Did you know? But-2-ene can be cis or trans, but but-1-ene cannot. Both have a carbon–carbon double bond; the difference is that one double-bonded carbon in but-1-ene carries two identical hydrogen atoms.
This original JC2 H2 Chemistry tutor guide teaches the two-different-groups test, the restricted π-bond rotation that makes geometrical isomerism possible, and how to distinguish real cis/trans pairs from merely redrawn or structurally different alkenes. It matches the Singapore–Cambridge 2027 syllabus 9476, which explicitly does not require E/Z nomenclature at H2.
All structures are conceptual paper examples. No experimental reaction, industrial preparation or laboratory manipulation is proposed. The immutable eduKateSG Clementi teaching reference and Hougang Chemistry library provide the lesson organisation.
EDUKATESG · HOUGANG · JC2 H2 CHEMISTRY
Choose your learning step
15 chapters · Alkene isomer matrix · Transfer practice · FAQs · 2027 syllabus
Full chapter index
- 1. Begin with the surprising difference between but-1-ene and but-2-ene
- 2. Understand why a double bond restricts free rotation
- 3. Apply the two-different-groups test to the first carbon
- 4. Apply the same test to the other carbon
- 5. Define cis and trans for straightforward matching groups
- 6. Use 1,2-dichloroethene to test structural reasoning
- 7. Distinguish geometrical from structural isomerism
- 8. Compare cis and trans physical properties without inventing a universal rule
- 9. Understand why cis–trans is not identical to chirality
- 10. Explain why ordinary single-bond rotation does not preserve a cis/trans label
- 11. Read drawings that have been rotated on the page
- 12. Use a two-column checklist for predicting whether cis/trans exists
- 13. Distinguish H2 cis/trans from optional E/Z nomenclature
- 14. Correct the most frequent false stereoisomer claims
- 15. Transfer the test to an unfamiliar carbon skeleton
1. Begin with the surprising difference between but-1-ene and but-2-ene
But-1-ene and but-2-ene have the same molecular formula C₄H₈, but the position of the double bond differs. In but-1-ene, CH₂=CHCH₂CH₃, the terminal double-bond carbon carries two identical hydrogen atoms. In but-2-ene, CH₃CH=CHCH₃, each double-bond carbon has one hydrogen and one methyl substituent.
That structural distinction has a stereochemical consequence. But-2-ene can exist as cis and trans forms, while but-1-ene cannot form a comparable pair at its C=C bond. The important reason is not merely that one double bond lies in the middle of the chain. It is the identities of the two substituents attached to each of the double-bonded carbon atoms.
The 2027 H2 Chemistry 9476 syllabus explicitly requires explaining cis–trans isomerism in alkenes in terms of restricted rotation due to the π bond. It also explicitly states that E/Z nomenclature is not required at H2. This guide therefore teaches the required cis–trans structure checks rather than making advanced priority nomenclature a prerequisite.
Students should develop a decision method that works on an unfamiliar structural formula. First find the C=C bond. Next inspect one double-bond carbon and list its two substituents. Then inspect the other carbon. Only after both pass the two-different-groups test should the student consider whether two spatial arrangements exist.
Student prompt: Why does but-2-ene form a cis–trans pair while but-1-ene does not?
What a correct answer shows: Both double-bond carbons in but-2-ene have two different substituents; one carbon in but-1-ene has two hydrogens.
2. Understand why a double bond restricts free rotation
A carbon–carbon double bond contains one σ bond and one π bond in the standard H2 bonding description. The σ interaction lies along the line connecting the nuclei, while the π interaction involves sideways overlap of suitable p orbitals above and below the main bonding framework.
Rotation about an ordinary carbon–carbon single σ bond can often occur without destroying the σ overlap, although conformational barriers may exist. In contrast, free twisting around the C=C bond would disrupt the sideways p-orbital overlap that produces the π bond. This introduces a substantial energy barrier to interconversion between the two alkene geometries.
As a result, two differently arranged alkene groups can be sufficiently persistent to represent distinct geometrical isomers under ordinary conditions. They are not simply two snapshots of a molecule that rotates freely through every possible angle. This is why cis- and trans-but-2-ene are different chemical species with the same bond connectivity.
The high-level explanation should avoid saying that the C=C atoms are physically welded in a perfect unchanging line. Molecules vibrate and can undergo reactions, but the π bond makes ordinary free rotation very different from that of a single bond. The chemical barrier, not absolute immobility, is the relevant concept.
Student prompt: What bonding feature prevents unrestricted rotation around a C=C bond?
What a correct answer shows: Rotating would disrupt the π overlap associated with the double bond.
3. Apply the two-different-groups test to the first carbon
Look at one end of an alkene double bond and identify its two attached substituents other than the carbon at the opposite end of the double bond. If the two substituents are identical, changing their spatial positions cannot produce a distinct geometric arrangement at that end.
For the terminal carbon of but-1-ene, the two groups are H and H. They are identical, so the double bond fails the stereochemical test immediately. Even if the other double-bond carbon has two different groups, the two arrangements at the terminal end cannot be distinguished as cis and trans.
For a double-bond carbon that has CH₃ and H, the groups differ. That passes the test at that carbon but does not yet prove the whole alkene can form a pair; the second double-bond carbon must also pass. A tutor should not stop after examining the left half of a structural drawing.
Students sometimes confuse the double-bond partner itself with a substituent that should be counted. The test applies to the two groups bonded to each sp² carbon in addition to its bond with the other alkene carbon. This precise definition removes a common source of false positives.
Student prompt: Why does a CH₂= terminal group fail the cis–trans test?
What a correct answer shows: Its double-bond carbon has two identical hydrogen substituents.
4. Apply the same test to the other carbon
The second alkene carbon must also have two distinct substituents for ordinary geometrical isomerism across the double bond. In but-2-ene, both carbon centres have one H and one CH₃, so the structure meets the criterion at both ends.
Consider 2-methylbut-2-ene, which can be represented as (CH₃)₂C=CHCH₃. One double-bond carbon is attached to two CH₃ groups. Although the other carbon carries H and CH₃, the repeated methyl groups on the first carbon make cis–trans isomerism impossible for that particular C=C bond.
This illustrates why a longer or more substituted alkene does not automatically have more stereoisomers. Increasing carbon count may introduce new structural possibilities, but the stereochemical test is local to the substituents directly attached to the two alkene carbons.
A useful parent-facing tutoring question is to ask the child to write a pair of substituents beside each alkene carbon. If either pair consists of identical groups, stop and explain why no distinct cis–trans pair arises. This is more reliable than trying to visualise a complicated molecule before checking the rule.
Student prompt: Why can 2-methylbut-2-ene not form a cis–trans pair?
What a correct answer shows: One of its C=C carbons carries two identical methyl groups.
5. Define cis and trans for straightforward matching groups
For but-2-ene, the two methyl groups provide a simple visual reference. In cis-but-2-ene, the methyl groups lie on the same side of the double-bond arrangement. In trans-but-2-ene, they lie on opposite sides. The hydrogens necessarily show the matching opposite pattern.
The terms same side and opposite side refer to the relative spatial arrangement across the C=C unit when represented in an appropriate planar alkene drawing. They are not labels for whether a substituent appears near the top of the printed page after an arbitrary rotation of the entire picture. Rotating the paper does not convert one isomer into the other.
An equivalent reference in 1,2-dichloroethene is the two chlorine atoms: cis when the chlorine substituents lie on the same side, trans when they lie on opposite sides. The starting structures must first pass the two-different-groups criterion on both alkene carbons.
With four different substituents and no identical comparison group across the double bond, simple cis/trans labels may be ambiguous. The more general E/Z system handles such cases through priority rules, but its nomenclature is explicitly not required by H2 9476. Tutors may identify it as optional enrichment rather than demand it for every H2 problem.
Student prompt: What is the cis arrangement of but-2-ene?
What a correct answer shows: The two CH₃ substituents are on the same side of the C=C arrangement.
6. Use 1,2-dichloroethene to test structural reasoning
1,2-Dichloroethene can be represented as CHCl=CHCl. Each double-bond carbon carries one hydrogen and one chlorine atom, so both carbon ends have different substituents. Thus the structure can give cis and trans forms, with the two chlorines on the same or opposite sides, respectively.
Contrast 1,1-dichloroethene, CH₂=CCl₂. One double-bond carbon has two hydrogens and the other has two chlorines. Neither end satisfies the two-different-groups requirement. Therefore the molecule does not possess the corresponding cis–trans pair, even though its formula also contains two chlorine atoms.
These isomers have related formulas, but their connectivity differs before one even discusses the cis–trans relationship. A tutor should name structural isomerism for the different placement of chlorine atoms, and geometrical isomerism for the two spatial arrangements of 1,2-dichloroethene.
Students should be able to prove both conclusions from the four substituent slots without relying on molecular models or memorised images. This is especially valuable when a formula is drawn in a zigzag arrangement that can be rotated on the page.
Student prompt: Which can give cis/trans forms: CHCl=CHCl or CH₂=CCl₂?
What a correct answer shows: CHCl=CHCl; each alkene carbon has H and Cl, whereas CH₂=CCl₂ has repeated groups at both ends.
7. Distinguish geometrical from structural isomerism
Structural isomers share the same molecular formula but differ in how their atoms are bonded. But-1-ene and but-2-ene are position isomers because the C=C bond occurs at different positions along the carbon chain. Cis- and trans-but-2-ene, however, share the same bonding connectivity and differ in spatial arrangement.
The distinction helps prevent students from counting but-1-ene as the trans version of but-2-ene. They do not represent the same connectivity. A chemical name cannot be assigned simply from the position of methyl groups on a drawing if the position of the double bond has also changed.
Another example arises from the dichloroethene pair. 1,1-dichloroethene and 1,2-dichloroethene differ in chlorine placement along the alkene carbon framework, so they are constitutional or structural isomers. Within 1,2-dichloroethene, cis and trans differ only geometrically.
The 2027 H2 syllabus includes both structural and cis–trans isomerism, so a good examination preparation exercise should require students to name the correct category before listing possible compounds. More isomers are not necessarily better if the proposed structures duplicate the same molecule after a simple rotation.
Student prompt: Are but-1-ene and cis-but-2-ene geometrical isomers?
What a correct answer shows: No. They differ in C=C position, so their connectivity differs.
8. Compare cis and trans physical properties without inventing a universal rule
Distinct geometrical isomers can have different physical properties because their overall shapes and molecular dipoles differ, affecting intermolecular interactions and packing. The differences may include boiling point, melting point and solubility under suitable conditions. Identical molecular formulas do not guarantee identical physical behaviour.
Consider cis- and trans-1,2-dichloroethene. Their C–Cl bond dipoles have different spatial arrangements: the cis form generally has a larger net molecular dipole, while the trans form is more symmetrical, allowing substantial cancellation. This can help explain differences in the liquids’ physical behaviour.
However, a claim that cis always has a higher boiling point or that trans always has a higher melting point for every molecule would be unsound. Properties depend on molecular structure, shape, packing and the full collection of intermolecular forces. A specific numerical comparison needs measured or supplied data.
A tutor can ask learners to infer qualitative polarity from a correctly drawn structure and then test the inference against an explicitly supplied property table. The goal is to connect geometry with physics while respecting the fact that one simple dipole argument is not a universal predictor of all thermodynamic properties.
Student prompt: Can cis and trans forms have different boiling points despite having the same formula?
What a correct answer shows: Yes, but the direction and magnitude depend on molecular interactions and cannot be assumed universally.
9. Understand why cis–trans is not identical to chirality
Cis–trans isomerism across an alkene arises because different substituent arrangements cannot be interconverted by free rotation around a π-bond-containing double bond. Chirality, in familiar H2 examples, involves a molecule lacking a plane of symmetry and possessing non-superimposable mirror-image arrangements, often around a tetrahedral carbon with four different groups.
Ordinary cis- and trans-but-2-ene are not a pair of non-superimposable mirror images of one another. They are geometrical isomers with distinct spatial arrangements, but the familiar pair is not a pair of enantiomers. Confusing these categories leads to incorrect claims about optical rotation.
The H2 9476 syllabus separately covers enantiomerism and the optical behaviour of chiral molecules. The term diastereomers is explicitly not required at H2, even though advanced stereochemistry uses that terminology for many non-mirror-image stereoisomers. Tutors should not force unfamiliar category jargon into a cis/trans explanation when a simpler correct description is sufficient.
In a written answer, define which stereochemical feature is responsible for the pair. If the structure has a C=C with valid substituents, focus on restricted rotation. If the question involves a tetrahedral centre with four different groups and optical activity, perform the separate chiral-centre and symmetry analysis.
Student prompt: Are cis- and trans-but-2-ene an enantiomeric mirror-image pair?
What a correct answer shows: No. They are geometrical isomers with the same connectivity but different substituent arrangements.
10. Explain why ordinary single-bond rotation does not preserve a cis/trans label
Take a simple alkane with an ordinary C–C single bond. Many different conformations can be reached by rotation about that σ bond without breaking its covalent connectivity. They may differ in energy, but under ordinary conditions often interconvert rapidly. Simply drawing substituents above or below such a freely rotating single bond does not automatically establish distinct isolable cis and trans compounds.
A carbon–carbon double bond restricts this rotation because its π overlap must be disrupted to turn the substituent arrangement through the required angle. This creates a more substantial barrier and makes the two geometrical configurations distinguishable in the appropriate molecule.
The student should therefore inspect the bond order before writing labels. A straight line between two carbon atoms in a structural drawing usually represents a single bond; a double line represents C=C. If an isomerism task omits the multiple bond, the relevant geometric constraint may not be present.
This explanation is not a claim that all single-bond conformations are identical in energy or that all double bonds can never isomerise. Photochemical processes and other reactions can change geometrical isomers, but those require appropriate energetic or chemical conditions beyond freely rotating an ordinary σ bond.
Student prompt: Why is drawing two orientations about a freely rotating C–C single bond not enough to prove distinct cis–trans alkenes?
What a correct answer shows: The defining constraint in alkenes is restricted rotation around the π-containing C=C bond.
11. Read drawings that have been rotated on the page
Suppose a printed cis-but-2-ene structure is rotated 180° on the page. The methyl groups that originally appeared at the top of the page may now be at the bottom, but they remain on the same side relative to the double bond. The molecule’s geometry has not changed.
A student who labels isomers by the absolute direction of a page arrow will change the answer incorrectly after rotation. The correct reference is the positions of comparable substituents relative to the C=C axis. If both lie on the same side in one equivalent drawing, the molecule remains cis after any overall rotation.
Mirror reflections require more care: reflecting a diagram can alter the viewpoint but does not always produce a new chemical species. For simple cis/trans alkenes, one must identify whether the reflection preserves or exchanges the relative substituent arrangement. Redrawing a diagram from another orientation is not a chemical reaction.
To teach this, present two pictures of the same but-2-ene geometry rotated differently, plus one actual trans arrangement. Ask students to group equivalent depictions before counting distinct isomers. Counting pictures rather than chemically distinct arrangements is a common examination error.
Student prompt: Does rotating a cis-but-2-ene diagram on paper convert it to trans?
What a correct answer shows: No. Overall rotation preserves the substituents’ relative placement across the double bond.
12. Use a two-column checklist for predicting whether cis/trans exists
Create a column for the left alkene carbon and a column for the right alkene carbon. In each, list the two groups directly attached to that carbon apart from the other alkene carbon. If either column contains two identical groups, the molecule cannot display the usual cis/trans pair at that C=C bond.
For CH₃CH=CHCH₃, the pairs are (CH₃,H) and (H,CH₃), so both pass. For CH₂=CHCH₂CH₃, the terminal pair is (H,H), so it fails. For (CH₃)₂C=CHCH₃, one pair is (CH₃,CH₃), so it also fails.
Only once both sides pass should the student draw alternative relative configurations. When matching comparison groups occur across the double bond, label whether they are on the same or opposite side. The drawing needs consistent bonds and no silent change in connectivity between the two candidates.
This checklist also helps with unfamiliar substituted alkenes that are drawn as condensed structural formulas. Instead of guessing from a complicated name, the learner translates each C=C carbon’s two attachments into a short list. The final cis/trans verdict follows from those local lists.
Student prompt: What is the shortest reliable yes/no test for alkene geometrical isomerism?
What a correct answer shows: Each C=C carbon must have two different substituents; otherwise no cis/trans pair arises at that bond.
13. Distinguish H2 cis/trans from optional E/Z nomenclature
The H2 2027 syllabus explicitly specifies that E/Z nomenclature is not required for the cis–trans isomerism outcome. Therefore a tutor should not imply an H2 candidate must apply Cahn–Ingold–Prelog priority rules or assign an E/Z label to every proposed alkene stereoisomer.
The more general E/Z system is valuable for structures where the four substituents are not conveniently described by simple identical-group cis/trans comparisons. Students who progress to H3 Chemistry will encounter E/Z nomenclature as an explicit molecular stereochemistry outcome. That is a legitimate enrichment pathway but a separate requirement.
For the core H2 case, cis- and trans-but-2-ene can be described fully by methyl groups on the same or opposite sides. It is appropriate to practise that chemistry with accurate drawings and a restricted-rotation explanation, then treat E/Z as optional further reading if the learner requests extension.
This division of scope also avoids a naming pitfall. A student may know E/Z terminology from another source and try to force it into a teacher’s simpler cis/trans task, introducing unnecessary priority mistakes. A clear exam answer uses the nomenclature the question and syllabus actually require.
Student prompt: Must a 2027 H2 Chemistry student use E/Z notation for the alkene outcome?
What a correct answer shows: No. The 9476 syllabus explicitly states E/Z nomenclature is not required.
14. Correct the most frequent false stereoisomer claims
Error one is calling but-1-ene cis because the terminal hydrogens are drawn both above the line. The drawing can be changed without creating another distinct structure because one double-bond carbon has identical H groups. Error two is calling 2-methylbut-2-ene cis/trans even though one alkene carbon has two methyl groups.
Error three is interpreting but-1-ene and but-2-ene as a cis/trans pair. They are structural position isomers and do not share the same C=C connectivity. Error four counts rotating a printed cis structure as a new isomer. A page rotation is not the chemical bond rotation required to interconvert configurations.
Error five assumes cis/trans pairs must have identical melting and boiling points because they have the same molecular formula. Distinct three-dimensional arrangements can lead to different intermolecular attractions and packing, so their physical properties need not match. The direction cannot be predicted universally from the label alone.
The repair sequence is always the same: fix the structural formula, locate C=C, list two groups at each carbon, confirm the criterion, then compare relative arrangements. Students should explain one incorrect example as well as one correct example to prove they understand where the rule stops.
Student prompt: Which false claim is exposed by 2-methylbut-2-ene?
What a correct answer shows: That every alkene with a double bond and multiple substituents must have a cis/trans pair.
15. Transfer the test to an unfamiliar carbon skeleton
Consider pent-2-ene, CH₃CH=CHCH₂CH₃. At one C=C carbon the substituents are CH₃ and H; at the other they are CH₂CH₃ and H. Both ends have two different groups, so a geometrical pair exists. The two hydrogens can be used as the matching reference groups for a simple cis/trans comparison.
Now consider 2-methylpent-2-ene, where a double-bond carbon bears two identical CH₃ groups. That centre fails, so the alkene has no corresponding cis/trans pair despite having more carbon atoms than pent-2-ene. Carbon count is not the decisive condition.
A fictional multiple-choice question might offer but-1-ene, but-2-ene, 2-methylbut-2-ene and pent-2-ene. The correct candidates for cis/trans are but-2-ene and pent-2-ene. The student should justify each choice by its pair of substituent lists rather than a vague rule that internal double bonds always qualify.
Parents can ask one useful question at the end of a session: ‘What two groups are attached to each of the two alkene carbons?’ A child who can give those four groups and identify the failed pair is ready for more integrated organic-isomer questions.
Student prompt: Can pent-2-ene show cis/trans isomerism, and why?
What a correct answer shows: Yes. Both C=C carbons have H and a distinct carbon substituent, allowing two relative geometries.
Worked evidence: ten alkenes and isomer comparisons
This table starts with the four substituent slots at each double bond rather than a memorised picture. The distinction between structural and geometrical isomerism is retained throughout.
| Alkene or comparison | Two groups at each C=C carbon | Cis/trans? | Reason |
|---|---|---|---|
| Ethene CH₂=CH₂ | H/H and H/H | No | Both ends have repeated H |
| Propene CH₂=CHCH₃ | H/H and H/CH₃ | No | Terminal carbon has two H |
| But-1-ene CH₂=CHCH₂CH₃ | H/H and H/CH₂CH₃ | No | Terminal repeated H |
| But-2-ene CH₃CH=CHCH₃ | CH₃/H and H/CH₃ | Yes | Cis and trans by methyl placement |
| Pent-2-ene CH₃CH=CHCH₂CH₃ | CH₃/H and H/CH₂CH₃ | Yes | Hydrogens can be compared |
| 2-Methylbut-2-ene (CH₃)₂C=CHCH₃ | CH₃/CH₃ and H/CH₃ | No | Repeated methyl at one C |
| 1,2-Dichloroethene CHCl=CHCl | Cl/H and Cl/H | Yes | Cl groups same versus opposite sides |
| 1,1-Dichloroethene CH₂=CCl₂ | H/H and Cl/Cl | No | Repeated groups at both ends |
| But-1-ene vs but-2-ene | Different double-bond position | Structural isomers | Not a cis/trans pair |
| Cis vs trans but-2-ene | Same connectivity, changed geometry | Geometrical isomers | Restricted rotation about C=C |
The two-different-groups requirement applies to both carbons of the C=C unit. A molecule can satisfy it even if the carbon chain is asymmetric, but a simple cis/trans name should only be used when the relative groups are clearly defined. The more general E/Z priority nomenclature is optional extension, not required in H2 9476.
Four independent transfer decisions
- CH₂=CHCH₃: one carbon has H/H, so a cis/trans pair cannot be distinguished.
- CH₃CH=CHCH₂CH₃: both carbon centres have two different groups, so cis/trans is possible.
- (CH₃)₂C=CHCH₃: one end has two methyl groups, so no geometric pair arises at this double bond.
- CHCl=CHCl versus CH₂=CCl₂: the first passes both different-group tests, while the second fails; the two formulas must also be distinguished as different connectivity patterns.
Tutor checklist for stereochemistry revision
- Confirm H2 Chemistry 9476, 2027 Section 11.2 and the student’s actual examination year.
- Teach restricted rotation in terms of the π component of the carbon–carbon double bond.
- Require two substituent lists, one for each alkene carbon, before naming cis/trans.
- Use but-1-ene and but-2-ene to distinguish structural position isomers from geometric isomers.
- Check both cis and trans drawings preserve exactly the same atom connectivity.
- Explain that turning a drawing on paper does not interconvert the chemical isomers.
- Separate cis/trans from chiral-centre and optical-rotation concepts.
- Mark E/Z notation as an advanced optional extension because it is not required for H2 9476.
- Finish with an unfamiliar alkene and demand a written explanation, not merely a yes/no label.
Frequently asked questions
Why does cis/trans isomerism occur in alkenes?
The π component of the carbon–carbon double bond restricts free rotation, permitting distinct relative substituent arrangements when both double-bond carbons have different groups.
What is the two-different-groups rule?
Each double-bonded carbon must carry two different substituents. If either carbon has two identical groups, no distinct cis/trans pair arises at that C=C bond.
Does but-1-ene have cis/trans isomers?
No. Its terminal CH₂ carbon has two identical hydrogen substituents.
Does but-2-ene have cis/trans isomers?
Yes. Each double-bond carbon is attached to H and CH₃, giving arrangements with the two methyl groups together or opposite.
What is the difference between cis and trans?
For simple alkene structures with comparable groups, cis means matching groups on the same side of C=C and trans means on opposite sides.
Are but-1-ene and but-2-ene stereoisomers?
No. Their double-bond positions differ, making them structural position isomers.
Are E and Z required for 2027 H2 Chemistry?
No. SEAB syllabus 9476 explicitly notes that E/Z nomenclature is not required for the H2 alkene cis–trans learning outcome.
Do cis and trans always have the same boiling point?
No. Their different arrangements can produce different intermolecular properties, but the precise result requires evidence for the particular compound.
Continue with related Organic Chemistry learning
- JC2 H2 Chemistry | SN1 and SN2 Stereochemical Outcomes
- Sigma and Pi Bonds | Hougang H2 Chemistry
- Electrophilic Addition and Markovnikov’s Rule | Hougang H2 Chemistry
- Hougang Chemistry Tuition and Tutor Library
For tutorial support, bring the current H2 syllabus, examination year, and a student-drawn alkene isomer problem. Confirm class size, venue, teacher and availability directly with the tuition provider. This guide makes no promise of a grade or an available tutorial place.
Official syllabus and limits
- SEAB 2027 H2 Chemistry 9476, Section 11.2 Isomerism
- SEAB 2027 official A-Level syllabus listing
- eduKateSG immutable Clementi small-group teaching reference
- Hougang Chemistry Learning Library
SEAB’s 2027 H2 9476 syllabus explicitly requires cis–trans isomerism in alkenes explained through restricted π-bond rotation; E/Z nomenclature is specifically not required. The structural formulas, counterexamples and transfer questions in this article are original instructional representations, not official examination content.