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8873 / 2027
Polymers and Organic Chemistry

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)

01

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.

Five ways to describe lactic acid
RepresentationWhat it tells you
Empirical: CH2OThe simplest whole-number atom ratio, C:H:O = 1:2:1.
Molecular: C3H6O3The actual numbers of atoms in one molecule.
Structural: CH3CH(OH)CO2HThe middle carbon bears OH; the final carbon belongs to COOH. Brackets keep the connectivity unambiguous.
DisplayedEvery atom and bond is shown, including each C-H and O-H bond.
SkeletalCarbon 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.

COOH contains a C=O bond and a C-O-H group; it is not a carbon joined to 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.

Count a carbon at every unlabelled end or corner. Do not count the end labelled O or OH as another carbon.
Recognise the whole group, then name an example
Class and general structureExample and naming cue
Alkane: CnH2n+2 for an acyclic saturated hydrocarbonCH3CH3, ethane; suffix -ane.
Alkene: CnH2n for an acyclic hydrocarbon with one C=CCH2=CHCH3, propene; locate the C=C and use -ene.
Benzene: C6H6, a six-carbon aromatic ringUse 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 compoundCH3CH2Br, bromoethane; fluoro-, chloro-, bromo- or iodo- gives the substituent.
Alcohol: R-OH; CnH2n+2O for a saturated acyclic monohydric alcoholCH3CH(OH)CH3, propan-2-ol; number the OH position.
Aldehyde: R-CHO; CnH2nO for a saturated acyclic monoaldehydeCH3CHO, 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 monoketoneCH3COCH3, propanone; the carbonyl carbon is bonded to two carbon groups.
Carboxylic acid: R-COOH; CnH2nO2 for a saturated acyclic monocarboxylic acidCH3COOH, ethanoic acid; the COOH carbon is carbon 1.
Ester: R-COO-R'; CnH2nO2 for a saturated acyclic monoesterCH3COOCH2CH3, 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 groupsCH3CH2NH2, 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)-COOHH2NCH2COOH, 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.

Primary, secondary and tertiary classify the carbon carrying OH
Alcohol classCarbon groups attached to the OH-bearing carbonExample
PrimaryOne carbon group; methanol is conventionally treated with primary alcohols.CH3CH2OH
SecondaryTwo carbon groups.CH3CH(OH)CH3
TertiaryThree carbon groups.(CH3)3COH

Check your understanding

What is the name of CH3CH2COOCH3?

  1. Methyl propanoate
  2. Propyl methanoate
  3. Methyl ethanoate
Show the answer

Methyl propanoate

The CH3 on the oxygen gives methyl. CH3CH2COO- has three carbons including the carbonyl carbon, so it is propanoate.

02

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
  1. 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.
  2. Use the branched three-carbon skeleton with a methyl branch. OH can be on an end carbon: 2-methylpropan-1-ol, (CH3)2CHCH2OH.
  3. Or put OH on the central carbon: 2-methylpropan-2-ol, (CH3)3COH.
  4. 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.
Answer

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
  1. 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.
  2. 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.
  3. 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.
  4. 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.
  5. Methyl ethanoate gives methanol and ethanoate on alkaline hydrolysis. Ethyl methanoate gives ethanol and methanoate; propanoic acid undergoes neutralisation instead.
Answer

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.

Cis and trans forms share the same molecular formula and connectivity. They are not two more constitutional isomers.

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
  1. 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.
  2. But-2-ene has H and CH3 on each double-bonded carbon, so both cis and trans forms exist.
  3. But-1-ene and 2-methylpropene each contain a CH2 end of the double bond. Its two identical H substituents prevent cis-trans isomerism.
Answer

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?

  1. CH2=CClBr only, because one carbon carries two different halogens
  2. Both of them, because each has a C=C bond that cannot rotate
  3. CHCl=CHBr only, because each C=C carbon has two different groups
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.

03

Read shape from sigma and pi bonding

Describe the local geometry around a carbon before the whole molecule.

Three structures that establish the pattern
MoleculeCarbon bonding and local shapeAngles and freedom of movement
Ethane, CH3CH3Each 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=CH2Each 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, C6H6Each 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.

The ring and its six C-H sigma bonds lie in one plane. The delocalised pi electron density lies above and below that plane.

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.

Try it first. Then show the worked stepsHide the worked steps
  1. The two double-bonded carbons each have three sigma-bond directions. Their local arrangements are trigonal planar with angles near 120°.
  2. The CH3 carbon has four sigma-bond directions, so it is tetrahedral with angles near 109.5°.
  3. 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.
Answer

The alkene region is locally planar; the methyl carbon is tetrahedral.

Check your understanding

What shape is expected around the carbonyl carbon in CH3CHO?

  1. Trigonal planar, about 120 degrees, with one sigma and one pi bond in C=O
  2. Tetrahedral, about 109.5 degrees, as the carbon forms four bonds
  3. Linear, 180 degrees, as the C=O double bond pulls the groups apart
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.

04

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?

  1. Only the hexene decolourises it, by addition
  2. Both decolourise it, as both are hydrocarbons
  3. Neither changes the orange colour in the dark
Show the answer

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.

The specified reactions of ethane and ethene
Starting compound and conditionsEquationWhat changes
Ethane; oxygen, ignition2C2H6 + 7O2 → 4CO2 + 6H2OComplete 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 temperatureC2H6 + Cl2 → C2H5Cl + HClSubstitution replaces H by Cl. Further substitutions can occur, so one equation does not promise a single pure product.
Ethene; oxygen, ignitionC2H4 + 3O2 → 2CO2 + 2H2OComplete combustion.
Ethene; Br2 in CCl4CH2=CH2 + Br2 → CH2BrCH2BrAddition across C=C gives 1,2-dibromoethane. Bromine colour is discharged; no HBr is a product of this addition.
Ethene; H2, Ni catalyst, heatCH2=CH2 + H2 → CH3CH3Hydrogenation 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
  1. Locate the C=C between carbon 2 and carbon 3: CH3CH=CHCH3.
  2. Change that C=C to C-C and attach one Br to each of those carbons.
  3. The product is CH3CHBrCHBrCH3, 2,3-dibromobutane. No carbon is added or removed, and both Br atoms appear in the product.
Answer

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?

  1. It shows the compound is an alkane, since alkanes react quickly with bromine.
  2. Many alkenes, and some other compounds, decolourise bromine, so more evidence is needed.
  3. It proves the compound is ethene, since ethene is the alkene that reacts.
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.

05

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?

  1. Ethene, by elimination of HBr
  2. No reaction, as ethanol is the solvent
  3. Ethanol, by substitution of Br
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 map from bromoethane and ethanol
RouteReagents and essential conditionsBalanced change
Bromoethane → ethanolAqueous NaOH; heat, usually under reflux.CH3CH2Br + NaOH → CH3CH2OH + NaBr
Bromoethane → etheneNaOH in ethanol; heat.CH3CH2Br + NaOH → CH2=CH2 + NaBr + H2O
Ethanol → etheneConcentrated H3PO4 catalyst; heat.CH3CH2OH → CH2=CH2 + H2O
Ethanol → ethanoic acidAcidified K2Cr2O7 or acidified KMnO4; heat under reflux.CH3CH2OH + 2[O] → CH3COOH + H2O
Ethanol complete combustionOxygen; 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
  1. CH3CH2CH2Br → CH3CH2CH2OH using aqueous NaOH and heat: substitution replaces Br by OH.
  2. Heat the primary alcohol with acidified K2Cr2O7 or acidified KMnO4 under reflux to form CH3CH2COOH.
  3. Check that the carbon skeleton remains three carbons in both steps. Ethanolic NaOH would instead divert the first step towards propene.
Answer

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?

  1. Two ethanol molecules join together, giving ethene and a water molecule.
  2. An H atom replaces the OH group on the carbon, giving a new compound.
  3. Water is removed (OH from one carbon, H from the next) and a C=C forms.
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.

06

Link oxidation level with acid-base behaviour

Carbonyl chemistry changes C=O; acid-base chemistry transfers protons.

Aldehydes and ketones share reduction but differ in ordinary oxidation
Compound and changeReagent and conditionsEquation
Ethanal → ethanolLiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat.CH3CHO + 2[H] → CH3CH2OH
Propanone → propan-2-olLiAlH4, followed by aqueous work-up; or H2 with Ni catalyst and heat.CH3COCH3 + 2[H] → CH3CH(OH)CH3
Ethanal → ethanoic acidAcidified 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.

Acid and base reactions retain the organic skeleton
ReactionWhat the equation explains
CH3COOH + NaOH → CH3COONa + H2OOne mole of ethanoic acid neutralises one mole of hydroxide.
2CH3COOH + Na2CO3 → 2CH3COONa + H2O + CO2Two 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
  1. The starting carbonyl carbon is attached to CH3 and CH2CH3, so it is a ketone.
  2. Replace C=O by CH-OH while preserving both attached carbon groups: CH3CH(OH)CH2CH3.
  3. The product is butan-2-ol, a secondary alcohol. Both oxygen and all four carbons remain in the molecule.
Answer

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?

  1. On the carbon next to nitrogen, which takes the H+ from the acid.
  2. On the chloride ion, which takes the extra electrons from nitrogen.
  3. On nitrogen, as its lone pair bonds to H+ to give CH3CH2NH3+.
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.

07

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.

Two ways ethanoic acid makes a new covalent link
Link formedReagents and conditionsOrganic 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.

Track fragments rather than splitting a name. The ethyl group is attached through oxygen in the ester and through nitrogen in the amide.
Ethyl ethanoate hydrolysis: the medium changes the acid product
ConditionsEquationWhy this form appears
Aqueous acid and heatCH3COOCH2CH3 + H2O ⇌ CH3COOH + CH3CH2OHThe acidic mixture contains the carboxylic acid. Acid catalyses the reaction.
Aqueous NaOH and heatCH3COOCH2CH3 + NaOH → CH3COONa + CH3CH2OHAny acid formed is deprotonated by the alkali. The isolated acid-derived product in this medium is the carboxylate salt.
Ethanamide hydrolysis: include the nitrogen product
ConditionsNet ionic equationProduct check
Aqueous acid and heatCH3CONH2 + H2O + H+ → CH3COOH + NH4+The nitrogen product is protonated in acid; write ammonium, not free ammonia as the final major form.
Aqueous alkali and heatCH3CONH2 + OH- → CH3COO- + NH3The 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
  1. Identify the acid-derived fragment CH3CH2CO- and the O-attached methyl fragment.
  2. Breaking the ester link gives a propanoate ion and methanol in the alkaline mixture.
  3. Include Na+ as counter-ion: CH3CH2COOCH3 + NaOH → CH3CH2COONa + CH3OH. All four carbons are accounted for.
Answer

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?

  1. Ethanoic acid and ethanol, CH3CH2OH
  2. Ethanoic acid and ethylammonium ions, CH3CH2NH3+
  3. Ethanamide and ethane, CH3CONH2 and CH3CH3
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.

08

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.

Addition and condensation describe how chains form
FeatureAddition polymerisationCondensation polymerisation
Starting structuresTypically 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 changeThe C=C pi bond is replaced by single bonds connecting monomer units.New links form between functional groups, commonly ester or amide links.
Small productNo small molecule is eliminated in forming the addition chain.A small molecule is eliminated per link in the examples here, usually water.
Repeat unitContains 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.

Converting C=C to a backbone does not remove CH3 or move it into the main chain. Brackets and n identify the repeat unit.

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.

Try it first. Then show the worked stepsHide the worked steps
  1. Remove OH from the left molecule's COOH and H from the right molecule's NH2, accounting for H2O overall.
  2. Join the left carbonyl carbon to the right nitrogen: H2NCH2CONHCH(CH3)COOH.
  3. 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.
Answer

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?

  1. CH3-CH2Cl, chloroethane
  2. CHCl=CHCl, dichloroethene
  3. CH2=CHCl, chloroethene
Show the answer

CH2=CHCl, chloroethene

Restore the C=C between the two backbone carbons. Chlorine stays on its carbon. The polymer is PVC.

09

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?

  1. They stay intact; weaker interactions break
  2. They break, giving free amino acids
  3. They become stronger as the chain unfolds
Show the answer

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.

Interactions that help stabilise protein shape
InteractionStructural originHow conditions can disturb it
Hydrogen bondingAttractions 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 attractionsPermanent 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 linkagesAttraction 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.

Denaturation changes the fold. Hydrolysis instead breaks peptide bonds and shortens the covalent 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.

Use the shape argument in familiar observations
ObservationExplanation
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
  1. In A, changed noncovalent interactions destroy the original fold and allow new associations. Long covalently bonded chains can remain.
  2. In B, C-N bonds of peptide links are cleaved by hydrolysis. The original long chain is broken into amino-acid-derived molecules.
  3. In the acidic final mixture, amino groups are mainly protonated, so show the appropriate positive charges when drawing products.
Answer

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?

  1. The acid hydrolyses the peptide bonds, which then pull the two charged groups apart.
  2. The -COO- gains H+ to become -COOH, so its charge and the attraction are lost.
  3. The acid adds to the -NH3+ group, making it -NH4 2+, which repels the -COO-.
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.

10

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?

  1. The branched form, as branches tangle and hold chains together
  2. The branched form, because its chains are longer
  3. The straight-chain form, as its chains pack closely
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).

Softening overcomes attractions between suitable separate chains. A cross-linked network must suffer chemical damage before it can flow in that way.
Describe the consequence, not just the category
PropertyLinear thermoplastic: poly(ethene)Cross-linked thermoset: poly(diallyl phthalate)
Softening and remouldingCan 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.
RigidityChains 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.
StrengthResistance 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
  1. Branches hinder chains from approaching in a regular arrangement. Sample A therefore tends to have less efficient packing.
  2. Sample B can form more closely packed regions and more effective interchain contacts.
  3. 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.
Answer

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?

  1. It can be remoulded, but collection, sorting, contamination and cost decide if it is.
  2. Thermoplastics break down when heated, so they cannot be remoulded into new items.
  3. Thermoplastics are cross-linked, so they cannot be melted down in a recycling plant.
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.

11

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.

Recognise the important links and side groups
PolymerCondensed repeat-unit representationFeature to inspect
PET, poly(ethylene terephthalate)[-O-CH2-CH2-O-C(=O)-C6H4-C(=O)-]nEster 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)-]nAmide links provide N-H donors and carbonyl O acceptors for interchain hydrogen bonding.
PVA, poly(vinyl alcohol)[-CH2-CH(OH)-]nMany OH side groups can hydrogen-bond with water.
PVC, poly(vinyl chloride)[-CH2-CHCl-]nC-Cl bonds are polar, but the chain lacks the repeated OH groups that provide strong hydration in PVA.
PP, poly(propene)[-CH2-CH(CH3)-]nA 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
  1. X contains ester links, which can undergo alkaline hydrolysis and cleave the chain.
  2. Y is a polyalkene-type hydrocarbon structure with no corresponding hydrolysable link, so it is the stronger chemical-resistance candidate.
  3. 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.
Answer

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?

  1. No. Its many OH groups form hydrogen bonds with water.
  2. Yes. Covalent compounds do not dissolve in water.
  3. No. PVA is a protein, so it dissolves by forming peptide links with water.
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.

12

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.

Separate four different end-of-life ideas
ProcessWhat changesLimit to remember
ReuseThe same object serves again with little chemical change.Suitability depends on its condition, cleaning and intended use.
Mechanical recyclingMaterial 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 hydrolysisChemical bonds are broken to recover smaller molecules or feedstock.Reagents, separation and energy are still needed; a chemically possible route is not automatically economical.
BiodegradationBiological 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.

Three dimensions of a recycling decision
DimensionBenefits to examineCosts or constraints to examine
EconomicRecover 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.
EnvironmentalReduce 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.
SocialCleaner 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
  1. The higher recovery fraction is a benefit: more material is retained per initial batch.
  2. 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.
  3. Consider whether collection and sorting are practical for users and workers. State which evidence supports the recommendation and which missing data could change it.
Answer

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?

  1. Ester links cannot be hydrolysed, so PET lasts forever in any conditions.
  2. PET has no ester links; it is an addition polymer with a carbon backbone.
  3. Ester links can be hydrolysed, but in tightly packed PET this is very slow.
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 style

This 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.

  1. (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]
  2. (b) Draw the repeat unit of poly(ethene) and name the type of polymerisation. [2]
  3. (c) Explain why poly(ethene) is hardly broken down in the environment. [1]
  4. (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]
  5. (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]
  6. (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]
  7. (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]
  8. (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.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Homologous seriesA 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) isomersCompounds with the same molecular formula but different structural formulae.
PolymerA macromolecule built up from many monomers, with an average relative molecular mass of at least 1000 or at least 100 repeat units.
Addition polymerisationMonomers containing C=C bonds join together to form a polymer, with no other product formed.
Condensation polymerisationMonomers with two functional groups join together to form a polymer, with a small molecule such as water lost at each link.
ThermoplasticA polymer of linear chains held by intermolecular forces. It softens on heating, can be remoulded and recycled.
Thermosetting polymerA polymer with covalent cross-links between chains. It does not soften on heating (it decomposes), so it cannot be remoulded; it is rigid.
Read a reaction before naming its product
  1. Locate the group

    Mark C=C, C-X, OH, C=O, COOH, COO or CONH.

  2. Read the conditions

    Aqueous vs ethanolic NaOH, oxidant vs reductant, and acid vs alkali hydrolysis change the answer.

  3. Preserve the skeleton

    Change only the reacting bonds; include small products, charge and counter-ions as needed.

  4. Check the result

    Conserve atoms and charge, then name the new functional group and product.

High-value distinctions
PairDeciding point
Constitutional / cis-transDifferent connections / same connections with restricted C=C arrangement.
Ester / amideCarbonyl C attached to O / N. Acid + amine requires DCC for the specified amide route, not simple acid-base mixing.
Acid / alkali hydrolysisAcid medium retains COOH and protonates amines; alkali gives COO- and unprotonated amine/ammonia products.
Denaturation / hydrolysisLoss of the specific fold / cleavage of covalent peptide links.
Thermoplastic / thermosetSeparate 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
  1. 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

  2. 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

  3. 9(c) Describe constitutional isomerism.

    • Same molecular formula; different connectivity.

    Change connectivity or fixed spatial arrangement

  4. 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.

    Change connectivity or fixed spatial arrangement

  5. 9(e) Deduce isomers from a known molecular formula.

    • Systematic skeleton and position changes; distinguish constitutional and cis-trans counts.

    Change connectivity or fixed spatial arrangement

  6. 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.

    Read shape from sigma and pi bonding

  7. 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

  8. 9(h) Recognise polymers as macromolecules built from monomers.

    • Average relative molecular mass at least 1000 or at least 100 repeat units.

    Turn monomers into chains

  9. 9(i) Distinguish addition and condensation polymers.

    • Monomer functionality, bond formation, repeat units and presence or absence of small-molecule elimination.

    Turn monomers into chains

  10. 9(j) Describe proteins as condensation polymers of alpha-amino acids.

    • Peptide bonds are amide links; monomer groups and side chains remain correctly connected.

    Turn monomers into chains

  11. 9(k) Describe protein hydrolysis.

    • Aqueous acid or aqueous alkali and heat; peptide-bond cleavage and amino-acid-derived products.

    Change a protein shape without cutting its backbone

  12. 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.

    Change a protein shape without cutting its backbone

  13. 9(m) Explain protein denaturation.

    • Temperature extremes and pH changes disrupt structure-stabilising interactions; link to enzyme behaviour in 7(i).

    Change a protein shape without cutting its backbone

  14. 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.

    Change a protein shape without cutting its backbone

  15. 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.

    Packing and crosslinks control how a plastic moves

  16. 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

  17. 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

  18. 9(r) Relate polyalkene inertness to difficult biodegradation.

    • Robust carbon backbone; connect to alkane reactivity in 9(g)(i).

    Follow a plastic beyond its first use

  19. 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.

    Follow a plastic beyond its first use

  20. 9(t) Evaluate recycling in the context of finite material resources.

    • Economic, environmental and social factors.

    Follow a plastic beyond its first use