Distinguish changes in connectivity from changes in spatial arrangement, and apply CIP priorities to decide whether an alkene is E or Z.
Same molecular formula, different connectivity
Structural isomers have the same molecular formula but different structural formulae: the atoms are connected differently. Before classifying a pair, count every atom in both molecules. Similar-looking names do not establish isomerism.
| Type | What changes? | Example pair |
|---|---|---|
| Chain | Arrangement of the carbon skeleton | Pentane and 2-methylbutane: C₅H₁₂ |
| Position | Location of a group or multiple bond on the same skeleton | Pentan-1-ol and pentan-2-ol: C₅H₁₂O |
| Functional group | Type of functional group | Propanal and propanone: C₃H₆O |
Stereoisomerism and the two E/Z conditions
Stereoisomers have the same structural formula, with the same connectivity, but different arrangements of atoms in space. E/Z isomerism needs restricted rotation about C=C and two different substituents attached to each carbon of that double bond.
Each carbon at C=C has approximately trigonal planar geometry, with angles near 120°. Ordinary rotation about the double bond would disrupt its π bond. Compare the two groups on the left carbon with each other, and separately compare the two on the right. The molecule does not need four different groups overall.
Pent-1-ene cannot show E/Z because its terminal carbon has two H atoms. Pent-2-ene can: one double-bond carbon has H and CH₃; the other has H and CH₂CH₃. Optical isomerism is a separate later topic.
CIP: choose one priority group at each end
At each alkene carbon, compare the atomic numbers of the two atoms directly attached to it. Higher atomic number gives higher priority. Do not rank by the group’s size, total mass or alphabetical order. Compare the two ends independently.
If the first atoms are the same, compare the atoms bonded to each of those atoms, excluding the alkene carbon you came from. Arrange each list in descending atomic number and compare at the first difference. Continue outwards only if necessary. For these simple examples, ethyl outranks methyl: its list C,H,H beats H,H,H.
Once one priority group is identified at each end, Z has those groups on the same side of C=C; E has them on opposite sides. In the special case where both carbons also carry H, matching non-H groups can make cis/trans intuitive, but E/Z uses priorities and applies more widely.
Diagram placeholder
E and Z arrangements with priority labels
Labels to include:
- planar C=C
- left: Br above and CH₃ below
- right Z: Cl above and ethyl below
- right E: ethyl above and Cl below
- priority groups Br and Cl
Draw two C=C structures with four bonds extending outwards. Keep the left groups fixed, then exchange the right-hand positions between the two drawings. Label the drawing with Br and Cl above the bond Z, and the drawing with Br above and Cl below E. Swapping both ends would merely reorient the same arrangement.
Isomerism can change physical properties
Different spatial arrangements can change the way bond dipoles combine and the way molecules pack. In Z-1,2-dichloroethene the C–Cl bond dipoles give a net molecular dipole; in the E isomer the symmetric arrangement makes them cancel. Both have London forces, but the Z isomer also has permanent dipole–dipole attractions and has a higher boiling temperature.
Do not turn this example into a universal rule that every Z isomer has a higher boiling or melting point. Molecular shape, polarity and crystal packing all matter. Use the structures or data supplied in the question.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. Classify pentan-2-one and pentan-3-one as isomers.Show answer
They are position isomers: both have C₅H₁₀O and the same straight carbon skeleton, but the ketone carbonyl is at a different position.
Q2. Give the condensed formula of a functional-group isomer of butanal.Show answer
Butan-2-one, CH₃COCH₂CH₃, is one answer. Both have C₄H₈O; one is an aldehyde and the other a ketone.
Q3. Can CH₃C(CH₃)=CHCH₂CH₃ show E/Z isomerism? Explain.Show answer
No. The left carbon of the double bond has two identical methyl groups. Restricted rotation alone is insufficient.
Q4. One alkene carbon carries H and Br. The other carries CH₃ and CH₂CH₃. The Br and ethyl groups are opposite. Assign E or Z.Show answer
Br has priority over H. Ethyl has priority over methyl after comparing C,H,H with H,H,H. The priority groups are opposite, so the isomer is E.
Q5. A student draws a straight chain of five carbons, then bends it and calls the result another C₅H₁₂ isomer. Explain the error and name a genuine alternative.Show answer
Bending a chain without changing connectivity does not create a structural isomer. A genuine alternative is 2-methylbutane, CH₃CH(CH₃)CH₂CH₃; another is 2,2-dimethylpropane, C(CH₃)₄.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- Chemrevise: Introduction to organic chemistry — Coverage checklist, pp1–9. Explanations and practice on this site are original.
- AQA 7405 specification — 3.3.1: nomenclature, reaction mechanisms and isomerism.
- AQA June 2022 AS Paper 2 mark scheme — Q03, pp15–18: formula representations and isomers.
- AQA June 2022 AS Paper 2 examiner report — Q03, pp3–4: drawing and naming isomers.
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