Take but-2-ene, CH₃CH=CHCH₃. Its two methyl groups can sit on the same side of the double bond or on opposite sides. These two arrangements are two different compounds, with different boiling points, and neither turns into the other at room temperature. They have the same formula and the same atoms joined in the same order. Only their arrangement in space differs.
This page names that difference. The words cis and trans handle the simple cases, on double bonds and on rings. Where they stop working, the E/Z system takes over. It ranks groups with the same priority rules you used for R and S in R/S configuration.
The Bonding section of Foundations called these compounds cis-2-butene and trans-2-butene. Those are older-style names for the same compounds. This page writes but-2-ene, the current style used in The parent chain.
Why these isomers exist at all
Try to turn cis-but-2-ene into the trans isomer. You would have to spin one alkene carbon half a turn around the C=C axis, carrying its methyl from one side to the other.
The pi bond stops that turn. Each alkene carbon has a p orbital, and the pi bond is the side-by-side overlap of the two. Twist one carbon a quarter turn and its p orbital ends up at right angles to the other. The overlap is gone, and so is the pi bond. Breaking it costs about 65 kcal/mol, far more than collisions supply at room temperature. The Bonding section of Foundations shows these orbitals in detail.
A single bond has no such problem. Its overlap is the same all the way around the bond axis, so its two ends spin freely. That is why butane has no cis and trans forms, while but-2-ene does.
A ring holds groups in place in a different way. Take 1,2-dimethylcyclohexane. Its ring bonds twist, and the ring flips between chairs all the time. But the ring always has two faces, a top and a bottom. A methyl on the top face could reach the bottom face only if a ring bond broke, so each methyl stays on the face it started on.
Cis and trans on a ring
Draw 1,2-dimethylcyclohexane as a flat hexagon and put each methyl on a wedge or a hash. A wedge, the solid bond that widens toward the group, puts the group above the plane of the ring. A hash, the dashed bond, puts it below.
If both methyls sit on wedges, both are on the top face. Two groups on the same face are cis. If one sits on a wedge and the other on a hash, they are on opposite faces, and that is trans. Two hashes would also be cis, with both groups on the bottom face.
The test never asks how far apart the groups are, so it works the same way on 1,3- and 1,4-dimethylcyclohexane. The locants say which carbons carry the groups, and the wedges and hashes say which face each group is on.
Now draw cis-1,2-dimethylcyclohexane as a chair, the shape from Alkanes & Conformations. One methyl is axial and one is equatorial, and both point up. Flip the ring. Every axial position becomes equatorial and every equatorial position becomes axial, so the two methyls trade roles. Both of them still point up. The compound is still cis.
The trans isomer behaves the same way. In one chair both methyls are equatorial, with one pointing up and one down. After the flip both are axial, still one up and one down. A ring flip can change the energy, because groups move between axial and equatorial. It never moves a group to the other face, so it never turns cis into trans.
How cis and trans translate into axial and equatorial depends on the spacing, because axial bonds alternate up and down around the ring. On any one carbon, the equatorial bond tilts the opposite way to the axial bond. On C1 and C3 the axial bonds point the same way. On C1 and C2, and on C1 and C4, they point opposite ways. That gives the table below. The 1,3 row is the reverse of the other two. Which chair is lower in energy is worked out in Conformational analysis. For naming, only the faces matter.
| Ring positions | cis | trans |
|---|---|---|
| 1,2- | one axial, one equatorial | both equatorial or both axial |
| 1,3- | both equatorial or both axial | one axial, one equatorial |
| 1,4- | one axial, one equatorial | both equatorial or both axial |
Cis and trans on a double bond, and where they run out
Back to but-2-ene.
The words work here because each carbon has exactly one group that is not hydrogen, so "the substituent" on each carbon is clear. That is the condition for using cis and trans on an alkene: each alkene carbon carries one hydrogen and one other group.
Now look at 3-methylpent-2-ene, CH₃–CH=C(CH₃)–CH₂CH₃. The right-hand alkene carbon carries a methyl and an ethyl, and no hydrogen. However you draw it, one of those two groups is on the same side as the left-hand methyl and the other is not. So “cis or trans?” has no answer.
3-Methylpent-2-ene still exists as two stereoisomers, and they are different substances. The E/Z system names them.
E/Z: rank each carbon on its own
Take CH₃–C(Br)=CH–CH₃, drawn with its two methyl groups on opposite sides of the double bond. Anyone who looks only at the methyls would call it trans.
Step 1: rank the two groups on C2. C2 carries Br and CH₃. Compare the atoms bonded directly to C2: bromine, atomic number 35, against carbon, 6. Br is the higher-priority group.
Step 2: rank the two groups on C3. C3 carries CH₃ and H. Carbon beats hydrogen, so CH₃ is the higher-priority group. The bromine on C2 plays no part in this step.
Step 3: compare the two higher-priority groups. The Br on C2 and the CH₃ on C3 are on the same side of the double bond. Same side is Z, so the compound is (Z)-2-bromobut-2-ene.
Nothing is wrong with the drawing or the name. "Trans" describes the methyls, while E/Z describes the higher-priority groups. Here those are different groups, because Br outranks the methyl on C2.
That example is the whole procedure:
- Rank the two groups on one alkene carbon by CIP priority.
- Rank the two groups on the other alkene carbon the same way, without looking across the double bond.
- Compare the two higher-priority groups. If they are on the same side, the alkene is Z, from the German zusammen, "together". If they are on opposite sides, it is E, from entgegen, "opposite".
The ranking is the CIP system from R/S configuration, used unchanged. There is no step that points a group away from you, because an alkene carbon is flat and has no tetrahedral center to look through. E or Z goes in parentheses at the front of the name, just as R or S does.
When the first atoms tie
Often both groups on an alkene carbon attach through carbon. Then the first atoms tie, and you move one atom further out, exactly as in R/S configuration. The next example is the comparison students most often get wrong.
3-(Chloromethyl)-4-methylpent-2-ene is CH₃–CH=C(CH₂Cl)–CH(CH₃)₂. C2 carries a methyl and a hydrogen. C3 carries a chloromethyl group, –CH₂Cl, and an isopropyl group, –CH(CH₃)₂.
Rank C2. Methyl against hydrogen: the methyl wins.
Rank C3. Both groups attach through carbon, so the first atoms tie. List the three atoms bonded to each of those carbons, highest first. The chloromethyl carbon holds (Cl, H, H), and the isopropyl carbon holds (C, C, H). Compare the first terms: chlorine, 17, beats carbon, 6. The comparison stops there, and –CH₂Cl is the higher-priority group.
Compare. In the drawing below, the C2 methyl and the C3 chloromethyl both point up. The isomer drawn is (Z)-3-(chloromethyl)-4-methylpent-2-ene.
Notice what did not decide it. The isopropyl group has three carbons to the chloromethyl group's one, and it still lost. CIP compares the two sets term by term and stops at the first difference. It never counts atoms, adds up atomic numbers or judges size.
A double bond in a group counts twice
Now compare a vinyl group, –CH=CH₂, with an isopropyl group, –CH(CH₃)₂. Here too the first atoms tie, since each group is joined to the alkene through a carbon.
R/S configuration gave the rule for a multiple bond: count the doubly bonded partner twice. The second copy is a duplicate atom, a placeholder written in brackets that has nothing attached to it. A triple-bonded partner counts three times: once for real and twice as duplicates.
First atom out. The vinyl group's attachment carbon is doubly bonded to a CH₂. Count that CH₂ carbon twice, once for real and once as a duplicate, and add the hydrogen. The set is (C, C, H). The isopropyl carbon really does carry two methyls and a hydrogen, so its set is (C, C, H) too. The two groups tie.
Next atom out. Follow the highest-ranked branch of each group. On the vinyl side, the real CH₂ is doubly bonded back to the first carbon, so it carries a duplicate as well, and its set is (C, H, H). On the isopropyl side, each methyl holds (H, H, H). Carbon beats hydrogen, so vinyl outranks isopropyl.
The duplicate atoms carry nothing, so they lose any comparison in which they are explored. They matter only in the set where they first appear.
Two double bonds, two descriptors
Hexa-2,4-diene, CH₃–CH=CH–CH=CH–CH₃, has two double bonds. Every carbon of both double bonds carries a hydrogen and a carbon, so both bonds are stereogenic, and each needs its own letter.
C2=C3. On C2, the C1 methyl beats H. On C3, C4 beats H. In the drawing, C1 and C4 are on opposite sides, so this bond is E.
C4=C5. On C4, C3 beats H. On C5, the C6 methyl beats H. C3 and C6 are on the same side, so this bond is Z.
Place the letters. Each letter takes the locant of the lower-numbered carbon of its double bond. Both go inside one pair of parentheses at the front of the name, in locant order, the way (2R,3S) is written for two stereocenters. As numbered here, the name is (2E,4Z)-hexa-2,4-diene. A name that leaves out either letter is incomplete.
Check the numbering. This chain reads the same from either end, so it can be numbered either way. IUPAC breaks the tie by giving Z the lower locant. Numbered from the other end, the Z bond becomes C2=C3, so the preferred name is (2Z,4E)-hexa-2,4-diene.
Here E matches trans only because a carbon group beats H on every alkene carbon.
Where these isomers sit among the stereoisomers
Cis and trans isomers, and E and Z isomers, are diastereomers: stereoisomers that are not mirror images of each other. The mirror image of cis-but-2-ene is cis-but-2-ene again, never the trans isomer. Diastereomers explains why such a pair, unlike a pair of enantiomers, has different melting points, boiling points and dipole moments.
A double bond is a stereogenic unit, a part of a molecule that gives rise to stereoisomers. It is not a stereocenter, because in this course that word means a single tetrahedral atom. Both count when you count stereoisomers.
Hexa-2,4-diene shows how the counting works. It has two stereogenic double bonds, so you might expect 2² = 4 stereoisomers. But (2E,4Z) and (2Z,4E) are the same molecule read from opposite ends, so there are only three: (2E,4E), (2Z,4Z) and (2Z,4E).
What the geometry costs, physically
The trans isomer of but-2-ene is lower in energy than the cis isomer, by about 1 kcal/mol. In the cis isomer the two methyls sit on the same side and crowd each other. In the trans isomer they are held apart. In Alkenes & Alkynes you will measure this difference by hydrogenating both isomers to the same alkane and comparing the heat each releases.
The general rule is about size, not priority: the isomer with its two bulkier groups on opposite sides is usually the more stable one. The bulkier groups are often also the higher-priority groups, so the E isomer is often the more stable one. But CIP priority is set by atomic number, not by size, so the two rankings do not always match.
Geometry also changes how molecules pack into a crystal. A trans double bond keeps a long chain nearly straight, while a cis double bond puts a bend in it. Straight chains stack closely, so trans isomers usually melt higher. That is why trans fats are solid at temperatures where the cis fats they came from are liquid. Trans fats form when a vegetable oil is partly hydrogenated. Hydrogen removes some of its double bonds, and some of the rest turn trans.
Polarity goes the other way: here the cis isomer is the more polar one. In cis-1,2-dichloroethene both C–Cl bonds point to the same side, so their dipoles add up to a net dipole. In the trans isomer they point in opposite directions and cancel exactly, so trans has no net dipole.
Five mistakes to avoid
What carries forward
From here on, every alkene that has a descriptor is named with it, so the assignment needs to become automatic. E2 eliminations in Substitution & Elimination usually favor the E alkene. In Alkenes & Alkynes, additions to a double bond are described as syn or anti, and reductions of alkynes give E or Z alkenes. Same face or opposite faces on a ring is the core of many exam questions about substituted cyclohexanes. The counting rule joins what you already know about stereoisomers: stereocenters and stereogenic double bonds both count, and 2n+m is the ceiling.