OCR GCSE Chemistry A: Organic chemistry (J248)

Sub-topic C6.2 Organic chemistry sits in Topic C6 Global challenges of OCR’s Chemistry A (Gateway Science) GCSE (9–1) specification, code J248, Version 4.0 (August 2026). This guide teaches outcomes C6.2a to C6.2q, maths outcome CM6.2i and both practical suggestions (use of models; making nylon). Statements C6.2e and C6.2f are Higher Tier only; every other statement can be tested on either tier. Topic C6 is assessed on Paper 2 (Foundation) and Paper 4 (Higher).

Every J248 topic is on the OCR GCSE Chemistry course hub. Tick off outcomes on a printed checklist for J248. Use a free diagnostic (about 10 minutes) to see which areas need work first.

Statements in C6.2

Reference	Skills	Tier
C6.2a–c	Recognise, name and draw the four series; predict products of their reactions	Both tiers
C6.2d, C6.2g	Addition polymerisation; monomer to polymer and back	Both tiers
C6.2e–f	Condensation polymerisation and making a condensation polymer	Higher Tier only
C6.2h–i	Natural polymers; functional groups decide reactions	Both tiers
C6.2j–o	Crude oil, fractional distillation, fuels, feedstock, cracking	Both tiers
C6.2p–q	Chemical cells and fuel cells	Both tiers
CM6.2i	Turn 3D structures into 2D drawings and back	Both tiers

Homologous series and functional groups (C6.2a, C6.2i)

A homologous series is a family of organic compounds that:

• share a general formula and differ from the next member by CH₂;
• have the same functional group (the atom or group that gives the series its reactions);
• show a steady trend in physical properties, such as boiling point rising with chain length.

Series	Name ending	Functional group	General formula
Alkanes	-ane	none (C–C single bonds only)	CₙH₂ₙ₊₂
Alkenes	-ene	C=C double bond	CₙH₂ₙ
Alcohols	-ol	–OH	CₙH₂ₙ₊₁OH
Carboxylic acids	-oic acid	–COOH	CₙH₂ₙ₊₁COOH

Statement C6.2i follows: hexan-1-ol reacts like ethanol because both carry –OH. Chain length changes physical properties; the functional group sets the chemistry.

Worked example. Which series does each belong to: C₆H₁₂, C₃H₇COOH, C₈H₁₈, C₂H₅OH?

• C₆H₁₂: 2 × 6 = 12 hydrogens fits CₙH₂ₙ, so alkene.
• C₃H₇COOH ends in –COOH: carboxylic acid (butanoic acid, four carbons).
• C₈H₁₈: 2 × 8 + 2 = 18 fits CₙH₂ₙ₊₂: alkane (octane).
• C₂H₅OH ends in –OH: alcohol (ethanol).

Naming and drawing the first four members (C6.2b, CM6.2i)

The stem gives the number of carbons: meth- 1, eth- 2, prop- 3, but- 4. The alkenes start at two carbons, because a C=C bond needs two.

Carbons	Alkane	Alkene	Alcohol	Carboxylic acid
1	methane CH₄	—	methanol CH₃OH	methanoic acid HCOOH
2	ethane C₂H₆	ethene C₂H₄	ethanol C₂H₅OH	ethanoic acid CH₃COOH
3	propane C₃H₈	propene C₃H₆	propanol C₃H₇OH	propanoic acid C₂H₅COOH
4	butane C₄H₁₀	butene C₄H₈	butanol C₄H₉OH	butanoic acid C₃H₇COOH
5	—	pentene C₅H₁₀	—	—

The first four alkenes are ethene, propene, butene and pentene. In some members the functional group can sit in more than one place; a number shows where (but-1-ene, propan-1-ol). The carbon of –COOH counts in an acid’s name.

A displayed formula shows every atom and every bond as a line. Butanoic acid:

    H   H   H   O
    |   |   |   ‖
H — C — C — C — C — O — H
    |   |   |
    H   H   H

Check that each carbon has four bonds, each hydrogen one and each oxygen two.

Three dimensions to two (CM6.2i). A carbon with four single bonds is tetrahedral (about 109.5°), but a displayed formula draws 90° angles. Likewise diamond is a 3D network with each carbon bonded to four others, while one layer of graphite’s hexagons can be drawn flat. The suggested molecular models help: build butane from a kit, rotate it, then draw it flat.

Reactions you must predict (C6.2c)

Combustion. Complete combustion of any alkane, alkene or alcohol gives carbon dioxide and water.

Worked example. Balance the complete combustion of butan-1-ol, then find the mass of carbon dioxide from 3.70 g. Use C = 12, H = 1, O = 16.

1. 4 carbons give 4CO₂; 10 hydrogens give 5H₂O.
2. Oxygen on the right is 8 + 5 = 13; one is in the alcohol, so 6O₂: C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O
3. Mr = 48 + 10 + 16 = 74, so moles = 3.70 ÷ 74 = 0.0500 mol.
4. CO₂ = 4 × 0.0500 = 0.200 mol; mass = 0.200 × 44 = 8.80 g.

Addition of bromine. Bromine adds across the C=C bond: but-1-ene + Br₂ → 1,2-dibromobutane, C₄H₈Br₂. Orange bromine water turns colourless. Alkanes have no C=C bond, so the bromine water stays orange. This is the test for an alkene.

Addition of hydrogen. Heating an alkene with hydrogen over a nickel catalyst gives the alkane with the same carbon chain: C₃H₆ + H₂ → C₃H₈ (propene to propane).

Oxidation of alcohols. Warming an alcohol with acidified potassium manganate(VII) turns its –CH₂OH end into –COOH. The purple solution turns colourless. Using [O] for oxygen from the oxidising agent:

C₃H₇OH + 2[O] → C₂H₅COOH + H₂O (propan-1-ol to propanoic acid)

The acid keeps the alcohol’s carbon count. The same rules apply to “other given members” such as hex-1-ene (C6.2i).

Addition polymers (C6.2d, C6.2g)

In addition polymerisation, many small unsaturated molecules (monomers) join and the C=C bond in each opens up to link to its neighbours. There is only one product: the polymer. The specification uses the representation [repeat unit]ₙ.

Monomer to polymer. Change the C=C to a single bond, add a bond out of each side, bracket it and write n.

But-1-ene, CH₂=CH–CH₂CH₃, gives poly(but-1-ene):

  H   H
  |   |
—[C — C]ₙ—
  |   |
  H   C₂H₅

Polymer to monomer. Find the pair of backbone carbons inside the brackets, remove the bonds that stick out and put a double bond between them. A repeat unit –[CF₂–CF₂]– comes from tetrafluoroethene, CF₂=CF₂, and the polymer is poly(tetrafluoroethene), PTFE.

Worked example. One chain of PTFE has Mr 250 000. How many monomers joined? Use C = 12, F = 19.

1. Mr of C₂F₄ = 24 + 76 = 100.
2. Nothing is lost in addition polymerisation, so n = 250 000 ÷ 100 = 2500.

Condensation polymers (C6.2e, Higher Tier only)

In condensation polymerisation, monomers with two functional groups each react so that a small molecule, often water, forms at every link.

• Polyester: a diol (two –OH groups) plus a dicarboxylic acid (two –COOH groups). Each link is an ester link and releases H₂O.
• Polyamide (such as nylon): a diamine (two –NH₂ groups) plus a dicarboxylic acid. Each link is an amide link and releases H₂O.

Each monomer needs at least two functional groups, one for each side of the chain; with one, the chain stops.

Block diagrams hide the carbon chain. For a polyester from HO–●–OH and HOOC–■–COOH:

n HO–●–OH + n HOOC–■–COOH → –[O–●–O–OC–■–CO]ₙ– + 2n H₂O

Each repeat unit holds one of each monomer and two ester links, so every repeat unit added to a growing chain releases two small molecules. Addition polymerisation gives one product; condensation gives the polymer and a small molecule at the same time.

Making nylon (C6.2f, Higher Tier only)

The suggested practical is the “nylon rope trick”:

1. Put 1,6-diaminohexane dissolved in water into a small beaker.
2. Pour hexanedioyl dichloride in cyclohexane gently down a glass rod so it floats as a top layer. Do not stir.
3. Nylon forms as a film where the layers meet. Lift it with tweezers and wind it slowly onto a glass rod; more keeps forming.
4. Wash the nylon with water before handling it.

The acid is used as its dichloride, so the small molecule is hydrogen chloride, not water. Wear gloves and eye protection (both reactants are corrosive) and keep flames away from the flammable cyclohexane; use a fume cupboard.

Natural polymers (C6.2h)

• DNA is a polymer of four different monomers called nucleotides. Each is named after the base it contains: adenine (A), thymine (T), cytosine (C) and guanine (G).
• Starch, cellulose and glycogen are polymers built from sugars such as glucose.
• Proteins are polymers built from amino acids.

Crude oil and fractional distillation (C6.2j–C6.2n)

Crude oil is a mixture of hydrocarbons. Its fractions are largely alkanes, CₙH₂ₙ₊₂ (C6.2l).

Describe (C6.2j). Heated crude oil vapour enters near the bottom of a column with a temperature gradient: hot at its base, cooler towards its top. Vapours rise and each compound condenses on the tray where the temperature falls below its boiling point. Fractions are drawn off at different heights; the largest molecules leave at the bottom.

Fraction (top to bottom)	Main use
Refinery gases	bottled gas for heating and cooking
Petrol	car fuel
Kerosene	aircraft fuel
Diesel	lorry, bus and train fuel
Fuel oil	ships and some power stations
Bitumen	road surfaces and roofing

Explain (C6.2k). Larger molecules have stronger intermolecular forces, which need more energy to overcome, so they have higher boiling points and condense lower down. Covalent bonds inside the molecules are not broken.

Why it matters (C6.2m, C6.2n). Crude oil is a main source of hydrocarbons and the feedstock for the petrochemical industry (plastics, solvents, detergents, medicines). Transport, heating and much electricity generation depend on hydrocarbons. It formed over millions of years and is used far faster than it forms, so it is finite.

Cracking (C6.2o)

Fractional distillation gives more long-chain molecules than people want, and too few short ones. Cracking breaks long alkanes into shorter alkanes and alkenes. Conditions: vaporise the alkane and pass it over a hot catalyst; catalytic cracking runs at about 500 °C with zeolite catalysts. In the lab, liquid paraffin on mineral wool is heated, with the vapour passed over hot aluminium oxide.

Why: shorter alkanes are in demand as fuels such as petrol; alkenes are feedstock for polymers.

Worked example. Complete: C₁₅H₃₂ → C₉H₂₀ + 2 ____.

Carbon left: 15 − 9 = 6, so 3 each; hydrogen left: 32 − 20 = 12, so 6 each. The product is C₃H₆, propene.

Cells and fuel cells (C6.2p, C6.2q)

A chemical cell keeps giving a potential difference only until its reactants run out. A fuel cell is fed continuously with fuel and oxygen, so it keeps working while they flow in.

In the hydrogen/oxygen fuel cell with an acidic electrolyte:

• negative electrode: H₂ → 2H⁺ + 2e⁻
• positive electrode: O₂ + 4H⁺ + 4e⁻ → 2H₂O
• overall: 2H₂ + O₂ → 2H₂O

Worked example. What mass of water forms from 4.0 g of hydrogen? 4.0 ÷ 2 = 2.0 mol H₂, giving 2.0 mol H₂O, which is 2.0 × 18 = 36 g.

Advantages	Disadvantages
Only water is produced at the point of use	Hydrogen is flammable and hard to store
No carbon dioxide from the cell itself	Much hydrogen is made from fossil fuels
Runs as long as fuel is supplied	Fuel cells are expensive
Quiet, few moving parts	Few refuelling stations

Other fuel cells use fuels such as methanol: easier to store as a liquid, but it releases carbon dioxide. For a given use, weigh storage, cost, emissions and running time, then reach a judgement.

Common errors

• Giving C₂H₅OH when a displayed formula is asked for.
• Forgetting the carbon in –COOH when naming an acid.
• Saying bromine water “goes clear”. Describe the colour: it becomes colourless.
• Drawing an addition polymer with the C=C still in the chain.
• (Higher Tier) Missing the small molecule in condensation polymerisation.

Next steps

Background reading: the C2.3 materials guide for allotropes, the bonding guide for intermolecular forces and the C3.1 reactions guide for balancing. Then use the C6.2 revision notes and the C6.2 practice set.

Official syllabus

Published by OCR: Version 4.0 (August 2026) of specification J248, for the GCSE (9–1) in Chemistry A, Gateway Science, first assessed in 2018. Content taken from it: sub-topic C6.2 Organic chemistry, statements C6.2a–C6.2q, maths outcome CM6.2i, and the practical suggestions “Use of models” and “Making nylon”.