Organic Chemistry · Alkenes: Structure and Reactivity
Alkene Stereochemistry and the E,Z Designation
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In 30 seconds
The cis/trans system of the previous topic works for disubstituted alkenes but breaks down with three or four substituents: "same side" and "opposite side" no longer have an obvious pair of identical groups to compare. The general solution is the E,Z designation (German entgegen, "opposite," and zusammen, "together").
The E,Z system uses the Cahn–Ingold–Prelog (CIP) priority rules — the same rules you will apply to tetrahedral stereocenters in Chapter 5. At each alkene carbon you rank the two substituents by atomic number; then you compare the two higher-priority substituents: same side of the double bond = Z; opposite sides = E. The resulting (E) or (Z) prefix completes the alkene name. This topic walks through the rules step by step and shows why the system is unambiguous where cis/trans is not.
Why this matters
Geometry decides biological activity: isomers with different shapes meet different receptors. In the vitamin A family, all-trans-retinoic acid (tretinoin) and its 13-cis isomer (isotretinoin) are both prescribed drugs, used for different conditions precisely because their shapes differ. More dramatically, the visual pigment rhodopsin contains 11-cis-retinal The cis form of the vitamin-A aldehyde bound in rhodopsin. Full entry →; a photon isomerizes that double bond to the all-trans form, and this single geometric change is the first chemical event in vision. In the clinic and the lab, a name without an (E) or (Z) prefix is an incomplete — and dangerous — specification.
The college version
Core Concepts
Why cis/trans is not enough
Cis/trans compares two "like" groups. In a trisubstituted alkene such as 2-bromo-2-butene (CH3–C(Br)=CH–CH3), two isomers exist — each alkene carbon still has two different substituents — but there is no pair of identical groups to compare, so "cis" and "trans" are ambiguous. The E,Z system removes the ambiguity: rank substituents by fixed rules and compare the top-ranked group at each carbon.
CIP rule 1: higher atomic number wins
At each alkene carbon, assign priority 1 to the substituent whose directly attached atom has the higher atomic number, and priority 2 to the other. In 1-bromo-1-chloropropene (C(Br)(Cl)=CH–CH3), the carbon on the left bears Br (atomic number 35) and Cl (17): Br is priority 1, Cl priority 2. The carbon on the right bears CH3 and H: carbon (Z = 6) outranks hydrogen (Z = 1), so CH3 is priority 1 and H is priority 2. Note that priority is decided by atomic number, not by size, mass, or steric bulk.
CIP rule 2: break ties by moving along the chain
If the two directly attached atoms are identical (e.g., two carbons), compare the atoms attached to them, one by one. In CH3–CH=C(CH2CH3)–CHO, the right-hand alkene carbon carries CH2CH3 and CHO, both starting with carbon. Next atoms: the CH2 carbon is bonded to C, H, H; the carbonyl carbon of CHO is bonded to O, O, H (rule 3). Oxygen (Z = 8) outranks carbon, so CHO outranks CH2CH3. Keep moving outward until a difference appears.
CIP rule 3: count multiple bonds twice (or three times)
A double bond counts as two attachments of the bonded atom; a triple bond counts as three. So the carbonyl carbon of –CHO is scored as C bonded to O, O, H — the C=O counts as two oxygens. Without this rule, aldehydes would tie with saturated groups starting with the same atom.
The decision: E or Z
Once priorities are assigned at both alkene carbons, compare the two priority-1 groups: same side = Z (zusammen, together); opposite sides = E (entgegen, opposite). The prefix goes in front of the name with the double-bond locant: (Z)-2-pentene, or (2Z)-pent-2-ene.
E/Z and cis/trans agree for disubstituted alkenes
For a disubstituted alkene like 2-pentene, the two priority-1 groups are exactly the two alkyl groups that cis/trans compares, so E = trans and Z = cis. The systems agree where both apply, and E,Z extends the assignment where cis/trans cannot go.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| E | trans | They agree only for disubstituted alkenes; a trisubstituted (Z) isomer is not "cis" in any useful sense. |
| Z | cis | Same limitation: Z means the two priority-1 groups are together, whatever they are. |
| Priority by atomic number | priority by size or bulk | A tert-butyl group is huge but starts with carbon; it loses to a directly attached bromine (35 > 6). |
| Tie at the first atom | end of the ranking | Ties are broken atom-by-atom along the chain; the CHO vs CH2CH3 example shows you must keep going. |
| Single-bond scoring | double-bond scoring | –CHO counts as C–O,O,H (double bond = two attachments), which is why it outranks –CH2CH3. |
| (E)-2-pentene | (Z)-2-pentene | Different compounds with different physical properties — the prefix is part of the name, not decoration. |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Two kids are standing on opposite sides of a wide river with a bridge (the double bond). To decide who is "across from" whom, you first rank each side by a rule — heavier coin wins. Then you look at the two winners: if they stand on the same riverbank, it’s Z ("together"); if they stand on opposite banks, it’s E ("opposite"). The rule works even when there are three or four kids, which is why it beats the old same-side/opposite-side game.
Worked example
Example 1 — Assign (E)/(Z) to 2-pentene. Structure: CH3–CH=CH–CH2–CH3. At C2 (the left alkene carbon), substituents are CH3 and H: CH3 is priority 1. At C3, substituents are CH2CH3 and H: CH2CH3 is priority 1. In the isomer with CH3 and CH2CH3 on the same side of the double bond, the two priority-1 groups are together → (Z)-2-pentene. In the isomer with them on opposite sides → (E)-2-pentene. (Equivalently, Z = cis-2-pentene and E = trans-2-pentene here, since the compound is disubstituted.)
Example 2 — Assign (E)/(Z) to 1-bromo-1-chloropropene. Structure: C(Br)(Cl)=CH–CH3. Left carbon: Br (Z = 35) beats Cl (Z = 17) → Br priority 1. Right carbon: CH3 (Z = 6) beats H → CH3 priority 1. Br and CH3 on the same side → (Z)-1-bromo-1-chloropropene; opposite sides → (E)-1-bromo-1-chloropropene. No "identical group" was needed — something cis/trans could not manage.
Example 3 — A tie-break with double bonds: 2-ethylbut-2-enal. Structure: CH3–CH=C(CH2CH3)–CHO. At the carbon bearing CH3 and H, CH3 wins. At the carbon bearing CH2CH3 and CHO, both start with carbon: the CHO carbon is bonded to O, O, H (C=O counts twice), the CH2 carbon to C, H, H — oxygen wins, so CHO is priority 1. CH3 and CHO on the same side → (Z)-2-ethylbut-2-enal; opposite → (E)-2-ethylbut-2-enal.
Example 4 — Why cis/trans fails but E/Z succeeds. In 2-bromo-2-butene, CH3–C(Br)=CH–CH3, each carbon has two different substituents and two isomers exist, but no identical pair exists, so "cis" and "trans" are meaningless. CIP handles it: left carbon, Br (35) > CH3 (6) → Br priority 1; right carbon, CH3 > H. Compare Br and CH3: same side = (Z)-2-bromo-2-butene; opposite = (E)-2-bromo-2-butene.
Key takeaways
- E,Z works for all alkenes, including tri- and tetrasubstituted ones; cis/trans works cleanly only for disubstituted alkenes.
- CIP priorities: (1) higher atomic number wins; (2) break ties atom-by-atom along the chain; (3) count multiple bonds as 2–3 single attachments.
- Compare the two priority-1 substituents: same side = Z, opposite sides = E.
- E ≠ always trans and Z ≠ always cis; the terms coincide only for disubstituted alkenes.
- A complete name carries the prefix with the double-bond locant: (E)-2-pentene.
- Atomic number decides priority — never size or bulk.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why does the cis/trans system fail for trisubstituted alkenes, and what replaces it?
Show answer
With three or four substituents there is no pair of identical groups to place "on the same side," so cis/trans is ambiguous. The E,Z system ranks every substituent by the CIP rules and compares only the two priority-1 groups, so it assigns a unique label to every alkene.
List the three CIP rules in order, and state which one makes –CHO outrank –CH2CH3.
Show answer
(1) Higher atomic number wins; (2) break ties by comparing the next atoms along each chain; (3) count multiple bonds as two or three single attachments. Rule 3 makes –CHO (scored C–O,O,H) outrank –CH2CH3 (C–C,H,H).
Assign (E) or (Z): 2-pentene with the two ethyl-type groups on the same side.
Show answer
Same side for the two priority-1 groups = Z: the isomer is (Z)-2-pentene (which is also cis-2-pentene, because the compound is disubstituted).
A student claims "Z is just another word for cis." Is that correct? Explain.
Show answer
Not generally. Z and cis coincide only for disubstituted alkenes. In tri- or tetrasubstituted alkenes a Z isomer need not resemble a "cis" arrangement of anything in particular — the priority-1 groups are together, but no identical pair exists.
In 1-bromo-1-chloropropene, which substituent has priority 1 at the bromine-bearing carbon, and why?
Show answer
Bromine. Both Br (Z = 35) and Cl (Z = 17) are directly attached, and 35 > 17, so bromine is priority 1 by CIP rule 1.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- E (entgegen)
- Configuration in which the two priority-1 substituents are on opposite sides of the double bond.
- Z (zusammen)
- Configuration in which the two priority-1 substituents are on the same side of the double bond.
- Cahn–Ingold–Prelog (CIP) rules
- The priority-ranking rules (atomic number, tie-breaking, multiple-bond counting) used to assign E/Z and R/S.
- priority 1 substituent
- The substituent at an alkene carbon whose directly attached atom has the higher atomic number (or wins the tie-break).
- configurational prefix
- The (E)- or (Z)- label placed at the start of the name, with the double-bond locant.
- 11-cis-retinal
- The cis form of the vitamin-A aldehyde bound in rhodopsin.
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