Full chapter
Organic Chemistry
Recognise molecular families and connect structures, reactions, fuels and materials.
O-Level 6092 (2026) / SEC G3 K324 (2027)
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.
- Heat the oil
Many components vaporise.
- Vapour rises through the column
The column is hotter at the bottom and cooler at the top.
- Condense different fractions
Higher-boiling, generally larger molecules condense lower down; smaller, more volatile molecules travel higher.
- 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
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.
| Name | Molecular formula | Condensed structure |
|---|---|---|
| Methane | CH4 | One C with four C-H bonds |
| Ethane | C2H6 | CH3-CH3 |
| Propane | C3H8 | CH3-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.
Displayed methane
Complete displayed structure with 1 carbon, 4 hydrogen atoms: CH4. Every bond and hydrogen is shown.
Displayed ethane
Complete displayed structure with 2 carbon, 6 hydrogen atoms: C2H6. Every bond and hydrogen is shown.
Displayed propane
Complete displayed structure with 3 carbon, 8 hydrogen atoms: C3H8. Every bond and hydrogen is shown.
Displayed butane
Complete displayed structure with 4 carbon, 10 hydrogen atoms: C4H10. Every bond and hydrogen is shown.
Displayed branched C4 alkane
Complete displayed structure with 4 carbon, 10 hydrogen atoms: C4H10. Every bond and hydrogen is shown.
| Structure | Condensed representation |
|---|---|
| Unbranched butane | CH3-CH2-CH2-CH3 |
| Branched C4H10 | CH3-CH(CH3)-CH3 |
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
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.
| Connectivity | Condensed structure |
|---|---|
| But-1-ene | CH2=CH-CH2-CH3 |
| But-2-ene | CH3-CH=CH-CH3 |
| Branched C4H8 | CH2=C(CH3)2 |
Worked example
Expand an alkene without adding too many hydrogens
Draw propene from CH2=CH-CH3.
- Join three carbon atoms. Put the double bond between the first two; this uses two of each participating carbon's four bonds.
- 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.
- Count the result: C3H6, with four bonds at each carbon.
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.
Displayed propene
Complete displayed structure with 3 carbon, 6 hydrogen atoms: C3H6. Every bond and hydrogen is shown.
Displayed but-1-ene
Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.
Displayed but-2-ene
Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.
Displayed branched C4 alkene
Complete displayed structure with 4 carbon, 8 hydrogen atoms: C4H8. Every bond and hydrogen is shown.
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.
| Reagent or process | Essential conditions | Product/change |
|---|---|---|
| Oxygen: complete combustion | Ignition, sufficient oxygen | C2H4 + 3O2 -> 2CO2 + 2H2O |
| Aqueous bromine | Room conditions; no UV required | C2H4 + Br2 -> C2H4Br2 |
| Hydrogen | Nickel catalyst, heat | C2H4 + H2 -> C2H6 |
| Addition polymerisation | Suitable temperature, pressure and catalyst | Many ethene molecules form poly(ethene) |
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
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.
| Name | Condensed structure |
|---|---|
| Methanol | CH3-OH |
| Ethanol | CH3-CH2-OH |
| Propan-1-ol | CH3-CH2-CH2-OH |
| Propan-2-ol | CH3-CH(OH)-CH3 |
| Structure | Condensed representation |
|---|---|
| Butan-1-ol | CH3-CH2-CH2-CH2-OH |
| Butan-2-ol | CH3-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.
Displayed ethanol
Complete displayed structure with 2 carbon, 6 hydrogen and 1 oxygen atoms: C2H6O. Every bond and hydrogen is shown.
Displayed propan-1-ol
Complete displayed structure with 3 carbon, 8 hydrogen and 1 oxygen atoms: C3H8O. Every bond and hydrogen is shown.
Displayed propan-2-ol
Complete displayed structure with 3 carbon, 8 hydrogen and 1 oxygen atoms: C3H8O. Every bond and hydrogen is shown.
Displayed butan-1-ol
Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.
Displayed butan-2-ol
Complete displayed structure with 4 carbon, 10 hydrogen and 1 oxygen atoms: C4H10O. Every bond and hydrogen is shown.
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.
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.
- Feedstock
Glucose solution, obtainable from sugarcane sugars.
- Conditions
Yeast supplies enzymes; keep warm and exclude oxygen. Excessive heat damages enzymes.
- Reaction
C6H12O6(aq) -> 2C2H5OH(aq) + 2CO2(g).
- 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: 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
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.
| Name or connectivity | Condensed formula |
|---|---|
| Methanoic acid | H-C(=O)-OH |
| Ethanoic acid | CH3-C(=O)-OH |
| Propanoic acid | CH3-CH2-C(=O)-OH |
| Butanoic acid | CH3-CH2-CH2-C(=O)-OH |
| Branched C4 acid | (CH3)2CH-C(=O)-OH |
Worked example
Draw the whole carboxyl group
Draw propanoic acid and account for all three carbons.
- Draw a three-carbon chain. The end carbon is part of -COOH; do not add a fourth carbon for this group.
- 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.
- 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.
C3H6O2. In methanoic acid the carboxyl carbon instead bonds to H because there is no carbon chain attached.
Displayed methanoic acid
Complete displayed structure with 1 carbon, 2 hydrogen and 2 oxygen atoms: CH2O2. Every bond and hydrogen is shown.
Displayed ethanoic acid
Complete displayed structure with 2 carbon, 4 hydrogen and 2 oxygen atoms: C2H4O2. Every bond and hydrogen is shown.
Displayed propanoic acid
Complete displayed structure with 3 carbon, 6 hydrogen and 2 oxygen atoms: C3H6O2. Every bond and hydrogen is shown.
Displayed butanoic acid
Complete displayed structure with 4 carbon, 8 hydrogen and 2 oxygen atoms: C4H8O2. Every bond and hydrogen is shown.
Displayed branched C4 carboxylic acid
Complete displayed structure with 4 carbon, 8 hydrogen and 2 oxygen atoms: C4H8O2. Every bond and hydrogen is shown.
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.
| Partner | Balanced reaction |
|---|---|
| Sodium hydroxide | CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l) |
| Sodium carbonate | 2CH3COOH(aq) + Na2CO3(aq) -> 2CH3COONa(aq) + H2O(l) + CO2(g) |
| Magnesium | 2CH3COOH(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
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-.
| Starting family | C1 | C2 | C3 | C4 |
|---|---|---|---|---|
| Unbranched acid part | Methanoate | Ethanoate | Propanoate | Butanoate |
| Alcohol part, -OH at chain end | Methyl | Ethyl | Propyl | Butyl |
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.
- Split at the bond between the single-bonded O and the alcohol-derived carbon group.
- CH3CH2CO- is the three-carbon acid fragment: propanoic acid.
- -CH3 attached to the ester O comes from methanol.
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
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.
- Start with the alkene
Identify the two carbon atoms of C=C.
- Change the double bond to a single bond
Give each of those carbons a new bond to the next unit.
- Preserve every attached group
Do not detach or lose hydrogens, chlorine atoms or side chains.
- 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.
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.
Worked example
Recover a monomer
A polymer repeat unit is [-CH2-CHCl-]n. Deduce its monomer.
- Identify the two backbone carbons.
- Remove their links to neighbouring repeat units and restore a C=C double bond.
- Keep the chlorine attached to the same carbon.
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
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.
| Polymer | A simplified repeating pattern | Linkage |
|---|---|---|
| Nylon | [-NH-A-NH-C(=O)-B-C(=O)-]n | Amide: -C(=O)-NH- between neighbouring units |
| Terylene | [-O-A-O-C(=O)-B-C(=O)-]n | Ester: -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.
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.
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?
- The diol has two alcohol ends and the diacid two carboxylic-acid ends, so repeated joining can build a long chain.
- Each ester link eliminates water and connects -O-A-O- with -C(=O)-B-C(=O)-.
- With a diamine instead, -NH-A-NH- replaces the oxygen-containing segment; amide links form.
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.
- Add water under acidic conditions
Acid acts as a catalyst; heating can speed the process.
- Break ester links
Water contributes H and OH across the broken linkage.
- 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.
- Mark every -C(=O)-O- ester link, including the one at the repeat boundary.
- 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.
- Restore H to each alcohol-derived oxygen. Segment A has two alcohol ends; segment B has two acid ends.
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
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.
| Route | What happens | Trade-off |
|---|---|---|
| Physical recycling | Sort and clean plastic, cut it into pieces, then melt it into pellets; poly(ethene) is an example | Retains polymer molecules but mixed materials or contamination can reduce product quality |
| Chemical recycling by cracking | Break plastic molecules into smaller hydrocarbons that can be used as fuel | Can process molecules differently, but requires energy; burning the fuel still releases carbon dioxide |
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.
| Dimension | Questions that matter |
|---|---|
| Social | Is collection accessible? Are sorting instructions understandable? Are workers and nearby communities affected? |
| Economic | What do collection, transport, sorting and processing cost? Is there a market for the recovered material? |
| Environmental | How 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?
- The first may use less processing but depends on clean sorted input and a useful output market.
- The second may accept different waste but adds transport and cracking energy.
- Compare actual quantities, energy sources, yields and avoided virgin production.
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
Quick revision
Revisit the essentials, then return to an explanation when you need it.
- Alkane
Combustion; chlorine substitution under UV; larger molecules can crack.
- Alkene
Bromine/hydrogen addition and addition polymerisation.
- Ethanol
Fermentation makes it; combustion releases energy; oxidation forms ethanoic acid.
- Polymer
Match the repeat unit to its monomer; distinguish physical and chemical recycling.
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
11.1(a) Identify non-renewable fuel sources
- Natural gas mainly methane
- Crude oil
11.1(b) Explain crude-oil fractionation
- Hydrocarbon mixture
- Fractions
- Competing fuel and chemical feedstock uses
11.1(c) Recognise renewable biofuel
- Bioethanol from sugarcane
11.1(d) Compare carbon impacts of fuels
- Plant-growth uptake offsets burning emissions
- Compare fossil-carbon release
- Sustainability depends on wider inputs
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
11.2(b) Describe alkanes
- Saturated hydrocarbons
- CnH2n+2
11.2(e) Describe alkane reactions
- Methane combustion
- Chlorine substitution with UV
- Generally unreactive otherwise
11.2(f) Describe alkenes
- Unsaturated hydrocarbons
- CnH2n
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
11.2(i) Distinguish saturation
- Molecular structure
- Aqueous bromine test
11.2(k) Explain polyunsaturated food molecules
- More than one C=C
11.2(l) Explain margarine manufacture
- Hydrogen addition to vegetable oils
- Solid product
- Nickel catalyst and heat
11.2(c) Draw and name alkanes
- Branched and unbranched C1-C4
- Names methane to butane
- Displayed-formula method
11.2(d) Identify isomerism
- Same molecular formula; different structural formulae
11.2(g) Draw and name alkenes
- Branched/unbranched C2-C4
- Ethene to butene
- Valency check
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
11.3(a) Recognise alcohol series
- -OH group
11.3(c) Describe alcohol reactions
- Combustion
- Oxidation to acids, exemplified by ethanol
11.3(e) Recognise carboxylic-acid series
- -CO2H/-COOH group
11.3(b) Draw and name alcohols
- Branched/unbranched C1-C4
- Methanol to butanol
- OH position
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
11.3(f) Draw and name carboxylic acids
- C1-C4 structures
- Methanoic to butanoic acid
11.3(g) Explain acid reactions
- Weak ionisation
- Carbonates, bases and suitable metals
11.3(h) Explain ethanol oxidation
- Atmospheric oxygen
- Acidified KMnO4
Alcohols: identify -OH and follow ethanolCarboxylic acids: recognise -COOH
11.3(i) Explain ester formation
- Acid plus alcohol
- Ethyl ethanoate example
- Catalyst and warming
11.3(j) Deduce ester names/formulae and precursors
- Unbranched C1-C4 acids and alcohols
- Forward and reverse deduction
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
11.4(b) Explain poly(ethene) formation
- Addition polymerisation of ethene
Addition polymers: open the double bond, keep the substituents
11.4(c) Identify poly(ethene) uses
- Plastic bags
- Clingfilm
Addition polymers: open the double bond, keep the substituents
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
11.4(g) Explain plastic pollution
- Non-biodegradability
- Disposal impacts
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
11.4(j) Evaluate recycling issues
- Social
- Economic
- Environmental
11.4(e) Recognise condensation polymers
- Nylon polyamide partial structure
- Terylene polyester partial structure
- No manufacture mechanisms
11.4(f) Identify man-made fibre uses
- Clothing, curtains, fishing line, parachutes, sleeping bags
11.4(i) Explain depolymerisation
- Polymers to monomers
- Acid-catalysed polyester hydrolysis
- No mechanism
- 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.