Stereoisomerism and E-Z Isomers Revision Notes | A Level AQA Chemistry

1. The Problem with Identical Formulae

You already know that structural isomers share a molecular formula but differ in how their atoms are connected. Now consider a more subtle situation: two molecules with identical molecular formulae and identical connectivity — yet they are genuinely different compounds with different physical properties. That is the territory of stereoisomerism.

The difference is not in which atoms are bonded together, but in where those atoms sit in space. This spatial distinction has real chemical consequences — different boiling points, different polarities, and in biological systems, dramatically different activities.

At A Level, the main type of stereoisomerism you need to master is E-Z isomerism. To understand why it exists, you need to look closely at what a carbon-carbon double bond actually prevents.

2. Why the Double Bond Locks Everything in Place

Picture a carbon-carbon single bond (C–C). The two carbon atoms can rotate freely around that bond axis — like two people holding hands and spinning. Substituents attached to each carbon can sweep through all possible positions relative to each other.

A carbon-carbon double bond (C=C) is fundamentally different. It consists of a sigma bond (end-on overlap) and a pi bond (sideways overlap of p orbitals above and below the plane). Rotating around the double bond would require breaking that pi bond — which costs around 260 kJ mol$^{-1}$ of energy. At room temperature, that rotation does not happen.

The result is restricted rotation: the four atoms directly attached to the two double-bond carbons are locked into a flat, planar arrangement. Whatever is on the left stays on the left; whatever is on the right stays on the right. This is the geometric origin of E-Z isomerism.

There is one further requirement before E-Z isomerism can exist: each carbon of the double bond must carry two different substituents. If both groups on one carbon are identical, swapping them makes no difference — there is only one possible arrangement, not two.

3. Introducing the CIP Priority Rules

Once you know a molecule can show E-Z isomerism, you need a reliable way to label which isomer is which. The old terms 'cis' and 'trans' work when both double-bond carbons share a common substituent (e.g., both carry H), but break down when all four substituents are different. The Cahn-Ingold-Prelog (CIP) priority rules solve this problem systematically.

The core idea is straightforward: look at the atom directly attached to each double-bond carbon and rank the two substituents by atomic number. Higher atomic number means higher priority. Then compare where the two higher-priority groups sit relative to each other across the double bond.

• Assign priority 1 (higher) and priority 2 (lower) to the two groups on each double-bond carbon, based on the atomic number of the atom directly attached.
• If the two higher-priority groups are on opposite sides of the double bond, the isomer is labelled E (from the German 'entgegen', meaning opposite).
• If the two higher-priority groups are on the same side of the double bond, the isomer is labelled Z (from the German 'zusammen', meaning together).
• If the directly attached atoms are the same element, move one bond further along each chain and compare again at that point.

Common atomic numbers you will use at A Level: H = 1, C = 6, N = 7, O = 8, F = 9, Cl = 17, Br = 35, I = 53. A bromine substituent always outranks a chlorine; a chlorine always outranks a carbon chain; a carbon chain always outranks hydrogen.

Worked example: Applying CIP Rules to But-2-ene

Assign E or Z configuration to the isomer of but-2-ene in which the two methyl groups (CH3) are on opposite sides of the double bond.

1. Carbon 2 carries: CH3 (left chain) and H Carbon 3 carries: CH3 (right chain) and H — But-2-ene is CH3-CH=CH-CH3. Each internal carbon of the double bond has one methyl group and one hydrogen attached.
2. CH3 group: directly attached atom is C (atomic number 6) H: atomic number 1 Priority order on C2: CH3 (priority 1) > H (priority 2) — Carbon (6) has a higher atomic number than hydrogen (1), so the methyl group is higher priority on this carbon.
3. CH3 group: directly attached atom is C (atomic number 6) H: atomic number 1 Priority order on C3: CH3 (priority 1) > H (priority 2) — The same logic applies to carbon 3. The methyl group is again higher priority than hydrogen.
4. Both CH3 groups (priority 1 on each carbon) are on opposite sides of the double bond. — The question states the methyl groups are on opposite sides. Higher-priority groups are therefore on opposite sides of the double bond.
5. Opposite sides → E configuration Name: (E)-but-2-ene — E comes from 'entgegen' (opposite). When the two higher-priority groups are on opposite sides, the label is E.

Answer: (E)-but-2-ene. The two methyl groups (higher priority on each carbon) are on opposite sides of the double bond.

Now consider the other isomer of but-2-ene, where the two methyl groups are on the same side. By the same CIP analysis, the higher-priority groups (both CH3) are on the same side — so this isomer is labelled (Z)-but-2-ene.

4. Drawing E and Z Isomers Correctly

In an exam, drawing E and Z isomers accurately is a skill that costs marks when done carelessly. The key is to represent the planar geometry of the double bond clearly, with substituents explicitly shown on the correct side.

The standard approach is to draw the C=C bond horizontally, then show the substituents on each carbon either above or below the bond. For the Z isomer of but-2-ene, both methyl groups appear on the same side (say, both above the double bond), with both hydrogens on the other side. For the E isomer, one methyl is above and one is below.

When drawing a more complex example — such as (Z)-1-bromo-1-chloropropene (CHBrCl=CHCH3) or a molecule with four different substituents — always assign CIP priorities first, then position the higher-priority groups deliberately before completing the rest of the structure. Rushing this step is the most common source of errors in exam drawings.

Worked example: Drawing the E and Z Isomers of 1-bromo-1-chloropropene

Draw the structural formulae of the E and Z isomers of 1-bromo-1-chloropropene (CHBrCl=CHCH3), clearly showing the geometry about the double bond.

1. Carbon 1: Br and Cl Carbon 2: CH3 and H — Carbon 1 carries two halogens; carbon 2 carries a methyl group and a hydrogen. All four substituents are different, so E-Z isomerism is possible.
2. Br: atomic number 35 Cl: atomic number 17 Priority order on C1: Br (priority 1) > Cl (priority 2) — Bromine (35) has a higher atomic number than chlorine (17), so Br is the higher-priority group on carbon 1.
3. CH3: directly attached atom is C (atomic number 6) H: atomic number 1 Priority order on C2: CH3 (priority 1) > H (priority 2) — Carbon (6) outranks hydrogen (1), so the methyl group is higher priority on carbon 2.
4. Place Br (priority 1 on C1) and CH3 (priority 1 on C2) on the SAME side of the double bond. Br CH3 \\ / C = C / \\ Cl H — Z means 'zusammen' (together) — the two higher-priority groups are on the same side. Here Br and CH3 are both above the double bond.
5. Place Br (priority 1 on C1) and CH3 (priority 1 on C2) on OPPOSITE sides of the double bond. Br H \\ / C = C / \\ Cl CH3 — E means 'entgegen' (opposite) — the two higher-priority groups are on opposite sides. Here Br is above and CH3 is below.

Answer: Z isomer: Br and CH3 on the same side of the C=C. E isomer: Br and CH3 on opposite sides of the C=C.

5. Applying CIP Rules When the First Atoms Are the Same

The straightforward cases involve substituents with different directly attached atoms — Br vs Cl, or C vs H. The trickier situation arises when both substituents on a carbon begin with the same element, such as an ethyl group (CH2CH3) and a methyl group (CH3), both starting with carbon.

When the directly attached atoms are the same, move one bond further along each chain and compare the atoms at that next position. For a methyl group, the carbon at position 1 is bonded to three hydrogens (H, H, H). For an ethyl group, the carbon at position 1 is bonded to one carbon and two hydrogens (C, H, H). Comparing the highest atom at this next shell: C (atomic number 6) outranks H (atomic number 1), so the ethyl group has higher priority than the methyl group.

This 'move along the chain' process continues until a point of difference is found. At A Level, you will rarely need to go beyond two shells, but the principle is the same at every level: compare the highest atomic number at the point of difference.

6. Recognising When E-Z Isomerism Is and Is Not Possible

A common exam task is to look at a series of alkenes and identify which ones can show E-Z isomerism. The test is always the same: does each carbon of the double bond carry two different substituents? If either carbon has two identical groups, the answer is no.

Notice that the presence of two different halogens on the same carbon (as in CHBrCl=) does not guarantee E-Z isomerism — you must check both carbons of the double bond independently.

7. Exam Technique and Common Pitfalls

E-Z isomerism questions appear in multiple formats at A Level: definition questions, drawing tasks, CIP priority assignments, and 'explain why' questions about restricted rotation. Each format has its own mark-winning requirements.

For a 'define stereoisomers' question, the mark scheme requires you to state both that the molecular formula is the same and that the connectivity (sequence of bonded atoms) is the same, with the difference being the spatial arrangement. Missing either component typically costs a mark.

For 'explain why E-Z isomerism arises' questions, the expected chain of reasoning is: the C=C double bond contains a pi bond formed by sideways overlap of p orbitals → rotation about the double bond would break the pi bond → this requires too much energy at room temperature → rotation is restricted → substituents are locked in fixed spatial positions → two distinct arrangements are possible if each carbon carries two different groups.