoms, including peptide C=O groups. | Heating and changed interactions with water can disrupt the pattern stabilising the folded arrangement. |
| Intermolecular-type attractions | Permanent dipole interactions and instantaneous dipole-induced dipole attractions act between suitable regions within a folded chain and between chains. | A changed arrangement alters the contacts and their collective stabilisation. These are not new peptide bonds. |
| Ionic linkages | Attraction between oppositely charged groups, such as -NH3+ and -COO-. | Changing pH can change whether these groups are charged, disrupting the original ionic attractions. |
Denaturation is the loss of the particular three-dimensional shape that a protein needs to work. High temperature can break the attractions that hold this shape. Extreme pH changes which groups carry a charge, so ionic attractions and other interactions change. The protein can unfold, refold wrongly or clump with other chains. Denaturation does not break the peptide bonds of the backbone.
The chain remains connected when its fold is disrupted
The top green line is a folded protein backbone stabilised by several dashed noncovalent contacts. The lower green line remains continuous after these contacts are disrupted. The drawing is schematic and does not represent a particular protein or the detailed atoms of a peptide chain.
Low temperature usually slows enzyme reactions because molecules move more slowly. It does not denature the enzyme. At too high a temperature, the enzyme can lose the shape of its active site. Then fewer enzyme-substrate complexes form, even though molecules move faster. Going back to the original temperature or pH does not always bring the activity back, for example if the protein has clumped.
| Observation | Explanation |
|---|---|
| Egg white becomes opaque and sets on heating. | Proteins lose their original folded arrangements. Exposed regions associate with other chains, forming a new extended network or aggregates that scatter light and trap water. This does not mean cooking has hydrolysed every protein into amino acids. |
| Vinegar causes milk to form curds. | Added acid changes protein charge and disrupts the balance maintaining dispersion. With reduced repulsion and changed interactions, proteins associate and precipitate into visible curds. The changed pH affects three-dimensional organisation and association; identifying the proteins by name is not required. |
Protein hydrolysis is a different chemical change. Heating with aqueous acid or aqueous alkali breaks the peptide links, and complete hydrolysis gives amino acids. In acid, the amino groups are protonated: H3N+-CH(R)-COOH. In alkali, the acid groups become carboxylate ions: H2N-CH(R)-COO-, with suitable counter-ions. The carbon chains and side groups of the amino acids are recovered.
Worked exampleDistinguish two treatments of a protein
Treatment A briefly heats a protein in water until it aggregates. Treatment B heats it with aqueous acid until its peptide links are completely hydrolysed. Compare the products.
Try it first. Then show the worked stepsHide the worked steps
- In A, changed noncovalent interactions destroy the original fold and allow new associations. Long covalently bonded chains can remain.
- In B, C-N bonds of peptide links are cleaved by hydrolysis. The original long chain is broken into amino-acid-derived molecules.
- In the acidic final mixture, amino groups are mainly protonated, so show the appropriate positive charges when drawing products.
Denaturation describes loss of the functional fold; complete hydrolysis describes cleavage of the chain into its monomer-derived products.
Check your understanding
Why can adding acid break an ionic link between -NH3+ and -COO- groups in a protein?
Show the answer
The -COO- gains H+ to become -COOH, so its charge and the attraction are lost.
The ionic link is an attraction between opposite charges. Protonating the carboxylate removes the negative charge, so the shape held by that link can change.
Packing and crosslinks control how a plastic moves
Explain a material property from its chain structure.
Make a guess. You are not marked.
Poly(ethene) can be made with straight chains or with many branched chains. Which form is denser and stronger?
Show the answer
The straight-chain form, as its chains pack closely
Straight chains pack closely and line up, so there are more attractions between chains and less empty space.
A polymer has strong covalent bonds along its backbone. Between separate chains there are weaker intermolecular attractions. A thermoplastic, such as straight-chain poly(ethene), softens when heating overcomes enough of these attractions for the chains to slide past each other. Cooling makes it solid again. This reshaping does not break the backbone into monomers.
A thermosetting polymer, such as cross-linked poly(diallyl phthalate), has covalent bonds between its chains, which form a network. These cross-links stop the chains moving. Once it has set, heating cannot separate the chains into a liquid. Strong heating breaks down the network instead. So it cannot be recycled by simply melting it again, as a suitable thermoplastic can.
Separate chains can slide; a covalent network cannot
The thermoplastic schematic has separate chains connected only by dashed intermolecular attractions. The thermoset schematic has solid covalent links joining chains into a network. It illustrates the network principle rather than the atomic structural formula of poly(diallyl phthalate).
| Property | Linear thermoplastic: poly(ethene) | Cross-linked thermoset: poly(diallyl phthalate) |
|---|---|---|
| Softening and remoulding | Can soften and be reshaped within a suitable temperature range; potentially recyclable by remelting. | Does not soften into a reusable melt after curing; decomposition occurs on sufficiently severe heating. |
| Rigidity | Chains can move relative to one another more readily. Flexibility depends on packing, crystallinity and conditions. | Crosslinks restrict movement and usually give greater rigidity for comparable structures. |
| Strength | Resistance to deformation depends on how well chains pack and interact. | Covalent connections resist chains being pulled apart or sliding; a rigid network can nevertheless be brittle. Do not equate strength with ability to bend. |
LDPE and HDPE are both poly(ethene), but their chains are arranged differently. LDPE has more branches, which stop the chains packing closely. Fewer close contacts between chains give a lower density and a softer, more flexible material, useful for plastic bags. HDPE has fewer branches, so its chains pack more closely. Stronger attractions between chains make it stiffer and harder, useful for plastic bottles. HDPE does not have stronger C-C bonds.
Worked exampleInfer a property from a changed structure
Two samples have the same backbone chemistry and similar average chain length. Sample A has more side branches; sample B has longer unbranched stretches. Predict which is likely to pack more closely.
Try it first. Then show the worked stepsHide the worked steps
- Branches hinder chains from approaching in a regular arrangement. Sample A therefore tends to have less efficient packing.
- Sample B can form more closely packed regions and more effective interchain contacts.
- For these otherwise comparable samples, B is expected to be denser and stiffer; A is expected to be softer and more flexible. The stated comparison controls other factors such as molecular mass.
Use chain shape → packing → attractions → property. Do not infer every property from the polymer name alone.
Check your understanding
Why does being a thermoplastic not guarantee that an item will be recycled?
Show the answer
It can be remoulded, but collection, sorting, contamination and cost decide if it is.
Thermoplastic describes what the material can do when heated. Whether recycling happens also depends on collection, sorting, additives, damage and demand.
Choose the material for the job
Functional groups explain creasing, water response and chemical resistance.
A good comparison of materials is a chain of reasoning. Name the difference in structure. Explain how that changes the interactions or reactions. Then link this to the use. Writing only "it is strong" or "it is polar" does not explain why one polymer is the better choice.
| Polymer | Condensed repeat-unit representation | Feature to inspect |
|---|---|---|
| PET, poly(ethylene terephthalate) | [-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]n | Ester links in the backbone; the two substituted positions of the aromatic ring are opposite one another. |
| Nylon 6,6 | [-NH-(CH2)6-NH-C(=O)-(CH2)4-C(=O)-]n | Amide links provide N-H donors and carbonyl O acceptors for interchain hydrogen bonding. |
| PVA, poly(vinyl alcohol) | [-CH2-CH(OH)-]n | Many OH side groups can hydrogen-bond with water. |
| PVC, poly(vinyl chloride) | [-CH2-CHCl-]n | C-Cl bonds are polar, but the chain lacks the repeated OH groups that provide strong hydration in PVA. |
| PP, poly(propene) | [-CH2-CH(CH3)-]n | A hydrocarbon backbone and methyl side groups; no ester link to hydrolyse in aqueous alkali. |
PET fabric creases slightly less than nylon 6,6 in the syllabus comparison. Nylon has amide groups that form many hydrogen bonds between chains. When the fabric is bent, especially when damp, some of these bonds break and new ones form in the new position, which holds the crease in. PET has polar ester groups but no N-H or O-H groups along its chain to donate hydrogen bonds, so this effect is smaller. PET still has attractions between chains. Neither fabric is fully crease-proof, and how the fabric is made and treated also matters.
PVA can dissolve in water and is used in eye drops. Its many OH groups form hydrogen bonds with water, so the chains can become surrounded by water and spread out. PVC is water-resistant and is used in raincoats. Its C-Cl bonds are polar, but being polar is not enough to dissolve. It does not have PVA's many OH groups to hydrogen bond with water, so water cannot pull its chains apart. The chlorine is covalently bonded, so it is not like a chloride ion in solution.
Choose PP rather than PET for strongly alkaline cleaning solutions. Alkali can hydrolyse the ester links in PET, cutting its chains into shorter pieces and weakening the container. PP has no ester links to hydrolyse, and its hydrocarbon chain resists this attack. This choice is about chemical resistance. Stiffness alone does not decide it.
Worked exampleUse supplied structure to recommend a material
Polymer X contains many neutral -C(=O)-O- links in its backbone. Polymer Y has a -CH2-CH(CH3)- backbone. Which is the better candidate for a reusable container holding concentrated aqueous alkali, and what additional evidence would help?
Try it first. Then show the worked stepsHide the worked steps
- X contains ester links, which can undergo alkaline hydrolysis and cleave the chain.
- Y is a polyalkene-type hydrocarbon structure with no corresponding hydrolysable link, so it is the stronger chemical-resistance candidate.
- For a real selection, still compare supplied data on temperature resistance, mechanical strength and long-term compatibility. The functional-group argument identifies a major risk but does not establish every performance property.
Y is preferred on the stated chemical criterion. A defensible answer names the ester link and the consequence of backbone cleavage.
Check your understanding
PVA dissolves in water. Must its repeat unit contain ions?
Show the answer
No. Its many OH groups form hydrogen bonds with water.
Solubility needs attractions to water that make up for the attractions broken. Hydrogen bonds from OH groups can do this without ions.
Follow a plastic beyond its first use
Chemical resistance helps in service but can become a disposal problem.
Poly(alkenes) have strong C-C backbones with mostly C-H bonds. Being so unreactive makes them useful for containers and waterproof materials. But it also makes many of them hard to biodegrade, because water and living things cannot easily cut those chains into small molecules. Breaking into small pieces is not the same as biodegrading completely.
Polyesters and polyamides have ester or amide links that can be hydrolysed. So they usually biodegrade more easily than poly(alkenes). Cutting these links shortens the chains and can give small molecules that living things can use. A link that can be hydrolysed does not guarantee fast breakdown everywhere. Temperature, water, how crystalline the polymer is, the chain structure and suitable microorganisms all affect the rate. A long-lasting PET bottle and a fast-degrading polyester both contain ester links.
| Process | What changes | Limit to remember |
|---|---|---|
| Reuse | The same object serves again with little chemical change. | Suitability depends on its condition, cleaning and intended use. |
| Mechanical recycling | Material is collected, sorted and processed into new items, commonly by remelting suitable thermoplastics. | Mixed polymers, contamination and repeated degradation can reduce useful quality. |
| Chemical recycling or controlled hydrolysis | Chemical bonds are broken to recover smaller molecules or feedstock. | Reagents, separation and energy are still needed; a chemically possible route is not automatically economical. |
| Biodegradation | Biological activity contributes to conversion into simpler substances under suitable conditions. | Do not assume an item will biodegrade quickly in ordinary soil, seawater or a landfill merely from a broad material label. |
Materials and the resources used to make them are limited. Recycling can reduce the need for new raw materials and cut waste, but judge the whole process. A good comparison follows the material and the energy through collection, sorting, transport, cleaning, processing and its next use. Do not stop at the recycling symbol.
| Dimension | Benefits to examine | Costs or constraints to examine |
|---|---|---|
| Economic | Recover usable material; reduce purchase of some fresh feedstock; support useful collection and processing work. | Collection, transport, sorting, cleaning, processing energy and demand for the recovered material affect viability. |
| Environmental | Reduce extraction of resources, waste accumulation and potentially the energy or emissions of making new material. | Washing uses water; transport and processing use energy; rejected contaminated material still needs treatment. Compare like-for-like outcomes. |
| Social | Cleaner surroundings, access to durable useful products, and opportunities for responsible collection and employment. | Convenience and affordability influence participation; workers and communities share the burdens of collection and waste handling. Clear instructions and suitable infrastructure matter. |
Worked exampleMake a balanced recommendation from data
A supplied study says process A recovers 80% of a sorted plastic stream, while process B recovers 95% but uses substantially more energy. Is B automatically better?
Try it first. Then show the worked stepsHide the worked steps
- The higher recovery fraction is a benefit: more material is retained per initial batch.
- Compare the quantity and quality of usable product, energy source, total energy, transport, cost and the fate of residues. A percentage alone omits these consequences.
- Consider whether collection and sorting are practical for users and workers. State which evidence supports the recommendation and which missing data could change it.
No. The decision must combine economic, environmental and social factors for the same functional outcome.
Check your understanding
"PET has ester links, so it will quickly disappear if thrown away." Why is this not justified?
Show the answer
Ester links can be hydrolysed, but in tightly packed PET this is very slow.
A possible reaction is not a fast one. The rate depends on conditions and on water reaching the links, and a dense, durable bottle can last a long time.
Exam-style question
Paper 2 Section B styleThis question draws on Chemical Bonding and Polymers and Organic Chemistry.
Poly(ethene) is made from ethene, CH2=CH2. Nylon 6,6 is made from hexane-1,6-diamine, H2N(CH2)6NH2, and hexanedioic acid, HOOC(CH2)4COOH. Both are used to make fibres and packaging.
- (a) Describe the bonding in the carbon-carbon double bond of ethene in terms of orbital overlap. State the shape around each carbon atom and the H-C-H bond angle. [3]
- (b) Draw the repeat unit of poly(ethene) and name the type of polymerisation. [2]
- (c) Explain why poly(ethene) is hardly broken down in the environment. [1]
- (d) Draw the link that joins the monomers in nylon 6,6. Name the link, the type of polymerisation and the small molecule that is lost. [3]
- (e) Nylon 6,6 has a higher melting point than poly(ethene) with chains of a similar length. Explain this in terms of the forces between the chains. [3]
- (f) Nylon 6,6 can be broken down by heating with dilute hydrochloric acid. Name this type of reaction and give the two organic products. [3]
- (g) Poly(ethene) softens when heated and can be remoulded, but some polymers char instead of softening. Explain this difference in terms of structure. [2]
- (h) Give one advantage and two limitations of recycling poly(ethene) bags into new plastic. [3]
[20]
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Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Term | Definition to write |
|---|---|
| Homologous series | A family of compounds with the same general formula and functional group, similar chemical properties, and a gradual change in physical properties as the molecules get larger. Successive members differ by CH2. |
| Structural (constitutional) isomers | Compounds with the same molecular formula but different structural formulae. |
| Polymer | A macromolecule built up from many monomers, with an average relative molecular mass of at least 1000 or at least 100 repeat units. |
| Addition polymerisation | Monomers containing C=C bonds join together to form a polymer, with no other product formed. |
| Condensation polymerisation | Monomers with two functional groups join together to form a polymer, with a small molecule such as water lost at each link. |
| Thermoplastic | A polymer of linear chains held by intermolecular forces. It softens on heating, can be remoulded and recycled. |
| Thermosetting polymer | A polymer with covalent cross-links between chains. It does not soften on heating (it decomposes), so it cannot be remoulded; it is rigid. |
- Locate the group
Mark C=C, C-X, OH, C=O, COOH, COO or CONH.
- Read the conditions
Aqueous vs ethanolic NaOH, oxidant vs reductant, and acid vs alkali hydrolysis change the answer.
- Preserve the skeleton
Change only the reacting bonds; include small products, charge and counter-ions as needed.
- Check the result
Conserve atoms and charge, then name the new functional group and product.
| Pair | Deciding point |
|---|---|
| Constitutional / cis-trans | Different connections / same connections with restricted C=C arrangement. |
| Ester / amide | Carbonyl C attached to O / N. Acid + amine requires DCC for the specified amide route, not simple acid-base mixing. |
| Acid / alkali hydrolysis | Acid medium retains COOH and protonates amines; alkali gives COO- and unprotonated amine/ammonia products. |
| Denaturation / hydrolysis | Loss of the specific fold / cleavage of covalent peptide links. |
| Thermoplastic / thermoset | Separate chains can move on warming / covalent crosslinks prevent ordinary remelting. |
Named materials: LDPE branches pack less closely than HDPE. Nylon 6,6 can form more hydrogen bonds between chains than PET. PVA OH groups bond with water, while PVC resists water. PP resists the ester hydrolysis that can damage PET in strong alkali.
Poly(alkenes) are unreactive, so they biodegrade with difficulty. Ester and amide links can be hydrolysed, but real rates depend on conditions. Recycling decisions must weigh economic, environmental and social effects, not only how much is recovered.
Scope
Learning outcomes
9. Polymers and Organic Chemistry (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
9(a) Read and name the required organic families and formula representations.
- (i) Hydrocarbons: alkanes, alkenes and benzene.
- (ii) Halogenoalkanes.
- (iii) Alcohols including primary, secondary and tertiary.
- (iv) Aldehydes and ketones.
- (v) Carboxylic acids.
- (vi) Esters.
- (vii) Amines.
- (viii) Amides.
- (ix) Amino acids, including aminoethanoic acid.
- Nomenclature, general and structural/displayed formulae; empirical, molecular and skeletal representations from the preamble.
Read and name an organic structureLink oxidation level with acid-base behaviour
9(b) Use organic reaction terminology.
- (i) Functional group.
- (ii) Addition, substitution and elimination.
- (iii) Condensation and hydrolysis.
- (iv) Oxidation and reduction; [O] and [H] notation accepted.
Read and name an organic structureAn alkane substitutes; an alkene addsConditions choose the reaction routeLink oxidation level with acid-base behaviourMake and break ester or amide links
9(c) Describe constitutional isomerism.
- Same molecular formula; different connectivity.
9(d) Explain cis-trans isomerism in alkenes.
- Restricted rotation due to the pi bond; each C=C carbon must carry different substituents.
- E/Z nomenclature is not required.
9(e) Deduce isomers from a known molecular formula.
- Systematic skeleton and position changes; distinguish constitutional and cis-trans counts.
9(f) Explain organic molecular shapes through sigma and pi bonding.
- (i) Ethane, ethene and benzene shapes.
- (ii) Bond angles and sigma/pi carbon-carbon bonding; hybridisation is not required.
- (iii) Predict analogous shapes and angles.
9(g) Apply the specified organic reactions, conditions and analogous product predictions.
- (i) Ethane: general low reactivity, combustion, chlorine substitution under UV at room temperature.
- (ii) Ethene: combustion, Br2 in CCl4, H2/Ni addition.
- (iii) Bromoethane: aqueous NaOH/heat substitution; ethanolic NaOH/heat elimination.
- (iv) Ethanol: combustion; acidified K2Cr2O7 or KMnO4/heat oxidation to acid; concentrated H3PO4/heat elimination.
- (v) Ethanal and propanone reduction with LiAlH4 or H2/Ni; aldehyde oxidation with acidified K2Cr2O7 or KMnO4/heat.
- (vi) Ethanoic acid with alkalis and carbonates; alcohol ester formation with concentrated H2SO4; ethylamine amide formation with DCC (DCC structure not required).
- (vii) Ethyl ethanoate and ethanamide hydrolysis with aqueous acid or alkali and heat; correct protonation/salt products.
- (viii) Ethylamine with aqueous acid gives an ammonium salt.
- Preamble: essential reagents/conditions and major products; analogous product predictions; infer reasonable purification/extraction needs without detailed practical procedures.
An alkane substitutes; an alkene addsConditions choose the reaction routeLink oxidation level with acid-base behaviourMake and break ester or amide links
9(h) Recognise polymers as macromolecules built from monomers.
- Average relative molecular mass at least 1000 or at least 100 repeat units.
9(i) Distinguish addition and condensation polymers.
- Monomer functionality, bond formation, repeat units and presence or absence of small-molecule elimination.
9(j) Describe proteins as condensation polymers of alpha-amino acids.
- Peptide bonds are amide links; monomer groups and side chains remain correctly connected.
9(k) Describe protein hydrolysis.
- Aqueous acid or aqueous alkali and heat; peptide-bond cleavage and amino-acid-derived products.
9(l) Explain interactions stabilising three-dimensional protein structure.
- Hydrogen bonds, intermolecular forces and ionic linkages.
- Specific structural levels, alpha helices and beta pleated sheets are not required.
9(m) Explain protein denaturation.
- Temperature extremes and pH changes disrupt structure-stabilising interactions; link to enzyme behaviour in 7(i).
9(n) Apply protein shape changes to real phenomena.
- Heating egg white and adding vinegar to milk; loss and formation of three-dimensional arrangements.
- Names of the proteins are not required.
9(o) Compare thermoplastic and thermosetting polymers.
- Linear poly(ethene) and cross-linked poly(diallyl phthalate).
- (i) Softening and capacity for recycling.
- (ii) Rigidity.
- (iii) Strength; explain from structure and bonding.
9(p) Explain all named polymer structure-property-use comparisons.
- (i) LDPE bags: softer/flexible; HDPE bottles: harder/stiffer.
- (ii) PET polyester fabric slightly less prone to creasing than nylon 6,6 polyamide.
- (iii) PVA water-soluble in eye drops; PVC water-resistant in raincoats.
- (iv) PP rather than PET containers for strongly alkaline cleaner because PET hydrolyses.
Packing and crosslinks control how a plastic movesChoose the material for the job
9(q) Predict polymer physical properties from structure.
- Reason through branching, packing, crosslinks and available intermolecular interactions.
Packing and crosslinks control how a plastic movesChoose the material for the job
9(r) Relate polyalkene inertness to difficult biodegradation.
- Robust carbon backbone; connect to alkane reactivity in 9(g)(i).
9(s) Recognise hydrolysis as a biodegradation route for polyesters and polyamides.
- Ester and amide backbone cleavage; connect to 9(g)(vii); distinguish possible hydrolysis from its condition-dependent rate.
9(t) Evaluate recycling in the context of finite material resources.
- Economic, environmental and social factors.