Isomerism — AQA A-Level Chemistry
Test yourself on Isomerism with AQA A-Level practice questions.
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Isomerism explained
Structural isomers have the same molecular formula but their atoms are joined in a different order.
Read the full explanation
There are three kinds and the exam expects you to name the one you have drawn. Chain isomers differ in the carbon skeleton: butane and 2-methylpropane are both C₄H₁₀, one straight and one branched. Position isomers keep the skeleton and move the functional group: propan-1-ol and propan-2-ol are both C₃H₈O. Functional group isomers have a different functional group altogether while keeping the same formula, so propan-1-ol and methoxyethane are again both C₃H₈O, and propanal and propanone are both C₃H₆O. Because of isomerism, a molecular formula often cannot identify a compound, though small molecules like CH₄ or C₂H₆ are unique.
Stereoisomerism.
Stereoisomers have the same structural formula, the same atoms joined to the same atoms, but a different arrangement in space. Two kinds appear on this course. E-Z isomerism arises where rotation is restricted, most often about a C=C double bond, so two groups fixed on the same side differ from two fixed on opposite sides. Optical isomerism arises where a carbon atom carries four different groups, making the molecule and its mirror image non-superimposable. The distinction from structural isomerism is worth being precise about: in structural isomers the connectivity changes, in stereoisomers only the geometry does. Square planar and octahedral transition metal complexes show the same phenomenon as cis-trans isomerism.
E–Z isomerism is a form of stereoisomerism and occurs as a result of restricted rotation about the planar carbon– carbon double bond.
A C=C double bond consists of a σ bond and a π bond, the π bond formed by sideways overlap of p orbitals above and below the axis. Rotating one end of the bond would break that overlap, so rotation is restricted and the attached groups stay where they are. The two carbon atoms and the four atoms attached to them all lie in one plane. That fixes two distinct arrangements whenever each doubly bonded carbon carries two different groups: but-2-ene exists as Z-but-2-ene, with the methyl groups on the same side, and E-but-2-ene, with them on opposite sides. A mark scheme for exactly this question rewards naming both isomers and then stating that there is no free rotation about the C=C.
Cahn–Ingold–Prelog (CIP) priority rules.
CIP rules decide which group on each end of a double bond takes priority, and therefore whether the isomer is E or Z. Look at the atom directly attached to each doubly bonded carbon and compare atomic numbers: the higher atomic number wins, so Br beats Cl, which beats O, then N, then C, then H. If the two attached atoms are the same element, move one atom further out along each branch and compare the sets of atoms you find, taking the first point of difference; that is how CH₂CH₃ outranks CH₃, because the set C, H, H beats H, H, H. An atom joined by a double bond is counted twice. Z means both higher priority groups on the same side and E means on opposite sides.
Students should be able to: define the term structural isomer draw the structures of chain, position and functional group isomers define the term stereoisomer draw the structural formulas of E and Z isomers apply the CIP priority rules to E and Z isomers.
This statement is the checklist for the whole isomerism topic, and C₄H₈ exercises all of it. As structural isomers you can draw but-1-ene and but-2-ene as position isomers, but-1-ene and 2-methylpropene as chain isomers, and cyclobutane as a functional group isomer with no double bond at all. As stereoisomers you can draw E-but-2-ene and Z-but-2-ene, which share a structural formula but differ in spatial arrangement. Applying CIP to but-2-ene is straightforward, because each doubly bonded carbon carries CH₃ and H, so the methyl group takes priority at both ends and Z has them on the same side. The definitions must be kept precise: structural isomers share a molecular formula, stereoisomers share a structural formula.
Your focus
- Define a structural isomer in terms of molecular formula and connectivity.
- Draw a pair of chain, a pair of position and a pair of functional group isomers for a given formula.
- Classify a given pair of structures as chain, position or functional group isomers.
Show all 15 objectives
- Define stereoisomerism and distinguish it from structural isomerism.
- Identify whether a given molecule can show E-Z isomerism, optical isomerism or neither.
- Draw two stereoisomers of a named compound so that the spatial difference is clear.
- Explain restricted rotation about a C=C in terms of the π bond formed by sideways p orbital overlap.
- Deduce whether a named alkene can show E-Z isomerism by inspecting the groups on each double bond carbon.
- Name both isomers of a given alkene and draw them showing the planar arrangement.
- Apply CIP rules to assign priorities to the two groups on each carbon of a C=C.
- Deduce whether a drawn alkene is the E or the Z isomer and name it in full.
- Resolve a priority tie by comparing the next set of atoms along each branch.
- Define structural isomer and stereoisomer in ways that clearly distinguish the two.
- Draw chain, position and functional group isomers for a given molecular formula such as C₄H₈.
- Apply CIP priority rules to label a drawn alkene as the E or the Z isomer.
Isomerism exam tips
Quick Revision Summary (Key Takeaway)
Isomerism in AQA A-Level Chemistry describes compounds with the same molecular formula but different arrangements of atoms, split into structural isomerism and stereoisomerism. Stereoisomerism encompasses E/Z geometrical isomerism caused by restricted C=C bond rotation and optical isomerism featuring non-superimposable mirror-image enantiomers around a chiral centre.
Topic Overview
Isomerism is a foundational topic in AQA A-Level Chemistry that explains how compounds sharing identical molecular formulae can possess drastically different physical, chemical, and biological properties. It is split into structural isomerism (chain, positional, and functional group) and stereoisomerism (geometrical E/Z and optical isomerism).
Mastering isomerism bridges organic reaction mechanisms and synthetic pathways, explaining why reaction conditions often lead to racemic mixtures or specific diastereomers. These principles are vital for understanding biochemistry and the pharmaceutical industry, where enantiomers can show vastly different pharmacological activities.
Key Concepts
- →Structural isomerism occurs when molecules possess identical molecular formulae but differ in the order in which atoms are connected (chain, position, and functional group isomers).
- →E/Z isomerism requires restricted rotation about a rigid C=C bond where each carbon in the double bond is bonded to two different atoms or groups.
- →The Cahn-Ingold-Prelog (CIP) priority rules assign high/low priority based on descending atomic number of directly bonded atoms; if identical on the same side, the molecule is Z (zusammen), and on opposite sides, it is E (entgegen).
- →Optical isomerism requires an asymmetric carbon atom (chiral centre) bonded to four different groups, producing two non-superimposable mirror-image enantiomers that rotate plane-polarised light in equal and opposite directions.
Marking Points
- one mark for the definition, compounds with the same molecular formula but a different structural formula
- one mark for a correctly drawn pair of chain, position or functional group isomers
- one mark for naming which of the three types the drawn pair illustrates
- one mark for showing that both structures have the same molecular formula
- one mark for defining stereoisomers as having the same structural formula but a different arrangement of atoms in space
- one mark for naming the sub-type, E-Z or optical, that the example shows
- one mark for identifying the structural feature responsible, a restricted C=C or a carbon with four different groups
- one mark for a drawing that makes the spatial difference unambiguous
- one mark for stating that there is restricted or no free rotation about the C=C double bond
- one mark for naming both isomers, such as Z-but-2-ene and E-but-2-ene
- one mark for stating that each carbon of the double bond must carry two different groups
- one mark for referring to the π bond or the planar arrangement as the reason rotation is restricted
- one mark for assigning priority at each carbon of the double bond by the atomic number of the attached atom
- one mark for resolving a tie by moving outwards to the next atoms and using the first point of difference
- one mark for the correct E or Z label deduced from the positions of the two higher priority groups
- one mark for the full name including the E or Z prefix and the correct locants
- one mark for defining structural isomers as having the same molecular formula but a different structural formula
- one mark for defining stereoisomers as having the same structural formula but a different arrangement of atoms in space
- one mark for each correctly drawn isomer in a required set, with the molecular formula unchanged
- one mark for an E or Z label justified by CIP priorities rather than by appearance
Examiner Tips
- 💡Count the atoms in both structures before claiming that they are isomers.
- 💡Name the type of isomerism even when the question only says draw, because a follow-up mark often needs it.
- 💡Redraw a suspicious pair with the longest chain horizontal, since identical molecules then become obvious.
- 💡Say same structural formula in the definition, because same molecular formula is not enough.
- 💡Check that each doubly bonded carbon carries two different groups before claiming E-Z isomerism.
- 💡Use clear wedge and dashed bonds when drawing optical isomers.
- 💡Write the phrase restricted rotation about the C=C, because that is the marking point.
- 💡Check both ends of the double bond for two different groups before you answer.
- 💡Give both isomer names when a question asks you to deduce names, not just one.
- 💡Circle the two atoms directly attached at each end before comparing anything else.
- 💡Write the priority you assigned next to each group so the examiner can follow the reasoning.
- 💡Remember that Z means together on the same side and E means opposite sides.
- 💡Learn the two definitions as a contrasting pair so that you never mix them up under pressure.
- 💡When asked for a fixed number of isomers, count the atoms in each one before you stop.
- 💡Show your priority assignments on the diagram whenever you apply CIP rules.
- 💡When drawing enantiomers in an exam, always use standard 3D wedge-and-dash notation around a tetrahedral central carbon, and draw a vertical dashed line representing the mirror plane.
- 💡Clearly state 'restricted rotation about the C=C double bond' rather than just 'there is a double bond' when explaining why E/Z isomerism occurs.
- 💡For CIP priority ties, trace along the longest path of highest atomic number atom by atom until the first point of difference is identified.
Common Mistakes
- drawing the same molecule twice, simply bent or rotated on the page
- drawing a structure with the wrong number of hydrogen atoms, so that it is not actually an isomer
- calling a pair position isomers when the carbon skeleton has changed as well
- offering E and Z forms as structural isomers, when they are stereoisomers
- defining stereoisomers as having the same molecular formula only, which also describes structural isomers
- claiming any alkene shows E-Z isomerism, when two identical groups on one carbon prevent it
- drawing two supposed stereoisomers that are in fact the same molecule rotated
- identifying a chiral centre on a carbon atom that carries two identical groups
- saying the double bond is too strong to rotate rather than that rotation would break the π overlap
- claiming but-1-ene shows E-Z isomerism, when one of its double bond carbons carries two hydrogen atoms
- swapping the labels so that groups on the same side are called E
- describing the two isomers as structural isomers rather than stereoisomers
- assigning priority by the size or mass of the whole group rather than by the atomic number of the attached atom
- using cis and trans labels when all four groups differ and CIP priorities are required
- comparing only the first atom and stopping, so failing to separate CH₃ from CH₂CH₃
- assigning Z because two groups look similar rather than because both are higher priority
- using the same definition for structural isomers and stereoisomers, which loses one of the two marks
- drawing a set of isomers in which one has a different molecular formula from the rest
- repeating the same structure in a list of isomers, drawn from a different angle
- labelling E and Z without assigning priorities, so guessing wrongly when the four groups all differ
- Thinking cis/trans is identical to E/Z notation; cis/trans only works when identical groups are present on both carbons, whereas CIP E/Z rules apply universally to any tri- or tetra-substituted alkene.
- Believing that an optical isomer rotates ordinary light; optical activity strictly refers to rotating the plane of plane-polarised light.
- Assuming functional group isomers share the same chemical properties; functional group isomers belong to distinct homologous series with entirely different functional groups (e.g. propanal and propanone).
Revision Plan
- 1Day 1-2: Review and practice drawing the three types of structural isomerism (chain, position, functional group) for molecules up to C6.
- 2Day 3-4: Master E/Z stereoisomerism by working through CIP priority rule assignments for complex multi-substituted alkenes.
- 3Day 5-6: Practise 3D wedge-and-dash diagrams for optical isomers, identifying chiral centres and explaining polarimetry and racemic mixtures.
- 4Day 7: Complete past AQA exam questions on mechanisms producing racemic mixtures (e.g. nucleophilic addition of HCN to aldehydes).
Exam Question Types
- 📋Identifying and drawing all structural isomers for a given molecular formula (e.g. C4H8O2).
- 📋Assigning E/Z configurations using CIP rules to substituted alkenes with justifications.
- 📋Explaining why a specific reaction mechanism (e.g. carbonyl nucleophilic addition) yields an optically inactive racemic mixture.
- 📋3D drawing of enantiomeric pairs showing non-superimposable mirror images around a chiral carbon.
Command Word Expectations (AQA)
Provide a reasoned explanation using correct scientific terminology (e.g. 'restricted rotation around C=C', 'planar intermediate attacked from both sides with equal probability').
Provide clear, unambiguous skeletal, displayed, or 3D stereochemical formulae (using wedges and dashes for chiral centres) that clearly show all bonds.
Use supplied information or chemical principles to arrive at an isomer structure or configuration without directly quoting the question.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: Explain whether pent-2-ene exhibits stereoisomerism. If so, draw and name both stereoisomers using CIP priority rules, clearly stating why they arise.
- 1.Step 1: Check requirements for stereoisomerism in alkenes: restricted rotation around the planar C=C double bond, and each carbon of the C=C bond must be attached to two different groups.
- 2.Step 2: Analyse carbon-2: bonded to -H (atomic number 1) and -CH3 (directly bonded C, atomic number 6). -CH3 has higher priority than -H.
- 3.Step 3: Analyse carbon-3: bonded to -H (atomic number 1) and -CH2CH3 (directly bonded C, atomic number 6). -CH2CH3 has higher priority than -H.
- 4.Step 4: Assign E/Z nomenclature: when the two higher priority groups (-CH3 and -CH2CH3) are on the same side of the double bond, it is (Z)-pent-2-ene. When on opposite sides, it is (E)-pent-2-ene.
Question: Butan-2-ol is formed by the reaction of aqueous sodium hydroxide with (±)-2-bromobutane via an SN1 mechanism. Explain whether the product solution will rotate plane-polarised light.
- 1.Step 1: Identify that the starting material (±)-2-bromobutane is a racemic mixture, or identify that an SN1 mechanism forms a planar carbocation intermediate: CH3-C(+)H-CH2CH3.
- 2.Step 2: Recognize that the planar carbocation intermediate can be attacked by the nucleophile (OH-) with equal probability from either side (above or below the plane).
- 3.Step 3: Conclude that attack from both faces in equal amounts generates an equimolar mixture of both enantiomers of butan-2-ol.
- 4.Step 4: State that this racemic mixture exhibits no net rotation because the equal and opposite rotations cancel out completely.