Assign cis/trans and E/Z configurations
A carbon–carbon double bond prevents free rotation, while a saturated ring can lock substituents on the same or opposite faces.
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The core idea
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Learning objectives
- Assign cis/trans and E/Z configurations
Test whether configuration is locked
A carbon–carbon double bond prevents free rotation, while a saturated ring can lock substituents on the same or opposite faces.
Use cis/trans when an unambiguous same-face comparison exists; use E/Z for an alkene after ranking substituents on each double-bond carbon.
Cis, trans, E and Z
Cis means the specified comparable groups are on the same side or ring face; trans means they are on opposite sides or faces.
For E/Z, each alkene carbon must carry two different substituents: Z places the two higher-CIP-priority groups together, and E places them opposite.
Apply CIP one alkene carbon at a time
Compare the atoms directly attached to one alkene carbon by atomic number; if tied, compare the next ordered set of attached atoms until the first difference, treating multiple bonds by duplicated-atom convention.
Do not choose groups by mass, chain length or visual size; after finding one winner on each alkene carbon, compare only those two winners across the double bond.
Keep ranking and naming as separate steps. Find the higher-priority group on the left alkene carbon, repeat on the right, and only then decide whether those two winners are together (Z) or opposite (E).
Assign CHCl=CBrF
On the left alkene carbon, Cl outranks H; on the right, Br outranks F because bromine has the higher atomic number.
If Cl and Br are drawn on opposite sides, the alkene is E-CHCl=CBrF.
Assign CHCl=CBrF when Cl is above the double bond and Br is below it.
Check your answer
A strong answer gives Cl > H, Br > F and E because the higher-priority groups are opposite.
Resolve a carbon-versus-carbon tie
Supplied structure: CH₃CH=C(Cl)CHO, drawn with CH₃ and Cl on the same side of the C=C bond.
On each alkene carbon, compare the atoms directly attached by atomic number, then use the positions of the two higher-priority groups. Visual size on the page is not a priority rule.
State the CIP winners and descriptor when CH₃ and Cl are drawn together in CH₃CH=C(Cl)CHO.
Check your answer
Left carbon: CH₃ (C) beats H. Right carbon: Cl (Z = 17) beats the carbonyl carbon (Z = 6), even though CHO looks larger. The two winners, CH₃ and Cl, are on the same side, so the alkene is Z.
Classify a substituted ring
Supplied chair: 1,2-dimethylcyclohexane with the C1 methyl axial-up and the C2 methyl axial-down.
Classify cis/trans from the up/down faces, not from axial/equatorial, and check what a chair flip would change.
Classify a 1,2-dimethylcyclohexane chair with C1 methyl axial-up and C2 methyl axial-down.
Check your answer
The methyls are on opposite faces (one up, one down), so this is trans-1,2-dimethylcyclohexane. A chair flip makes both equatorial but keeps them up and down, so it is still trans; a flip cannot turn trans into cis.
Do not equate drawn size with priority
CIP priority begins with atomic number at the first point of difference, not the apparent area of a skeletal group.
CH₂=CCl2 has no E/Z descriptor because each carbon of the double bond does not have two different substituents.
Better reasoning: ‘The longest carbon chain always has highest priority in an E/Z assignment.’
Check your answer
Reject it and compare atomic numbers outward only until the first difference.
Report an auditable descriptor
First prove that E/Z is defined, then write the priority comparison on each end and finally state same-side Z or opposite-side E.
For rings, label each substituent up or down before deciding cis/trans; next use four-group priority at tetrahedral stereocentres.
Show both local CIP comparisons before writing E or Z. If either alkene carbon has two identical groups, say that E/Z is not defined instead of forcing a label.
Give the four checks needed before assigning an alkene E/Z descriptor.
Check your answer
Your answer should include two different groups on each alkene carbon, a CIP winner on each end, their relative sides and the resulting E or Z name.
Assign cis/trans and E/Z configurations scientific representation
Every spatial relation is duplicated in text as same/opposite side or up/down, with named substituents and explicit priority comparisons.
About 5 minutes
Named alkene CIP example
- same-side winners → Z
Text alternative: Every spatial relation is duplicated in text as same/opposite side or up/down, with named substituents and explicit priority comparisons.