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Organic Chemistry

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Organic Chemistry

Recognise molecular families and connect structures, reactions, fuels and materials.

O-Level 6092 (2026) / SEC G3 K324 (2027)

01

Fuels, fractions and bioethanol

Follow carbon from a resource to useful products.

A hydrocarbon contains carbon and hydrogen only. Natural gas is mainly methane; crude oil is a mixture of hydrocarbons. Both are non-renewable because they form far more slowly than they are consumed. Burning a fuel releases energy, while the same hydrocarbon feedstock may also be valuable for manufacturing chemicals and plastics.

Fractionate crude oil
  1. Heat the oil

    Many components vaporise.

  2. Vapour rises through the column

    The column is hotter at the bottom and cooler at the top.

  3. Condense different fractions

    Higher-boiling, generally larger molecules condense lower down; smaller, more volatile molecules travel higher.

  4. Collect useful mixtures

    Each fraction contains hydrocarbons with a range of boiling points, not necessarily one pure compound.

As hydrocarbon molecules get larger, melting and boiling points generally rise overall, viscosity increases, and volatility decreases. Smaller fractions are useful fuels; larger fractions may be cracked into more sought-after smaller molecules. The amount supplied by fractional distillation need not match the demand for each fraction. Larger molecules generally have stronger intermolecular attractions, so more energy is needed to separate them and the liquid resists flow more. These are trends within comparable members, not a claim that every individual melting point rises smoothly.

Bioethanol can be made from sugars in a crop such as sugarcane and is renewable if the crop is regrown. Growing plants take in carbon dioxide, partly offsetting carbon dioxide released when the ethanol burns. Fossil fuels instead add carbon stored underground to the active carbon cycle. Biofuel is not automatically emission-free: cultivation, processing, transport and land-use change also affect its impact.

Check your understandingWhy is a crude-oil fraction not necessarily a pure substance?Think it through, then reveal the answer
Several hydrocarbons with similar boiling points can condense in the same temperature range. Fractional distillation groups them by volatility rather than separating every individual compound.
02

Alkanes: build a valid carbon skeleton

Use four bonds per carbon and one per hydrogen.

A homologous series has a common general formula, similar chemical properties and a gradual change in physical properties as molecule size and mass increase. Successive members differ by CH2. Alkanes are saturated hydrocarbons: all carbon-carbon bonds are single. For an unbranched or branched acyclic alkane, the general formula is CnH2n+2.

Shared alkane structures
NameMolecular formulaCondensed structure
MethaneCH4One C with four C-H bonds
EthaneC2H6CH3-CH3
PropaneC3H8CH3-CH2-CH3

To draw a displayed formula, draw the carbon-carbon connections first, then add individual C-H bonds until each carbon has four bonds. A condensed CH3 group stands for three separate hydrogens bonded to that carbon. Count atoms after drawing; moving the page or bending a chain does not create a new compound.

Four bonds around carbon

Methane has four single C-H bonds. Each ethene carbon has two C-H bonds and a double C=C bond, giving four bonds in total.

A double bond counts twice when checking carbon valency; flat drawings do not show molecular shape.

Displayed methane

Complete displayed structure with 1 carbon, 4 hydrogen atoms: CH4. Every bond and hydrogen is shown.

CH4. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed ethane

Complete displayed structure with 2 carbon, 6 hydrogen atoms: C2H6. Every bond and hydrogen is shown.

C2H6. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed propane

Complete displayed structure with 3 carbon, 8 hydrogen atoms: C3H8. Every bond and hydrogen is shown.

C3H8. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed butane

Complete displayed structure with 4 carbon, 10 hydrogen atoms: C4H10. Every bond and hydrogen is shown.

C4H10. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed branched C4 alkane

Complete displayed structure with 4 carbon, 10 hydrogen atoms: C4H10. Every bond and hydrogen is shown.

C4H10, just as in butane. The central carbon has three C-C bonds and one C-H bond; each outside carbon has three hydrogens.
Pure only
Pure: the two C4 alkane connectivities
StructureCondensed representation
Unbranched butaneCH3-CH2-CH2-CH3
Branched C4H10CH3-CH(CH3)-CH3
Pure only

Isomers have the same molecular formula but different structural formulae: their atoms are connected differently. These two C4H10 structures are isomers. A branch changes connectivity; simply drawing unbranched butane as a bent line does not.

Alkanes are generally unreactive, but burn and undergo substitution with chlorine in ultraviolet light. Complete combustion: CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l), showing cooled products. Limited oxygen can produce carbon monoxide and/or soot. Substitution: CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g), under UV. One H is replaced by Cl; further substitution is possible.

Check your understandingA saturated acyclic hydrocarbon contains three carbon atoms. What formula should it have, and how do you check a drawing?Think it through, then reveal the answer
C3H8, from CnH2n+2. Draw three connected carbons and fill remaining valencies with H; each C must have four bonds and the total hydrogen count must be eight.
03

Alkenes: a double bond creates new reactions

Distinguish substitution, addition and cracking.

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond. For acyclic molecules with one C=C bond, their general formula is CnH2n. Ethene is CH2=CH2; propene is CH2=CH-CH3. Each carbon still has four bonds in total.

Pure only
Pure: C4 alkene structures
ConnectivityCondensed structure
But-1-eneCH2=CH-CH2-CH3
But-2-eneCH3-CH=CH-CH3
Branched C4H8CH2=C(CH3)2

Worked example

Expand an alkene without adding too many hydrogens

Draw propene from CH2=CH-CH3.

  1. Join three carbon atoms. Put the double bond between the first two; this uses two of each participating carbon's four bonds.
  2. Carbon 1 needs two C-H bonds. Carbon 2 already has three bonds in total to carbons, so needs only one H. Carbon 3 needs three H.
  3. Count the result: C3H6, with four bonds at each carbon.
Answer

Use the displayed propene below as a check. A double bond counts twice for valency but still connects the same two carbon atoms.

Displayed ethene

Complete displayed structure with 2 carbon, 4 hydrogen atoms: C2H4. Every bond and hydrogen is shown.

C2H4. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed propene

Complete displayed structure with 3 carbon, 6 hydrogen atoms: C3H6. Every bond and hydrogen is shown.

C3H6. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed but-1-ene

Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.

C4H8. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed but-2-ene

Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.

C4H8. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed branched C4 alkene

Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.

CH2=C(CH3)2. The central carbon has a double bond and two single bonds, so it has no hydrogen.

Aqueous bromine distinguishes an alkene from an alkane under the usual test conditions without UV. The alkene decolourises orange/brown bromine water because bromine adds across C=C. Ethene forms CH2Br-CH2Br. The alkane does not show this addition reaction. The colour change is an observation; unsaturation is the inference.

Ethene reaction pathways
Reagent or processEssential conditionsProduct/change
Oxygen: complete combustionIgnition, sufficient oxygenC2H4 + 3O2 -> 2CO2 + 2H2O
Aqueous bromineRoom conditions; no UV requiredC2H4 + Br2 -> C2H4Br2
HydrogenNickel catalyst, heatC2H4 + H2 -> C2H6
Addition polymerisationSuitable temperature, pressure and catalystMany ethene molecules form poly(ethene)
Pure only

Pure: steam adds across C=C with an acid catalyst such as phosphoric acid, at elevated temperature and pressure: C2H4(g) + H2O(g) -> C2H5OH(g). The ethanol is condensed after the reaction. Exact numerical temperatures and pressures are not required.

Cracking breaks larger hydrocarbons into smaller molecules using strong heating, often with a catalyst. Products can include smaller alkanes, alkenes and hydrogen. Examples of possible balanced changes are C10H22 -> C8H18 + C2H4, and C2H6 -> C2H4 + H2. Actual cracking produces a mixture; use atom conservation to deduce a missing product.

Polyunsaturated food molecules contain more than one C=C bond. Hydrogenating unsaturated vegetable oils with hydrogen and a nickel catalyst reduces the number of double bonds, making the product more solid and useful in margarine manufacture. This is addition, not polymerisation.

Check your understandingC8H18 cracks to C6H14 and one other molecule. Deduce its formula.Think it through, then reveal the answer
Subtract atoms: two C and four H remain, so C2H4, ethene. Check both carbon and hydrogen totals rather than guessing from the word cracking.
04

Alcohols: identify -OH and follow ethanol

Structure connects preparation, combustion and oxidation.

Alcohols form a homologous series containing the -OH functional group. The group is covalently attached within a molecule; it is not a free hydroxide ion, so the presence of -OH does not make ethanol an alkali.

Unbranched C1-C3 alcohol structures
NameCondensed structure
MethanolCH3-OH
EthanolCH3-CH2-OH
Propan-1-olCH3-CH2-CH2-OH
Propan-2-olCH3-CH(OH)-CH3
Pure only
Pure: C4 alcohol connectivities
StructureCondensed representation
Butan-1-olCH3-CH2-CH2-CH2-OH
Butan-2-olCH3-CH(OH)-CH2-CH3
Branched, terminal -OH(CH3)2CH-CH2-OH
Branched, central -OH(CH3)3C-OH

Expand condensed structures by showing each bond to H separately and checking C has four bonds and O two. Moving -OH along the carbon chain changes the structure, even when the carbon skeleton remains unbranched.

Displayed methanol

Complete displayed structure with 1 carbon, 4 hydrogen and 1 oxygen atoms: CH4O. Every bond and hydrogen is shown.

CH4O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed ethanol

Complete displayed structure with 2 carbon, 6 hydrogen and 1 oxygen atoms: C2H6O. Every bond and hydrogen is shown.

C2H6O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed propan-1-ol

Complete displayed structure with 3 carbon, 8 hydrogen and 1 oxygen atoms: C3H8O. Every bond and hydrogen is shown.

C3H8O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.

Displayed propan-2-ol

Complete displayed structure with 3 carbon, 8 hydrogen and 1 oxygen atoms: C3H8O. Every bond and hydrogen is shown.

C3H8O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed butan-1-ol

Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.

C4H10O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed butan-2-ol

Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.

C4H10O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed branched C4 alcohol: terminal OH

Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.

C4H10O. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed branched C4 alcohol: central OH

Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.

(CH3)3C-OH. Three methyl groups and oxygen use all four bonds of the central carbon; do not add a hydrogen to that carbon.
Make ethanol by fermentation
  1. Feedstock

    Glucose solution, obtainable from sugarcane sugars.

  2. Conditions

    Yeast supplies enzymes; keep warm and exclude oxygen. Excessive heat damages enzymes.

  3. Reaction

    C6H12O6(aq) -> 2C2H5OH(aq) + 2CO2(g).

  4. Recovery

    Fractional distillation separates ethanol-rich liquid from the fermentation mixture.

Ethanol burns in sufficient oxygen: C2H5OH(l) + 3O2(g) -> 2CO2(g) + 3H2O(l), after cooling. It can also be oxidised to ethanoic acid by atmospheric oxygen, or by warming with acidified potassium manganate(VII). In shorthand, CH3CH2OH + 2[O] -> CH3COOH + H2O; [O] denotes oxygen supplied by an oxidising agent.

Use ethanol as the representative alcohol-to-acid conversion here. Do not assume that every possible alcohol structure gives a carboxylic acid: the position of -OH matters, and detailed alternative oxidation pathways are beyond these Secondary outcomes.

Pure only

Pure: ethanol can alternatively be made by catalytic addition of steam to ethene under the conditions given on the alkene page. Fermentation uses renewable plant sugars but gives a dilute mixture; the ethene route can operate continuously with a more concentrated product but commonly uses a fossil-derived feedstock.

Check your understandingWhy does fermentation exclude oxygen while oxidation to ethanoic acid involves an oxidising agent?Think it through, then reveal the answer
They are different conversions. Fermentation uses glucose to make ethanol under oxygen-limited conditions; oxidation changes ethanol into ethanoic acid by supplying oxygen through air or another oxidant.
05

Carboxylic acids: recognise -COOH

A functional group gives a family its characteristic chemistry.

Carboxylic acids contain the -COOH group, also written -CO2H. This includes a C=O bond and an O-H bond on the same carbon. Ethanoic acid is CH3COOH; it can form when ethanol is oxidised by oxygen in air or by warm acidified potassium manganate(VII). The carbon skeleton is retained in this conversion.

Pure only
Pure: C1-C4 acid structures
Name or connectivityCondensed formula
Methanoic acidH-C(=O)-OH
Ethanoic acidCH3-C(=O)-OH
Propanoic acidCH3-CH2-C(=O)-OH
Butanoic acidCH3-CH2-CH2-C(=O)-OH
Branched C4 acid(CH3)2CH-C(=O)-OH
Pure only

Worked example

Draw the whole carboxyl group

Draw propanoic acid and account for all three carbons.

  1. Draw a three-carbon chain. The end carbon is part of -COOH; do not add a fourth carbon for this group.
  2. Attach one oxygen by a double bond and a second oxygen by a single bond to that end carbon. Attach H to the singly bonded oxygen.
  3. Complete the remaining C-H bonds: the chain is CH3-CH2-C(=O)-OH. The carboxyl carbon already has four bonds and takes no H.
Answer

C3H6O2. In methanoic acid the carboxyl carbon instead bonds to H because there is no carbon chain attached.

Pure only

Displayed methanoic acid

Complete displayed structure with 1 carbon, 2 hydrogen and 2 oxygen atoms: CH2O2. Every bond and hydrogen is shown.

CH2O2. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed ethanoic acid

Complete displayed structure with 2 carbon, 4 hydrogen and 2 oxygen atoms: C2H4O2. Every bond and hydrogen is shown.

C2H4O2. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed propanoic acid

Complete displayed structure with 3 carbon, 6 hydrogen and 2 oxygen atoms: C3H6O2. Every bond and hydrogen is shown.

C3H6O2. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed butanoic acid

Complete displayed structure with 4 carbon, 8 hydrogen and 2 oxygen atoms: C4H8O2. Every bond and hydrogen is shown.

C4H8O2. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Displayed branched C4 carboxylic acid

Complete displayed structure with 4 carbon, 8 hydrogen and 2 oxygen atoms: C4H8O2. Every bond and hydrogen is shown.

C4H8O2. Count four bonds at every carbon, two at oxygen and one at hydrogen. The flat layout shows connectivity, not molecular shape.
Pure only

Pure: carboxylic acids are weak acids, partly ionising in water. They still react with suitable metals, bases and carbonates. Weak means partly ionised, not incapable of reaction. The salts of ethanoic acid contain CH3COO-, the ethanoate ion.

Pure only
Pure: characteristic ethanoic-acid reactions
PartnerBalanced reaction
Sodium hydroxideCH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)
Sodium carbonate2CH3COOH(aq) + Na2CO3(aq) -> 2CH3COONa(aq) + H2O(l) + CO2(g)
Magnesium2CH3COOH(aq) + Mg(s) -> (CH3COO)2Mg(aq) + H2(g)
Check your understandingWhat distinguishes the functional group in ethanol from the one in ethanoic acid?Think it through, then reveal the answer
Ethanol contains an alcohol -OH attached to a carbon chain. Ethanoic acid contains -C(=O)-OH: its -OH is part of the carboxyl group. Recognise the whole group, not just the letters OH.
06Pure only

Esters: join two fragments and keep track of their origins

The name has an alcohol part followed by an acid part.

Warm a carboxylic acid with an alcohol and a little concentrated sulfuric acid catalyst to form an ester and water. The reaction is reversible. Ethanoic acid and ethanol give ethyl ethanoate: CH3COOH + HOCH2CH3 reversibly form CH3COOCH2CH3 + H2O. The ester linkage is -C(=O)-O-.

Build an ester name
Starting familyC1C2C3C4
Unbranched acid partMethanoateEthanoatePropanoateButanoate
Alcohol part, -OH at chain endMethylEthylPropylButyl

The acid contributes the part ending in -oate, including its C=O carbon. The alcohol contributes the group attached through the ester oxygen. For an alcohol with -OH on a middle carbon, retain that attachment position when drawing the product; a numbered group such as propan-2-yl may be used when naming it.

Worked example

Name and reverse an ester

Identify the acid and alcohol that form CH3CH2COOCH3.

  1. Split at the bond between the single-bonded O and the alcohol-derived carbon group.
  2. CH3CH2CO- is the three-carbon acid fragment: propanoic acid.
  3. -CH3 attached to the ester O comes from methanol.
Answer

Methyl propanoate, made from propanoic acid and methanol. Count the carbonyl carbon when counting the acid carbons.

Check your understandingWhat ester forms from butanoic acid and ethanol? Give a condensed formula.Think it through, then reveal the answer
Ethyl butanoate: CH3CH2CH2COOCH2CH3. The name begins with the ethanol-derived ethyl group, although the condensed formula here starts with the acid-derived part.
07

Addition polymers: open the double bond, keep the substituents

A repeat unit shows the pattern within a long chain.

A polymer is a large molecule built from many smaller monomer molecules. Different monomers give different repeating units and properties. In addition polymerisation, the C=C bonds open to link monomers into a chain; no small molecule is eliminated. Poly(ethene) forms from ethene and is used in plastic bags and clingfilm.

Draw an addition polymer
  1. Start with the alkene

    Identify the two carbon atoms of C=C.

  2. Change the double bond to a single bond

    Give each of those carbons a new bond to the next unit.

  3. Preserve every attached group

    Do not detach or lose hydrogens, chlorine atoms or side chains.

  4. Show repetition

    Enclose a repeat unit in brackets, with backbone bonds extending through the brackets and n outside.

Ethene gives the repeat unit [-CH2-CH2-]n. Propene, CH2=CH-CH3, gives [-CH2-CH(CH3)-]n. The CH3 side group stays attached; it does not become a third carbon in the two-carbon backbone repeat.

Poly(ethene): bonds through the brackets

Two backbone carbon atoms have a single bond between them and open bonds passing through the brackets to neighbouring units. Each carbon has two hydrogen atoms.

The n means many repeats. The bonds crossing the brackets join neighbouring units; they are not additional hydrogen atoms. Restoring C=C between the two backbone carbons gives ethene.

Poly(propene): keep the side group

Two backbone carbon atoms have a single bond between them and open bonds passing through the brackets to neighbouring units. A complete CH3 side group remains attached to the second carbon.

The n means many repeats. The bonds crossing the brackets join neighbouring units; they are not additional hydrogen atoms. Restoring C=C between the two backbone carbons gives propene.

Worked example

Recover a monomer

A polymer repeat unit is [-CH2-CHCl-]n. Deduce its monomer.

  1. Identify the two backbone carbons.
  2. Remove their links to neighbouring repeat units and restore a C=C double bond.
  3. Keep the chlorine attached to the same carbon.
Answer

CH2=CHCl. An unfamiliar monomer can be deduced from bonding without memorising its name.

Check your understandingShould a poly(ethene) repeat unit retain the original C=C double bond?Think it through, then reveal the answer
No. The double bond changes to a single bond, allowing new bonds along the chain. Keeping both the double bond and the new chain bonds would give each carbon too many bonds.
08Pure only

Condensation polymers and depolymerisation

Recognise the linkage that is made or broken.

Condensation polymerisation joins monomers with suitable functional groups, eliminating small molecules such as water. Nylon is a polyamide, with -C(=O)-NH- links. Terylene is a polyester, with -C(=O)-O- links. Different polymers may differ in their monomer units, their linkages, or both.

Read partial chain structures
PolymerA simplified repeating patternLinkage
Nylon[-NH-A-NH-C(=O)-B-C(=O)-]nAmide: -C(=O)-NH- between neighbouring units
Terylene[-O-A-O-C(=O)-B-C(=O)-]nEster: -C(=O)-O- between neighbouring units

A and B stand for the intervening carbon-containing sections, not chemical element symbols. Follow the chain through the bracket boundary to find each linkage. These partial structures explain the polymer class without requiring industrial manufacture details or mechanisms.

Nylon: recognise the amide links

A partial chain with two carbonyl groups flanking carbon-containing segment B and NH groups flanking segment A. Open bonds at both edges continue the polymer.

An amide link is -C(=O)-NH-. Boxes A and B stand for the intervening carbon-containing sections, as in a partial structure; they are not element symbols.

Terylene: recognise the ester links

A partial chain with two carbonyl groups flanking carbon-containing segment B and O groups flanking segment A. Open bonds at both edges continue the polymer.

An ester link is -C(=O)-O-. Boxes A and B stand for the intervening carbon-containing sections, as in a partial structure; they are not element symbols.

Worked example

Deduce a condensation repeat unit

How would HO-A-OH and HOOC-B-COOH join, and how does replacing the alcohol with H2N-A-NH2 change the polymer?

  1. The diol has two alcohol ends and the diacid two carboxylic-acid ends, so repeated joining can build a long chain.
  2. Each ester link eliminates water and connects -O-A-O- with -C(=O)-B-C(=O)-.
  3. With a diamine instead, -NH-A-NH- replaces the oxygen-containing segment; amide links form.
Answer

The diol/diacid gives a polyester; the diamine/diacid gives a polyamide. To deduce monomers from the chain, cut the linkages and restore -OH at acid ends and H at alcohol/amine ends.

Nylon and Terylene are useful man-made fibres: applications include clothing, curtains, fishing line, parachutes and sleeping bags. Suitability depends on properties such as strength, durability and flexibility; knowing the material name is less useful than connecting a property to the job.

Hydrolyse a polyester
  1. Add water under acidic conditions

    Acid acts as a catalyst; heating can speed the process.

  2. Break ester links

    Water contributes H and OH across the broken linkage.

  3. Recover monomers

    Alcohol and carboxylic-acid groups are restored; extensive cleavage can return the original monomers.

For a chain made from HO-A-OH and HOOC-B-COOH, hydrolysis restores these two monomer types. This is depolymerisation: breaking a polymer into its monomers. It is different from merely melting the plastic, which leaves the long molecules largely intact.

Worked example

Recover both ends when a polyester is hydrolysed

A polyester repeat contains -O-A-O-C(=O)-B-C(=O)-, with the chain continuing at both ends. Deduce the two monomer types after complete acid-catalysed hydrolysis.

  1. Mark every -C(=O)-O- ester link, including the one at the repeat boundary.
  2. Break the bond between the carbonyl carbon and the single-bonded oxygen. Restore OH to each carbonyl carbon to form the two carboxylic-acid ends.
  3. Restore H to each alcohol-derived oxygen. Segment A has two alcohol ends; segment B has two acid ends.
Answer

HO-A-OH and HOOC-B-COOH. Water provides the H/OH; the acid is a catalyst. Cutting a C-C bond or returning only one end group would give the wrong monomers.

Check your understandingA chain contains -C(=O)-NH- repeatedly. Is it a polyester?Think it through, then reveal the answer
No: that is an amide linkage, so it is a polyamide such as nylon. A polyester has -C(=O)-O- links. Look at the atom next to the carbonyl group.
09

Evaluate plastic use and recycling

Compare what happens to the molecules and to the wider system.

Many plastics are non-biodegradable: natural biological processes do not break them down readily. Persistent waste can harm wildlife through ingestion or entanglement, accumulate in habitats and fragment into small pieces. Burning plastics releases carbon dioxide and, depending on the plastic and conditions, other pollutants. Disposal shifts a problem unless its consequences are considered.

Two recycling routes
RouteWhat happensTrade-off
Physical recyclingSort and clean plastic, cut it into pieces, then melt it into pellets; poly(ethene) is an exampleRetains polymer molecules but mixed materials or contamination can reduce product quality
Chemical recycling by crackingBreak plastic molecules into smaller hydrocarbons that can be used as fuelCan process molecules differently, but requires energy; burning the fuel still releases carbon dioxide
Pure only

Pure: chemical recycling can also use depolymerisation to recover chemical feedstock. Acid-catalysed polyester hydrolysis can regenerate monomers for making new material. Cracking to fuel and depolymerising to monomers have different products; neither should be described merely as melting.

Evaluate a proposed recycling system
DimensionQuestions that matter
SocialIs collection accessible? Are sorting instructions understandable? Are workers and nearby communities affected?
EconomicWhat do collection, transport, sorting and processing cost? Is there a market for the recovered material?
EnvironmentalHow much energy, transport and virgin resource use are avoided or added? What emissions and unrecovered waste remain?

Worked example

Compare two claims

One plan melts clean separated poly(ethene) locally; another ships mixed plastic far away for cracking. Is either always better?

  1. The first may use less processing but depends on clean sorted input and a useful output market.
  2. The second may accept different waste but adds transport and cracking energy.
  3. Compare actual quantities, energy sources, yields and avoided virgin production.
Answer

A justified decision uses evidence across the whole process, rather than assuming the word recycling guarantees a lower impact.

Check your understandingDoes melting poly(ethene) into pellets normally turn it back into ethene?Think it through, then reveal the answer
No. Physical recycling mainly changes shape while retaining long polymer molecules. Recovering monomers requires a chemical depolymerisation route, not ordinary melting.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Core reaction connections
  1. Alkane

    Combustion; chlorine substitution under UV; larger molecules can crack.

  2. Alkene

    Bromine/hydrogen addition and addition polymerisation.

  3. Ethanol

    Fermentation makes it; combustion releases energy; oxidation forms ethanoic acid.

  4. Polymer

    Match the repeat unit to its monomer; distinguish physical and chemical recycling.

Pure only

Pure: add C4 and branched structures, isomerism, ethene hydration, carboxylic-acid reactions, esters and condensation-polymer hydrolysis. Reagents and essential conditions matter; exact numerical process settings are not required.

Scope and references

Learning outcomes and sources

11. Organic Chemistry (6092 / K324). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map
  1. 11.1(a) Identify non-renewable fuel sources

    • Natural gas mainly methane
    • Crude oil

    Fuels, fractions and bioethanol

  2. 11.1(b) Explain crude-oil fractionation

    • Hydrocarbon mixture
    • Fractions
    • Competing fuel and chemical feedstock uses

    Fuels, fractions and bioethanol

  3. 11.1(c) Recognise renewable biofuel

    • Bioethanol from sugarcane

    Fuels, fractions and bioethanol

  4. 11.1(d) Compare carbon impacts of fuels

    • Plant-growth uptake offsets burning emissions
    • Compare fossil-carbon release
    • Sustainability depends on wider inputs

    Fuels, fractions and bioethanol

  5. 11.2(a) Describe homologous series

    • General formula
    • Similar chemistry
    • Gradual melting/boiling/viscosity change with size/mass

    Alkanes: build a valid carbon skeletonFuels, fractions and bioethanol

  6. 11.2(b) Describe alkanes

    • Saturated hydrocarbons
    • CnH2n+2

    Alkanes: build a valid carbon skeleton

  7. 11.2(e) Describe alkane reactions

    • Methane combustion
    • Chlorine substitution with UV
    • Generally unreactive otherwise

    Alkanes: build a valid carbon skeleton

  8. 11.2(f) Describe alkenes

    • Unsaturated hydrocarbons
    • CnH2n

    Alkenes: a double bond creates new reactions

  9. 11.2(h) Explain cracking

    • Alkenes and hydrogen products
    • Demand for smaller refinery molecules
    • Heating/catalyst conditions

    Alkenes: a double bond creates new reactionsFuels, fractions and bioethanol

  10. 11.2(i) Distinguish saturation

    • Molecular structure
    • Aqueous bromine test

    Alkenes: a double bond creates new reactions

  11. 11.2(k) Explain polyunsaturated food molecules

    • More than one C=C

    Alkenes: a double bond creates new reactions

  12. 11.2(l) Explain margarine manufacture

    • Hydrogen addition to vegetable oils
    • Solid product
    • Nickel catalyst and heat

    Alkenes: a double bond creates new reactions

  13. 11.2(c) Draw and name alkanes

    • Branched and unbranched C1-C4
    • Names methane to butane
    • Displayed-formula method

    Alkanes: build a valid carbon skeleton

  14. 11.2(d) Identify isomerism

    • Same molecular formula; different structural formulae

    Alkanes: build a valid carbon skeleton

  15. 11.2(g) Draw and name alkenes

    • Branched/unbranched C2-C4
    • Ethene to butene
    • Valency check

    Alkenes: a double bond creates new reactions

  16. 11.2(j) Describe ethene reactions

    • Combustion
    • Polymerisation
    • Addition of bromine, steam and hydrogen
    • Essential reagents/conditions

    Alkenes: a double bond creates new reactionsAddition polymers: open the double bond, keep the substituents

  17. 11.3(a) Recognise alcohol series

    • -OH group

    Alcohols: identify -OH and follow ethanol

  18. 11.3(c) Describe alcohol reactions

    • Combustion
    • Oxidation to acids, exemplified by ethanol

    Alcohols: identify -OH and follow ethanol

  19. 11.3(e) Recognise carboxylic-acid series

    • -CO2H/-COOH group

    Carboxylic acids: recognise -COOH

  20. 11.3(b) Draw and name alcohols

    • Branched/unbranched C1-C4
    • Methanol to butanol
    • OH position

    Alcohols: identify -OH and follow ethanol

  21. 11.3(d) Describe ethanol production

    • Catalysed steam addition to ethene
    • Glucose fermentation
    • Essential conditions

    Alcohols: identify -OH and follow ethanolAlkenes: a double bond creates new reactions

  22. 11.3(f) Draw and name carboxylic acids

    • C1-C4 structures
    • Methanoic to butanoic acid

    Carboxylic acids: recognise -COOH

  23. 11.3(g) Explain acid reactions

    • Weak ionisation
    • Carbonates, bases and suitable metals

    Carboxylic acids: recognise -COOH

  24. 11.3(h) Explain ethanol oxidation

    • Atmospheric oxygen
    • Acidified KMnO4

    Alcohols: identify -OH and follow ethanolCarboxylic acids: recognise -COOH

  25. 11.3(i) Explain ester formation

    • Acid plus alcohol
    • Ethyl ethanoate example
    • Catalyst and warming

    Esters: join two fragments and keep track of their origins

  26. 11.3(j) Deduce ester names/formulae and precursors

    • Unbranched C1-C4 acids and alcohols
    • Forward and reverse deduction

    Esters: join two fragments and keep track of their origins

  27. 11.4(a) Describe polymers and monomers

    • Large molecules from small units
    • Different monomers give different polymers
    • Different linkages

    Addition polymers: open the double bond, keep the substituentsCondensation polymers and depolymerisation

  28. 11.4(b) Explain poly(ethene) formation

    • Addition polymerisation of ethene

    Addition polymers: open the double bond, keep the substituents

  29. 11.4(c) Identify poly(ethene) uses

    • Plastic bags
    • Clingfilm

    Addition polymers: open the double bond, keep the substituents

  30. 11.4(d) Deduce addition-polymer structures

    • Monomer to repeat unit
    • Repeat unit to monomer

    Addition polymers: open the double bond, keep the substituentsCondensation polymers and depolymerisation

  31. 11.4(g) Explain plastic pollution

    • Non-biodegradability
    • Disposal impacts

    Evaluate plastic use and recycling

  32. 11.4(h) Compare recycling routes

    • Physical: poly(ethene) melted into pellets
    • Chemical: cracking into fuel
    • Depolymerisation to feedstock

    Evaluate plastic use and recyclingCondensation polymers and depolymerisation

  33. 11.4(j) Evaluate recycling issues

    • Social
    • Economic
    • Environmental

    Evaluate plastic use and recycling

  34. 11.4(e) Recognise condensation polymers

    • Nylon polyamide partial structure
    • Terylene polyester partial structure
    • No manufacture mechanisms

    Condensation polymers and depolymerisation

  35. 11.4(f) Identify man-made fibre uses

    • Clothing, curtains, fishing line, parachutes, sleeping bags

    Condensation polymers and depolymerisation

  36. 11.4(i) Explain depolymerisation

    • Polymers to monomers
    • Acid-catalysed polyester hydrolysis
    • No mechanism

    Condensation polymers and depolymerisation

  • 2026 Pure Chemistry 6092

    Official topic 11, pages 21-23. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.

  • 2027 Pure Chemistry K324

    Official topic 11, pages 21-23. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.

  • 2026 Combined Chemistry 5086 / 5088

    Official topic 11, pages 34-36. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.

  • 2027 Combined Chemistry K326 / K328

    Official topic 11, pages 34-36. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.

  • Grail: 6092 Chemistry Complete Notes, Version 1

    Background consultation: Chapters 19-21, small organic structures and polymer links, pp. 77-89; official Pure p. 23 polymer diagrams inspected separately. Teaching additions and examples are original; syllabus scope and chemistry independently checked.