Stereochemistry · Section 35 of 121

Cis/trans and E/Z

Practice this — interactive lesson

Two molecules can share a formula, share a connectivity map, share every stereocenter — and still be different substances, because some of their pieces sit on opposite sides of something that will not turn. That something is a pi bond, or a ring. This section is the naming half of that idea: when the words cis and trans are enough, when they quietly stop being enough, and how the E/Z system settles every case using nothing but the CIP ranking you already built for R and S.

Why these isomers exist at all

The mechanism was established back in Bonding, in Foundations: a sigma bond's overlap is cylindrically symmetric about the bond axis, so the two ends spin freely, while a pi bond's overlap is side-by-side and dies the moment one end is twisted. Rotating about a C=C therefore costs the whole pi bond, and nothing at room temperature pays that. The two ends are locked, and whatever is attached to them keeps the arrangement it was made with.

A ring does the same job by a different route. No individual bond in cyclohexane is locked — the ring flips and twists constantly — but the ring as a whole holds its atoms in a closed loop with two distinguishable faces, a top and a bottom. A substituent attached to the top face can only reach the bottom face by breaking a ring bond. So a ring, like a pi bond, converts a relative arrangement into a permanent structural identity.

That permanence is the point, and it is what the word configurational means. A configuration cannot be changed by rotating bonds; only by breaking and remaking them. A conformation — a chair, a gauche butane — changes freely and costs a few kcal/mol at most. Cis and trans isomers are separate compounds in separate bottles with separate boiling points, not two shapes of one compound.

Cis and trans on a ring

Draw the ring flat and use the wedge-and-dash convention: a wedge puts the substituent above the plane of the ring, a hash puts it below. Two substituents drawn on the same kind of bond — both wedges, or both hashes — are on the same face, and that is cis. One wedge and one hash means opposite faces, and that is trans.

Nothing in that test mentions how far apart the two groups are, and that is deliberate. It reads exactly the same way for a 1,2-, a 1,3- or a 1,4-disubstituted cyclohexane: find the two substituents, ask whether they are on the same face. cis-1,4-dimethylcyclohexane and cis-1,2-dimethylcyclohexane are both cis, and the locants say the rest.

both on a wedge — same faceCH₃CH₃C1C2cis-1,2-dimethylcyclohexaneone wedge, one hash — opposite facesCH₃CH₃C1C2trans-1,2-dimethylcyclohexaneWedge is above the ring, hash is below. Same face is cis, opposite faces trans.
Cis and trans on a ring, with nothing to compute. A wedge puts the group above the plane of the ring and a hash puts it below, so the question is only whether the two substituents are drawn on the same kind of bond. That reading is identical for a 1,2-, a 1,3- or a 1,4-disubstituted ring — the locants say how far apart the groups are, and the wedges say which face each one is on.A chair flip does not touch this. Flipping exchanges axial for equatorial at every carbon, which changes the energy of the conformer; the group that was above the ring is still above the ring when the flip is finished. Configuration survives a ring flip, position does not.
A ring flip does not change cis or trans. Flipping a chair swaps every axial substituent for equatorial and every equatorial for axial — that is conformational bookkeeping from Alkanes & Conformations, and it is worth knowing which chair is cheaper. But a group that was above the ring is still above the ring afterward. Which face a substituent is on survives the flip; which position it occupies does not.

Read against the chair drawings from that chapter, the translation is short: on a 1,2- or a 1,4-disubstituted ring, cis puts one group axial and one equatorial while trans puts both axial or both equatorial; on a 1,3-ring the two pairings swap over. Which of the two chairs is cheaper is conformational analysis and is argued there. For naming, the only question asked here is same face or opposite face.

both on a wedge — same faceCH₃CH₃C1C3cis-1,3-dimethylcyclohexaneone wedge, one hash — opposite facesCH₃CH₃C1C4trans-1,4-dimethylcyclohexanein a chaircistrans1,2-one axial, one equatorialboth equatorial (or both axial)1,3-both equatorial (or both axial)one axial, one equatorial1,4-one axial, one equatorialboth equatorial (or both axial)The 1,3- row is the one that trades places. Naming never uses this table — conformational analysis does.
The same wedge-and-hash reading on the two spacings the 1,2- figure did not draw. Two wedges are two groups on the same face, so the 1,3 compound on the left is cis however far apart its methyls are; a wedge and a hash are opposite faces, so the 1,4 compound on the right is trans. The locants change nothing about the test.The table underneath is a different question, and the row that catches people is the middle one. Axial directions alternate around the ring, so on carbons an even number apart (1,3-) the two axial positions point the same way and cis can be diequatorial, while on carbons an odd number apart (1,2- and 1,4-) they point opposite ways and cis is forced into axial/equatorial. None of that changes cis or trans — it only decides which conformer is cheap.

Cis and trans on an alkene, and where they run out

On a double bond the same two words work, with one condition attached: each alkene carbon must carry one hydrogen and one other group. Then "the substituents" is unambiguous — there is exactly one non-hydrogen group per carbon — and cis means they are on the same side of the C=C, trans that they are on opposite sides. But-2-ene is the standard pair.

Now put a second non-hydrogen group on one of the alkene carbons. In 3-methylpent-2-ene, CH₃–CH=C(CH₃)–CH₂CH₃, the right-hand alkene carbon carries a methyl and an ethyl; whichever way you draw it, one of them is on the same side as the methyl across the double bond and one is not. Asking whether the compound is cis or trans has no answer, because the question presupposes one substituent per carbon to talk about. Cis/trans has not given the wrong answer here — it has run out of definition, and the compound still has two distinct geometric isomers that need naming.

CH₃CH₃HHCH₃CH₃HHCH₃CH₃HCH₂CH₃cis-but-2-enetrans-but-2-ene3-methylpent-2-eneone H, one CH₃ per carbonsame test, other answerno H on the right carbonCis/trans needs one hydrogen and one other group on EACH carbon of the C=C.
The condition on cis and trans, and the case that violates it. In the first two panels each alkene carbon carries one hydrogen and one methyl, so “the substituent” means something and the two words divide the cases cleanly. In the third the right-hand carbon carries a methyl and an ethyl: one of them is cis to the left-hand methyl and the other is trans, so the question has no answer.The third compound is not an edge case without isomers — it has two, and they are different substances. What it lacks is a name for them in this vocabulary, which is exactly the gap E/Z was invented to fill.

E/Z: rank each carbon on its own

The general system replaces "the substituent" with "the higher-priority group," which every alkene carbon has, whatever it is carrying. The procedure is three lines long:

  1. Rank the two groups on one alkene carbon by CIP priority. Only those two. Nothing across the double bond enters this comparison.
  2. Rank the two groups on the other alkene carbon, the same way and just as independently.
  3. Compare the two winners. Same side of the double bond is Z (German zusammen, together); opposite sides is E (entgegen, opposite).

The ranking is the CIP system from the R/S section of this chapter, used unchanged and without the orientation step — there is no "point the lowest priority away," because there is no tetrahedral center to look through. Two parts of it do most of the damage on exams, and both are worth restating here even though neither is new.

Ties break at the first point of difference, not on size. When the two first atoms are the same element, list the three atoms attached to each, highest first, and compare term by term. A carbon carrying (Cl,H,H) beats one carrying (C,C,H), because chlorine beats carbon at the first term. The bulkier group loses that comparison routinely, and the instinct that says "isopropyl is bigger, so isopropyl wins" is wrong about half the time.

A multiple bond is counted as bonds to duplicate atoms. A doubly bonded partner is written twice, a triply bonded one three times, and each phantom duplicate carries no substituents of its own. That is what lets a vinyl group and an isopropyl group be compared at all, and it is where the comparison between them is decided.

Worked example — 2-bromo-2-butene, the one that looks trans and is Z

Take CH₃–C(Br)=CH–CH₃ drawn with its two methyl groups on opposite sides of the double bond: the arrangement anyone would call trans.

Step 1 — split the problem in two. C2 and C3 are ranked separately. The bromine on C2 is never weighed against the methyl on C3; that comparison is not part of the system and has no meaning.

Step 2 — rank C2. Its two groups are Br and CH₃. The first atoms settle it: bromine is atomic number 35, carbon is 6. Br is the higher priority.

Step 3 — rank C3. Its two groups are CH₃ and H. Carbon beats hydrogen, so the methyl wins.

Step 4 — compare the two winners. The bromine on C2 and the methyl on C3 point to the same side of the double bond. The compound is (Z)-2-bromo-2-butene.

Nothing is wrong with the drawing. "Trans" was reporting where the methyls are; E/Z reports where the priorities are, and here the two reports disagree because C2 carries a group that outranks its methyl.

drawn the way anyone would call "trans"BrHC1C2C3C4the two methyls, C1 and C4, are on opposite sidesbut rank each carbon separatelyOn C2: Br against CH₃Br wins — atomic number 35 beats 6nothing else is comparedpoints UPOn C3: CH₃ against HCH₃ wins — carbon beats hydrogennothing else is comparedpoints UPBoth winners on the same side:(Z)-2-bromo-2-buteneSame molecule, two labels: "trans" was reporting the methyls, E/Z reports the priorities.
One assignment, done in full. Each alkene carbon is ranked on its own two groups and nothing else — the bromine on C2 is never weighed against the methyl on C3. Here both winners finish above the double bond, so the alkene is Z even though the carbon skeleton is drawn trans.The error this drawing is built to prevent: comparing a group on one alkene carbon with a group on the other. That question has no meaning. E/Z asks two independent questions and then compares only the two answers.
Worked example — a tie broken one sphere out

3-(chloromethyl)-4-methylpent-2-ene: the C3 end of the double bond carries a –CH₂Cl group and an isopropyl group, and the C2 end carries a methyl and a hydrogen.

Rank C3. Both groups attach through carbon, so the first sphere ties and you move outward. The chloromethyl carbon holds (Cl, H, H); the isopropyl carbon holds (C, C, H). Compare highest to highest: chlorine, atomic number 17, against carbon, 6. –CH₂Cl wins at the first term, and the comparison stops there.

Rank C2. Methyl against hydrogen. The methyl wins.

Compare. The two winners are the –CH₂Cl on C3 and the methyl on C2, so whichever side the chloromethyl is drawn on is the side the methyl has to match for Z. In the drawing below both of them point up, on the same side of the double bond, so that isomer is (Z)-3-(chloromethyl)-4-methylpent-2-ene. Note what did not decide this: the isopropyl group has three carbons to chloromethyl's one, and it lost anyway. Term-by-term comparison at the first point of difference, never a head count and never a size estimate.

both branches start with carbonCH₃HCH₂ClCH(CH₃)₂C2C33-(chloromethyl)-4-methylpent-2-eneso move one sphere out and compare–CH₂Cl → (Cl, H, H)–CH(CH₃)₂ → (C, C, H)Highest against highest: Cl (17) beats C (6).First point of difference — stop there.–CH₂Cl is the higher priorityalthough isopropyl is the bigger groupdrawn as here, both winners are up: (Z)Size does not decide a CIP comparison. The first point of difference does.
A tie at the first atom, broken one sphere out. Both groups on C3 attach through carbon, so the first sphere says nothing and you list what each of those carbons holds. Chlorine appears at the head of one set and carbon at the head of the other, so the comparison is over at the first term — and the branch with three carbons in it loses to the branch with one. As drawn, the C2 methyl and the C3 chloromethyl are both up, so this particular isomer is (Z)-3-(chloromethyl)-4-methylpent-2-ene.This is the comparison students most often decide by eye. Bulk, mass and the number of atoms in a branch are all irrelevant to CIP; the sets are ordered high to low and read position by position, and the moment they differ the ranking is fixed.
Worked example — vinyl against isopropyl, decided by a phantom

Now an alkene carbon carrying a vinyl group (–CH=CH₂) and an isopropyl group (–CH(CH₃)₂). Both attach through carbon, so go outward.

First sphere out. The vinyl's attachment carbon is doubly bonded to a CH₂, so that partner is counted twice: the real carbon plus a duplicate. Its set is (C, C, H). The isopropyl's attachment carbon really does carry two methyls and an H, so its set is (C, C, H) as well. Exact tie — duplication is what makes the two comparable, and it does not by itself break anything.

Second sphere out. Explore the tied branches. On the vinyl side, the real terminal CH₂ is doubly bonded back to the first carbon, so it counts as (C, H, H). On the isopropyl side, each methyl is (H, H, H). Carbon beats hydrogen. Vinyl outranks isopropyl.

The phantom duplicates themselves are dead ends: they carry nothing, so they lose every comparison in which they are explored. Their whole job is to be counted once in the sphere where they appear.

vinyl –CH=CH₂HCCH₂(C)(C)first sphere: (C, C, H)that CH₂ then holds (C, H, H)so vinyl wins one sphere laterisopropyl –CH(CH₃)₂CCH₃CH₃Hfirst sphere: (C, C, H)each CH₃ holds only (H, H, H)Duplication creates the tie; the next sphere out breaks it. Vinyl beats isopropyl.
What a duplicated atom is for. The vinyl carbon really carries one carbon and one hydrogen, but its partner is doubly bonded, so CIP writes that partner in twice — once real, once as the parenthesized phantom — and the set becomes (C, C, H). That is the same set the isopropyl carbon genuinely holds, so the first sphere is an exact tie and the comparison has to go out one more.A phantom has no substituents of its own, so when the search reaches one it is a dead end and loses. Duplication is a bookkeeping device that makes a multiple bond comparable with a branch; it is not a way of scoring extra points.
Worked example — two double bonds, two descriptors

Hexa-2,4-diene, CH₃–CH=CH–CH=CH–CH₃, has two stereogenic double bonds, and each needs its own letter.

Assign C2=C3. On C2: the C1 methyl against H — methyl wins. On C3: the C4 chain carbon against H — C4 wins. Drawn with those two on opposite sides, this bond is E.

Assign C4=C5. On C4: C3 against H. On C5: the C6 methyl against H. Drawn with those two on the same side, this bond is Z.

Write the name. Each descriptor is prefixed by the locant of the lower-numbered carbon of its double bond, and the set goes in the front bracket in numerical order: (2E,4Z)-hexa-2,4-diene. The stereodescriptors sit in the same bracket, and in the same style, as the (2R,3S) descriptors from the R/S section — one label per stereogenic unit, each carrying its locant.

123456CH₃–CH=CH–CH=CH–CH₃one descriptor per double bondC2=C3: CH₃ and the C4 chainsit on opposite sidesEC4=C5: the C3 chain and CH₃sit on the same sideZ(2E,4Z)-hexa-2,4-dieneeach letter takes its bond’s lower locantTwo stereogenic double bonds, two letters, and both of them go in the name.
A diene needs a descriptor for every stereogenic double bond it has, and each one carries the locant of the lower-numbered carbon it spans. Both bonds here have one hydrogen and one carbon chain on each of their carbons, so the chain is the higher priority every time and E/Z agrees with trans/cis — which is the ordinary case, not a rule.The descriptors go inside one set of parentheses at the front of the name, in locant order, exactly as (2R,3S) does for two stereocenters. A name that omits one of them is incomplete, not merely informal.

Where these isomers sit among the stereoisomers

Cis and trans isomers — E and Z isomers — are diastereomers. They are stereoisomers, because they differ only in arrangement in space; and they are not enantiomers, because they are not mirror images of each other. Hold cis-2-butene up to a mirror and you get cis-2-butene back, never the trans isomer. That is the definition of diastereomers from earlier in this chapter, and it is the reason cis and trans isomers have different melting points, different boiling points and different dipole moments, while a pair of enantiomers has identical values for all three.

It also means a double bond can be a stereogenic unit without being a stereocenter. A stereocenter is a tetrahedral atom; a stereogenic double bond is a bond. Both create stereoisomers, and both get counted when you count.

A molecule with n stereocenters and m stereogenic double bonds has at most 2n+m stereoisomers. "At most" is doing real work: the formula is an upper bound, and it overcounts whenever a symmetry makes two of the arrangements the same molecule — meso compounds being the case this chapter already met. A double bond counts toward m only if it is stereogenic in the first place, which means each of its carbons carries two different groups.

What the geometry costs, physically

Geometry has consequences you can measure. In a pair of alkene isomers the E form is usually the more stable, because the two higher-priority groups are held apart instead of crowded on one side; the difference for 2-butene is about 1 kcal/mol. The E isomer also tends to melt higher, because a more symmetric, less bent molecule packs into a crystal lattice more efficiently — the classic industrial illustration is that the trans fats in a partially hydrogenated oil are solids at temperatures where the cis fats are liquid. Dipole moments run the other way: in cis-1,2-dichloroethene the two C–Cl dipoles have a component that adds, while in the trans isomer they cancel by symmetry, so cis is the polar one and trans has essentially no net dipole. A preview: the stability argument gets its evidence in Alkenes & Alkynes, where heats of hydrogenation let you measure how far apart two isomers really are by converting both to the same alkane and comparing the heat released.

The five mistakes this section exists to prevent

E is not a synonym for trans, and Z is not a synonym for cis. They agree only when the priorities happen to fall where the substituents do. Whenever an alkene carbon carries two groups that are not hydrogen, rank them and report the ranking — 2-bromo-2-butene is drawn trans and named Z.
Never rank across the double bond. The comparison is two groups on one carbon, then two groups on the other, then the two winners. Comparing the bromine on C2 with the methyl on C3 is not a harder version of the rule; it is a different question with no answer.
A terminal =CH₂ has no E/Z. If one alkene carbon carries two hydrogens — or two of anything identical — there is no way to tell one arrangement from the other, so the double bond is not stereogenic and takes no descriptor. 1-butene, 2-methyl-1-propene and every other terminal alkene are in this class, and so is any carbon carrying two identical groups, such as the (CH₃)₂C= end of 2-methyl-2-butene.
Configuration is not something a conformation can edit. Exchanging axial for equatorial changes which conformer is cheaper and nothing else, so if a problem claims a chair flip turned one isomer into its geometric partner, the second drawing is wrong somewhere.
Decide ties at the first point of difference. Order each branch's atoms highest to lowest and compare position by position; do not add atomic numbers, and do not count atoms. (O,H,H) beats (C,C,C), and –CH₂Cl beats isopropyl, in both cases on the first term alone.

What carries forward

Every alkene from here on is named with a descriptor when it has one, so the assignment has to be automatic rather than reconstructed. E2 eliminations in Substitution & Elimination are reported as giving predominantly the E alkene; additions across a double bond in Alkenes & Alkynes are reported as syn or anti and their products as E or Z; and the geometry of a ring substituent, same face or opposite face, is the whole content of a great many exam questions about substituted cyclohexanes. The counting rule joins the stereoisomer bookkeeping you already have: stereocenters and stereogenic double bonds both contribute, and 2n+m is the ceiling.