E/Z Stereoisomerism in Alkenes

Spec Mapping — OCR H432 Module 4.1.3 — Alkenes, covering stereoisomerism arising from restricted rotation around the C=C double bond, the requirement that each double-bond carbon carries two different substituents for E/Z isomers to exist, the application of the Cahn–Ingold–Prelog (CIP) priority rules to assign atomic-number priorities at each carbon (with the next-atom tie-break), the difference between the modern E/Z convention (entgegen / zusammen — based on CIP priorities) and the legacy cis/trans labels (which work only for simple symmetrical cases), and the link between E/Z geometry and physical properties such as m.p., b.p. and dipole moment (refer to the official OCR H432 specification document for exact wording).

Because rotation around the C=C double bond is restricted (Lesson 7 — breaking the π bond costs ~265 kJ mol⁻¹, far more than thermal energy at 300 K), the four groups attached to the two doubly-bonded carbons are locked in place. When each double-bond carbon carries two different substituents, the molecule can exist as two distinct stereoisomers — the E-isomer and the Z-isomer — which are different chemical compounds with different physical properties, different biological activities and (sometimes) different industrial uses. The classification is rigorously defined by the Cahn–Ingold–Prelog (CIP) priority rules, formulated by Robert Cahn (1899-1981), Christopher Ingold (1893-1970) and Vladimir Prelog (1906-1998; Nobel 1975). The CIP framework replaced the older cis/trans nomenclature, which is unambiguous only when the substituents are symmetrical. This lesson develops the rules, the geometry, the worked examples, and the biological/industrial significance of E/Z assignment — which is one of the few A-Level topics that genuinely rewards a clean systematic approach over rote memorisation.

Key Definition — Stereoisomers: compounds with the same molecular formula and the same structural formula (i.e., the same atom connectivity), differing only in the three-dimensional arrangement of atoms in space. E/Z (geometric) isomers are stereoisomers that arise specifically from restricted rotation around a C=C double bond when each double-bond carbon carries two different substituents. Z = high-priority groups on the zame zide; E = high-priority groups on opposite sides.

1. Stereoisomerism — A reminder

Unlike structural isomers (Lesson 3, which differ in atom connectivity), stereoisomers have their atoms joined in the same order. It is only the three-dimensional arrangement that differs. At A-Level you meet two principal types of stereoisomerism: E/Z isomerism (the geometrical / cis-trans type, arising from restricted rotation about C=C) — covered in this lesson — and optical isomerism (arising from a chiral carbon, which gives mirror-image enantiomers) — covered in A2.

The key insight is that spatial arrangement is part of the molecule's identity. Two E/Z isomers are different compounds. They can be isolated separately. They have different m.p.s, b.p.s, dipole moments, IR spectra, NMR spectra, and (often) biological activities. They cannot be interconverted at room temperature without breaking the π bond first.

2. The Origin of E/Z isomerism

Because the C=C bond cannot rotate (Lesson 7), the groups attached to the double-bond carbons stay on the same side or opposite sides of the bond permanently. If those groups are different, the two arrangements are non-superimposable — they are different molecules.

2.1 Conditions for E/Z isomerism

For a molecule to exhibit E/Z isomerism, both of the following must be true:

It must contain a C=C double bond (restricted rotation).
Each carbon of the C=C must be attached to two different groups.

If either carbon of the C=C is attached to two identical groups (e.g., two hydrogens), E/Z isomerism is not possible — flipping the molecule horizontally produces the same arrangement.

2.2 Which molecules show E/Z?

Molecule	Formula	E/Z?	Reason
Ethene	CH₂=CH₂	No	Each C has two identical H
Propene	CH₂=CHCH₃	No	Left C (CH₂=) has two identical H
But-1-ene	CH₂=CHCH₂CH₃	No	Left C has two identical H
(Z)-but-2-ene	CH₃CH=CHCH₃	Yes	Each C has one H and one CH₃
1,1-dichloroethene	CCl₂=CH₂	No	Left C has two identical Cl
1,2-dichloroethene	CHCl=CHCl	Yes	Each C has one H and one Cl
2-methylbut-2-ene	(CH₃)₂C=CHCH₃	No	Left C has two identical CH₃
2-methylpent-2-ene	(CH₃)₂C=CHCH₂CH₃	No	Left C has two identical CH₃
Hex-2-ene	CH₃CH=CHCH₂CH₂CH₃	Yes	Each C has H + alkyl, where alkyl differs (CH₃ vs C₃H₇)
Stilbene (PhCH=CHPh)	(C₆H₅)CH=CH(C₆H₅)	Yes	Each C has H + phenyl

graph TD
  A[Alkene with C=C] --> B{Each C=C carbon has<br/>two different groups?}
  B -->|Yes| C[E/Z isomerism possible]
  B -->|No| D[No E/Z isomerism]
  C --> E[Apply CIP rules<br/>to assign priority on each C]
  E --> F[High-priority groups<br/>on same side of C=C]
  F --> G[Z-isomer]
  E --> H[High-priority groups<br/>on opposite sides]
  H --> I[E-isomer]

3. The old way — cis and trans

Before the IUPAC E/Z system, geometrical isomers were named cis (from Latin "on the same side") or trans ("across, on opposite sides"):

cis — the two higher-priority groups (often the two alkyl groups or the two "main chain" groups) are on the same side of the C=C.
trans — they are on opposite sides.

For simple molecules like but-2-ene, cis/trans still works unambiguously:

cis-but-2-ene = (Z)-but-2-ene: both methyls on the same side.
trans-but-2-ene = (E)-but-2-ene: methyls on opposite sides.

But for alkenes with three or four different substituents, cis/trans becomes ambiguous — which group counts as "the chain"? The IUPAC E/Z system, based on rigorous CIP priorities, gives an unambiguous name in every case and has therefore replaced cis/trans in modern nomenclature. OCR will accept cis/trans for simple cases but expects E/Z for anything beyond.

4. The Cahn–Ingold–Prelog (CIP) priority rules

The IUPAC CIP rules assign priority to the groups attached to each C=C carbon.

4.1 Rule 1 — Atomic number

Compare the atoms directly attached to the double-bond carbon. The one with the higher atomic number has higher priority. Priority (highest → lowest for common atoms):

$\text{I (53)} > \text{Br (35)} > \text{Cl (17)} > \text{S (16)} > \text{P (15)} > \text{F (9)} > \text{O (8)} > \text{N (7)} > \text{C (6)} > \text{H (1)}$

Examples:

Br beats Cl, which beats C, which beats H.
O beats N, which beats C. So –OH (O is the first atom) beats –NH₂ (N first) beats –CH₃ (C first).
The whole comparison ignores the size of the substituent — only the first atom matters in Rule 1.

4.2 Rule 2 — Next atoms (if Rule 1 is a tie)

If the two atoms directly attached are the same, move to the next atoms outward along each group. Build a sorted-descending triplet (a,b,c) of the three atoms each next-out carbon is attached to. Compare the triplets lexicographically.

Example: compare –CH₂CH₃ (ethyl) and –CH₃ (methyl). Both have C directly attached, so Rule 1 ties.

Ethyl: next-out C is attached to (C, H, H).
Methyl: next-out C is attached to (H, H, H).
(C, H, H) > (H, H, H) in the first position. Ethyl beats methyl.

Another: compare –CH(CH₃)₂ (isopropyl) and –CH₂CH₂CH₃ (n-propyl). Both have C first. Next out:

Isopropyl C is attached to (C, C, H) — two methyl branches plus the implicit H.
n-propyl C is attached to (C, H, H) — one CH₂ branch plus two H.
(C, C, H) > (C, H, H). Isopropyl beats n-propyl.

4.3 Rule 3 — Double/triple bonds

A double bond to an atom counts as two single bonds to that atom (once for each carbon of the C=C, "phantom" atoms supplied as needed). A triple bond counts as three.

Examples:

An aldehyde –CHO has C attached to (O, O, H) for CIP purposes — the C=O double bond counts as two single bonds to O.
A carboxylic acid –COOH has C attached to (O, O, O) — the C=O is two O's, plus the O–H is one more O.
An alkene side-group like –CH=CH₂ has C attached to (C, C, H) — the C=C counts as two C's, plus the implicit H.
A nitrile –C≡N has the C attached to (N, N, N).

Consequence: –COOH (C attached to O, O, O) easily beats –CH₂OH (C attached to O, H, H) even though both contain one O atom further out, because the carboxylic acid's first C has three effective O neighbours.

4.4 Assignment of E or Z

Once priorities are assigned on both carbons of the C=C, look at the geometry:

Z isomer (from German zusammen = "together"): the two higher-priority groups are on the same side of the C=C.
E isomer (from German entgegen = "opposite"): the two higher-priority groups are on opposite sides of the C=C.

Memory aid: Z = Zame Zide.

5. Worked Examples

5.1 But-2-ene (the textbook case)

CH₃CH=CHCH₃. On each double-bond carbon, the attached groups are CH₃ and H. CH₃ (C is the first atom) beats H. So the high-priority groups are the two methyls.

Methyls on the same side → (Z)-but-2-ene ( = cis-but-2-ene), m.p. −139 °C, b.p. +4 °C.
Methyls on opposite sides → (E)-but-2-ene ( = trans-but-2-ene), m.p. −105 °C, b.p. +1 °C.

5.2 1,2-Dichloroethene

CHCl=CHCl. Each C has Cl and H. Cl (17) > H (1), so Cl is high-priority on both.

Cls on the same side → (Z)-1,2-dichloroethene (μ ≈ 1.9 D, polar, b.p. 60 °C).
Cls on opposite sides → (E)-1,2-dichloroethene (μ = 0 by symmetry, non-polar, b.p. 48 °C).

Note the b.p. ordering — the Z isomer has the higher dipole moment, larger permanent dipole–dipole forces, and the higher b.p. The E isomer has μ = 0 by symmetry (the two C–Cl bond dipoles cancel) — and that is the classic question OCR loves to set.

5.3 But-2-enedioic acid — maleic vs fumaric acid

HOOC–CH=CH–COOH. Each C carries –COOH and –H. COOH (C, with phantom O,O,O) beats H. So:

(Z) = maleic acid: m.p. 130 °C (anhydride forms easily because the two COOHs are close).
(E) = fumaric acid: m.p. 287 °C — much higher (better symmetric packing in the solid).

The dramatic m.p. difference is the classic example of the rule "E isomer packs better than Z, hence higher m.p." Fumaric acid is also a key intermediate in the Krebs (citric acid) cycle in cellular metabolism.

5.4 1-bromo-1-chloropropene (a tricky case)

CHBr=C(Cl)CH₃. Left C carries H and Br → high is Br. Right C carries Cl and CH₃ → high is Cl (Cl=17 > C=6). So the high-priority groups are Br and Cl.

Br and Cl on the same side → (Z).
Br and Cl on opposite sides → (E).

(Note that cis/trans is ambiguous here — which is "the main chain"? CIP gives a clear answer.)

5.5 But-2-enoic acid — the COOH tie-break

CH₃CH=CHCOOH. Left C: CH₃ and H, high is CH₃. Right C: COOH and H, high is COOH. CIP says COOH beats H (C attached to O,O,O beats nothing).

CH₃ and COOH on same side → (Z), m.p. 15 °C, b.p. 169 °C — also called isocrotonic acid.
CH₃ and COOH on opposite sides → (E), m.p. 72 °C, b.p. 185 °C — crotonic acid, the more stable isomer.

5.6 Retinal — Z/E in vision

11-cis-retinal in the retina absorbs a UV/visible photon, breaks the π bond temporarily, and rotates to 11-trans-retinal. The geometry change triggers the protein opsin to send a nerve signal — the primary photochemical event of vision. The π-bond-breaking energy (~265 kJ mol⁻¹) is supplied by a single photon at ~450 nm (~265 kJ mol⁻¹ — the wavelengths match perfectly). The 11-trans isomer subsequently isomerises back to 11-cis enzymatically (rhodopsin cycle).

6. Physical differences between E and Z isomers

Although E and Z isomers share the same molecular formula and atom connectivity, they have different shapes and therefore different physical properties:

Property	Z (cis-like)	E (trans-like)
Shape	Bent, less symmetric	Straight, more symmetric
Packing in solid	Less efficient (bent shape)	More efficient (straight shape)
Melting point	Lower	Higher
Dipole moment (if polar groups)	Usually larger	Usually smaller (zero by symmetry for some cases)
Boiling point	Slightly higher (bigger dipole → stronger dipole-dipole)	Slightly lower
Density	Similar	Similar
Reactivity	Often similar (same C=C)	Often similar (same C=C)

Lesson 8: E/Z Stereoisomerism in Alkenes