4.7 Organic chemistry — notes on one page

Every AQA GCSE Chemistry 8462 specification point in 4.7, read straight through: 12 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against AQA 8462 section 4.7. Review basis: the qualification registry sourced from the AQA GCSE Chemistry (8462) specification; registry verification recorded 17 July 2026.

How this checking works

4.7.1.1 · Crude oil, hydrocarbons and alkanes

Crude oil is a finite resource found in rocks. It formed from ancient biomass, mainly plankton, that was buried in mud.
Crude oil is a mixture containing many compounds; most are hydrocarbons, whose molecules contain carbon and hydrogen atoms only. Most crude-oil hydrocarbons are alkanes.
Their general formula is $\mathrm{C_nH_{2n+2}}$ , and the first four are methane, ethane, propane and butane.
To test an alkane formula, substitute its carbon count for $n$ and check the hydrogen count.
A common error is to call crude oil one compound rather than a mixture.

Worked example

Decide whether $\mathrm{C_5H_{12}}$ is an alkane.

1.Use the alkane formula $\mathrm{C_nH_{2n+2}}$ .
2.For $n=5$ , the hydrogen number is $2(5)+2=12$ .
3.The formula matches the general formula.

Answer: $\mathrm{C_5H_{12}}$ is an alkane.

Common mistakes

Don't fall into the trap of calling crude oil a single compound instead of a mixture of many compounds.
Don't fall into the trap of using $\mathrm{C_nH_{2n}}$ for an alkane instead of $\mathrm{C_nH_{2n+2}}$ .

Exam tip. When asked to identify an alkane, substitute the carbon number into $\mathrm{C_nH_{2n+2}}$ and check every atom.

4.7.1.2 · Fractional distillation and petrochemicals

Fractional distillation separates crude oil into fractions; each fraction contains hydrocarbons with similar numbers of carbon atoms and similar boiling points. Crude oil is heated so that most of it vaporises.
The column is hot at the bottom and cooler towards the top, so vapours condense at different heights.
Large molecules with higher boiling points condense lower in the column; smaller molecules with lower boiling points rise further before condensing.
Fractions provide fuels and petrochemical feedstock for solvents, lubricants, polymers and detergents.
Fractional distillation separates molecules; it does not crack them.

A fractionating column with a temperature gradient and condensation outlets for different molecular sizes.

Worked example

Explain why a small hydrocarbon is collected higher in a fractionating column than a large hydrocarbon.

1.Small hydrocarbons have lower boiling points.
2.They remain as vapour further up the column as the temperature falls.
3.They condense higher up when the temperature becomes low enough.

Answer: Its lower boiling point makes it condense in the cooler, upper part of the column.

Common mistakes

Don't fall into the trap of saying fractions separate because they have different densities rather than different boiling points.
Don't fall into the trap of reversing the column, so small molecules are said to condense low down where it is hottest.

Exam tip. For an “explain fractional distillation” question, follow one vapour from evaporation to condensation at the level below its boiling point.

4.7.1.3 · Properties of hydrocarbons

As hydrocarbon molecules become larger, their boiling points increase because the intermolecular forces between molecules become stronger. Increasing molecular size also increases viscosity and decreases flammability, influencing which fractions make convenient fuels.
Complete combustion releases energy and oxidises the carbon and hydrogen, producing carbon dioxide and water.
Balance combustion equations in the order carbon, hydrogen, then oxygen.
Do not confuse incomplete combustion products with the specified products of complete combustion.
In exam answers, connect each trend explicitly to increasing molecular size and distinguish complete combustion from incomplete combustion before choosing products.

Worked example

State the products when propane, $\mathrm{C_3H_8}$ , burns completely and write the balanced equation.

1.Complete combustion forms carbon dioxide and water.
2.Balance carbon to give $3\mathrm{CO_2}$ and hydrogen to give $4\mathrm{H_2O}$ .
3.Balance oxygen with $5\mathrm{O_2}$ .

Answer: $\mathrm{C_3H_8+5O_2\rightarrow3CO_2+4H_2O}$ .

Common mistakes

Don't fall into the trap of stating that larger hydrocarbons are more flammable, when flammability decreases as molecular size increases.
Don't fall into the trap of writing carbon monoxide or carbon as a product when the question specifies complete combustion.

Exam tip. In a trend question, state the direction for all three properties: boiling point and viscosity increase, while flammability decreases.

4.7.1.4 · Cracking and alkenes

Cracking breaks long-chain hydrocarbons into smaller, more useful molecules because demand for short-chain fuels is high. In catalytic cracking, a hydrocarbon is vaporised and passed over a hot catalyst; in steam cracking, hydrocarbon vapour is mixed with steam and heated strongly.
Cracking produces a mixture that includes shorter alkanes and alkenes.
Equations must conserve the number of carbon and hydrogen atoms.
Alkenes decolourise bromine water from orange to colourless and are useful for polymers and other chemicals.
Heating alone does not identify an alkene.

Worked example

Balance the cracking equation $\mathrm{C_{10}H_{22}\rightarrow C_6H_{14}+C_xH_y}$ and identify the second product type.

1.Conserve carbon: $x=10-6=4$ .
2.Conserve hydrogen: $y=22-14=8$ .
3. $\mathrm{C_4H_8}$ fits $\mathrm{C_nH_{2n}}$ , so it is an alkene.

Answer: $\mathrm{C_{10}H_{22}\rightarrow C_6H_{14}+C_4H_8}$ ; the second product is an alkene.

Common mistakes

Don't fall into the trap of describing fractional distillation as cracking, even though distillation only separates molecules.
Don't fall into the trap of giving orange to colourless as evidence for an alkane rather than for an alkene.

Exam tip. In a cracking equation, count carbon and hydrogen atoms on both sides before naming the alkene product.

4.7.2.1 · Structure and formulae of alkenes (chemistry only)

Alkenes are hydrocarbons containing a carbon-carbon double bond, C=C; this double bond makes them unsaturated.
Their homologous-series formula is $\mathrm{C_nH_{2n}}$ , so an alkene has two fewer hydrogen atoms than the alkane with the same carbon count.
The first four members are ethene, propene, butene and pentene, and each may be shown by a molecular or fully displayed structural formula.
When recognising an alkene, check both the double bond and the formula.
A common error is to apply the alkane formula $\mathrm{C_nH_{2n+2}}$ .

Worked example

Show that $\mathrm{C_4H_8}$ can be an alkene and name the relevant member of the series.

1.Use $\mathrm{C_nH_{2n}}$ .
2.For $n=4$ , $2n=8$ , so the formula matches.
3.The four-carbon alkene is butene.

Answer: $\mathrm{C_4H_8}$ matches the alkene formula and is butene.

Common mistakes

Don't fall into the trap of using the alkane formula $\mathrm{C_nH_{2n+2}}$ for an alkene.
Don't fall into the trap of calling a molecule unsaturated without showing or identifying its carbon–carbon double bond.

Exam tip. For “recognise an alkene”, check both $\mathrm{C_nH_{2n}}$ and the displayed C=C bond.

4.7.2.2 · Reactions of alkenes (chemistry only)

Alkene reactions are governed by the C=C functional group: addition places atoms across the double bond and leaves a single carbon-carbon bond.
Hydrogen adds in the presence of a nickel catalyst to form an alkane; steam adds with a phosphoric acid catalyst to form an alcohol.
Chlorine, bromine and iodine add across C=C to form saturated dihalogeno compounds; bromine water changes from orange to colourless.
Alkenes combust like other hydrocarbons but often burn with smoky flames in air because combustion is incomplete.
Do not leave C=C in an addition product.

Bromine adds across an alkene carbon–carbon double bond to form a saturated product.

Worked example

Describe the product when ethene reacts with bromine.

1.Identify the C=C functional group in ethene.
2.Open the double bond to make C–C.
3.Attach one bromine atom to each carbon.

Answer: 1,2-dibromoethane forms and bromine is decolourised.

Common mistakes

Don't fall into the trap of leaving the C=C double bond in the displayed product after an addition reaction.
Don't fall into the trap of naming bromine-water decolourisation but giving the colour change as colourless to orange.

Exam tip. For an addition product, replace C=C by C–C and attach one incoming atom or group to each carbon.

4.7.2.3 · Alcohols (chemistry only)

Alcohols contain the -OH functional group; the first four are methanol, ethanol, propanol and butanol. They burn in air, react with sodium and are oxidised to carboxylic acids.
Their solubility in water decreases as the carbon chain becomes longer.
Ethanol is made by fermenting a sugar solution with yeast under warm conditions without oxygen; fermentation also produces carbon dioxide.
Alcohols are used as fuels and solvents.
Balanced equations are required for their combustion, but not for their other reactions in this specification.

Worked example

State the conditions needed to produce ethanol by fermentation of sugar solution.

1.Add yeast to the sugar solution.
2.Keep the mixture warm.
3.Exclude oxygen so fermentation occurs.

Answer: Yeast, warm conditions and no oxygen.

Common mistakes

Don't fall into the trap of saying fermentation needs oxygen, when yeast ferments sugar without oxygen.
Don't fall into the trap of assuming all four alcohols are equally soluble in water instead of recognising that solubility decreases with chain length.

Exam tip. For fermentation conditions, state yeast, a warm temperature and absence of oxygen.

4.7.2.4 · Carboxylic acids (chemistry only)

Carboxylic acids contain the -COOH functional group; the first four are methanoic, ethanoic, propanoic and butanoic acid. They dissolve in water to form acidic solutions and react with carbonates to make a salt, water and carbon dioxide.
Higher tier: AQA restricts "explain why carboxylic acids are weak acids in terms of ionisation and pH" to Higher tier; naming the first four carboxylic acids and describing their reactions is examined on both tiers.
A carboxylic acid reacts with an alcohol to form an ester and water; ethanol with ethanoic acid forms ethyl ethanoate.
Recognise the -COOH group before naming the molecule.
A common error is to identify it as the -OH group of an alcohol.

Worked example

Predict the observation and products when ethanoic acid reacts with calcium carbonate.

1.Use the acid–carbonate reaction pattern.
2.The products are a salt, water and carbon dioxide.
3.Escaping carbon dioxide produces effervescence.

Answer: Fizzing occurs; calcium ethanoate, water and carbon dioxide form.

Common mistakes

Don't fall into the trap of identifying -COOH as the -OH functional group of an alcohol.
Don't fall into the trap of saying a weak acid is dilute rather than explaining that it only partially ionises in water (Higher tier).

Exam tip. For an acid–carbonate reaction, name all three product types and link effervescence to carbon dioxide.

4.7.3.1 · Addition polymerisation (chemistry only)

In addition polymerisation, many alkene monomers join to form one long-chain polymer molecule. The monomer's C=C bond opens to form C-C bonds linking neighbouring repeating units; no small molecule is produced.
The repeating unit contains the same atoms as the monomer and is shown in brackets with continuation bonds and a subscript $n$ .
To recover the monomer from a repeating unit, place C=C between the two backbone carbons.
Do not leave a double bond in the addition polymer.
Questions may require drawing the repeating unit from a displayed monomer or reversing the process, so every side group must be preserved and bonds must pass through both brackets.

General addition polymerisation of a substituted alkene into a bracketed repeating unit.

Worked example

Write the repeating unit formed from propene, $\mathrm{CH_2{=}CHCH_3}$ .

1.Use the two double-bonded carbons as the backbone.
2.Change C=C to C–C and retain the $\mathrm{CH_3}$ side group.
3.Add brackets, continuation bonds and subscript $n$ .

Answer: $[\mathrm{-CH_2-CH(CH_3)-}]_n$ .

Common mistakes

Don't fall into the trap of keeping the C=C double bond inside the addition-polymer repeating unit.
Don't fall into the trap of omitting the continuation bonds through the brackets or moving a side group to the wrong backbone carbon.

Exam tip. Copy every atom from one monomer, change C=C to C–C, then add brackets, continuation bonds and $n$ .

4.7.3.2 · Condensation polymerisation (chemistry only) (HT only)

Higher tier: condensation polymerisation requires monomers with two functional groups, allowing each monomer to join at both ends of a growing chain.
When a link forms, a small molecule such as water is usually eliminated; the polymer therefore does not contain every atom from the monomers.
A diol and a dicarboxylic acid form a polyester because -OH groups react with -COOH groups to make ester links.
Read the repeating unit across the new links and preserve the remaining carbon chains.
A common error is to treat condensation as C=C addition.

Worked example

Higher tier: explain why a diol and a dicarboxylic acid can form a polyester.

1.Each monomer has two functional groups, so it can join at both ends.
2.-OH and -COOH groups react to form ester links.
3.Each link forms with elimination of a small molecule such as water.

Answer: Repeated condensation forms a polyester with ester links and releases water.

Common mistakes

Don't fall into the trap of describing condensation polymerisation as opening a C=C bond.
Don't fall into the trap of drawing monomers with only one functional group, so they cannot join at both ends to make a chain.

Exam tip. Higher tier: identify both functional groups, the new link and the small molecule eliminated.

4.7.3.3 · Amino acids (chemistry only) (HT only)

Higher tier: an amino acid contains two different functional groups in one molecule: an amino group, -NH₂, and a carboxyl group, -COOH. Glycine is H₂NCH₂COOH.
Amino and carboxyl groups on different molecules undergo condensation to form peptide links. Each peptide link, -CONH-, forms with the loss of water, and repeated condensation produces a polypeptide.
Different amino acids can occur in the same chain, producing proteins.
Do not describe this as addition polymerisation because no C=C bond opens.
Questions may require drawing the product formed when amino acids join, so the -CONH- link and every atom in the water removed must be accounted for.

Worked example

Higher tier: three amino acids join into one unbranched chain. State the numbers of peptide links and water molecules formed.

1.Three molecules require two joining reactions to make one chain.
2.Each joining reaction forms one peptide link.
3.Each condensation also releases one water molecule.

Answer: Two peptide links and two water molecules.

Common mistakes

Don't fall into the trap of calling the -CONH- connection an ester link rather than a peptide link.
Don't fall into the trap of saying four amino acids form four peptide links; an unbranched chain of four has three links.

Exam tip. Higher tier: for $n$ amino acids joining into one chain, expect $n-1$ peptide links and $n-1$ water molecules.

4.7.3.4 · DNA (deoxyribonucleic acid) and other naturally occurring polymers (chemistry only)

DNA is a large molecule essential for life because it encodes genetic instructions for the development and functioning of organisms and viruses.
Most DNA consists of two polymer chains arranged as a double helix and built from four different monomers called nucleotides.
Proteins are polymers of amino acids, while starch and cellulose are polymers made from glucose monomers.
Match each natural polymer to its own monomer type.
A common error is to describe DNA as a protein or to say that amino acids are its monomers.

A simplified DNA double helix made from two nucleotide polymer chains.

Worked example

Match DNA, protein and starch to their monomer types.

1.DNA is built from four different nucleotides.
2.Proteins are built from amino acids.
3.Starch is built from glucose.

Answer: DNA–nucleotides; protein–amino acids; starch–glucose.

Common mistakes

Don't fall into the trap of calling DNA a protein or saying its monomers are amino acids.
Don't fall into the trap of saying starch and cellulose have different monomer types, when both are polymers of glucose.

Exam tip. In a matching question, keep the pairs distinct: DNA–nucleotides, proteins–amino acids, starch/cellulose–glucose.