go to the AUS-e-TUTE homepage

cis-trans Isomers (Geometric Isomers) of Alkenes Chemistry Tutorial

Key Concepts

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

Identifying cis-trans Isomers in Alkene Molecules

Cis-trans isomers occur if there is:

  1. restricted rotation, or rigidity, as in the presence of a C=C bond

    AND

  2. 2 different groups on the left-hand side of the C=C bond and 2 different groups on the right-hand side of the C=C bond.

Identifying cis-trans Isomers Example: 1,2-dibromoethane

Will 1,2-dibromoethane have both cis and trans isomers?

First, we need to draw the possible structures for this molecule:

    H
|
  H
|
   
Br-C-C-Br
    |
H
  |
H
   
    Br
|
  H
|
   
H-C-C-Br
    |
H
  |
H
   
    H
|
  H
|
   
H-C-C-Br
    |
Br
  |
H
   
    H
|
  H
|
   
H-C-C-H
    |
Br
  |
Br
   
    H
|
  Br
|
   
H-C-C-H
    |
Br
  |
H
   
    Br
|
  Br
|
   
H-C-C-H
    |
H
  |
H
   

but, because the carbon-carbon single bond, C-C, is NOT rigid, there is rotation around the C-C bond and in reality all of these possible structures exist in dynamic equilibrium with each other:

    H
|
  H
|
   
Br-C-C-Br
    |
H
  |
H
   
    Br
|
  H
|
   
H-C-C-Br
    |
H
  |
H
   
    H
|
  H
|
   
H-C-C-Br
    |
Br
  |
H
   
    H
|
  H
|
   
H-C-C-H
    |
Br
  |
Br
   
    H
|
  Br
|
   
H-C-C-H
    |
Br
  |
H
   
    Br
|
  Br
|
   
H-C-C-H
    |
H
  |
H
   

since none of these structures exist permanently, we can not refer to cis and trans isomers of 1,2-dibromoethane.

Only a molecule that contains rigidity, a C=C for example, can display cis-trans isomerism.

Identifying cis-trans Isomers Example: 1,1,2-tribromoethene

Will 1,1,2-tribromoethene have both cis and trans isomers?

First, draw the possible structures for this molecule:

Br           Br
  \       /  
    C=C    
  /       \  
Br           H
Br           H
  \       /  
    C=C    
  /       \  
Br           Br
 
Br           Br
  \       /  
    C=C    
  /       \  
H           Br
H           Br
  \       /  
    C=C    
  /       \  
Br           Cl
Rotating the molecule 180° in space makes the first 2 structures identical.   Rotating the molecule 180° in space makes the second 2 structures identical.

rotating this molecule 180° in space
Br           H
  \       /  
    C=C    
  /       \  
Br           Br
turns it into this molecule
H           Br
  \       /  
    C=C    
  /       \  
Br           Br
so all 4 of the structures we have drawn are actually exactly the same!

1,1,2-dibromoethene does NOT display cis-trans isomerism.

Only a molecule with rigidity AND 2 different groups on the left-hand side of the C=C and 2 different groups on the right-hand side of the C=C can display cis-trans isomerism.

Identifying cis-trans Isomers Example: 1,2-dibromoethene

Will 1,2-dibromoethene have both cis and trans isomers?

First, draw the possible structures for this molecule :

Br           Br
  \       /  
    C=C    
  /       \  
H           H
Br           H
  \       /  
    C=C    
  /       \  
H           Br
H           H
  \       /  
    C=C    
  /       \  
Br           Br
H           Br
  \       /  
    C=C    
  /       \  
Br           H

if we rotate the whole molecule in space, we end up with only two different structures:

2 identical structures:
both Br's on same side of the double bond
  2 identical structures:
Br's on different sides of the double bond
Br           Br
  \       /  
    C=C    
  /       \  
H           H
H           H
  \       /  
    C=C    
  /       \  
Br           Br
 
Br           H
  \       /  
    C=C    
  /       \  
H           Br
H           Br
  \       /  
    C=C    
  /       \  
Br           H
cis-isomer   trans-isomer

because the C=C is rigid, you can't rotate the carbon atoms around the double bond.

You can NOT turn a cis-isomer into a trans-isomer because you can't rotate the carbon atoms around the double bond.

So these two structures, the cis-isomer and the trans-isomer, are NOT in dynamic equilibrium with each other and they are permanently different.

1,2-dibromoethene does display cis-trans isomerism because:

  1. 1,2-dibromoethene has rigidity: a C=C
  2. there are 2 different groups on the left-hand side of the C=C (an H and a Br)
    and
    2 different groups on the right-hand side of the C=C (an H and a Br)

Do you know this?

Join AUS-e-TUTE!

Play the game now!

Naming cis-trans Isomers

Before you can name a cis-trans (geometric) isomer you must know its structure.

Naming the cis-trans isomer of an alkene is then a two step process:

  1. Name the alkene:

    (similar to naming haloalkanes but using the "ene" suffix)

    (i) number the longest carbon chain containing the double bond

    (ii) carbon with the functional group is given the lowest possible number

    (iii) alkyl groups and halogen atoms are written in alphabetical order

    order of citation: bromo chloro ethyl fluoro iodo methyl

    ⚛ Note: if more than 1 of any particular alkyl group or halogen atom is present, use the prefixes:

    di for two (dichloro)

    tri for three (trimethyl)

    tetra for four (tetrafluoro)

    ⚛ commas are used between numbers and numbers (1,2)

    ⚛ hyphens are used between numbers and letters (2-)

  2. attach a prefix: cis or trans:

    ⚛ attach the prefix cis- to the name of the alkene if the groups are on the same side of the double bond

    ⚛ attach the prefix trans- to the name of the alkene if the groups are on different sides of the double bond (across from each other)

Cis-trans Isomer Naming Example: alkene with alkyl groups

Name the molecule shown below:

H3C           CH3
  \       /  
    C=C    
  /       \  
H           H

  1. Name the alkene: number the longest carbon chain including the double bond (shown in blue)

    H3C1           4CH3
      \       /  
        C2=C3    
      /       \  
    H           H

    Name the alkene: but-2-ene (or 2-butene)

    ⚛ 4 carbon atoms in the longest chain = but

    ⚛ double bond = en

    ⚛ no other functional groups present = e (butene)

    ⚛ double bond between carbons 2 and 3, use the lowest number = but-2-ene (or 2-butene)

  2. Add the prefix cis- because both CH3 groups are on the same side of the double bond:

    cis-but-2-ene (or cis-2-butene)

Naming cis-trans Isomers Example: halogenated alkene

Name the molecule shown below:

Cl           H
  \       /  
    C=C    
  /       \  
H           Cl

  1. Name the alkene: number the longest carbon chain (shown in blue)

    Cl           H
      \       /  
        C1=C2    
      /       \  
    H           Cl

    (i) name the longest carbon chain: ethene

    2 carbon atoms in the longest chain = eth

    double bond = en

    no other functional groups present = e (ethene)

    (ii) identify and name halogens: 1,2-dichloro

    Cl atoms = chloro

    2 × Cl atoms = dichloro

    1 Cl is bonded to carbon 1, the other is bonded to carbon 2: 1,2-dichloro

    (iii) halogen atom part of the name is attached as a prefix to the name of the longest carbon chain:

    1,2-dichloroethene

  2. Add the prefix trans- to the name because the chlorine atoms are on different sides of the double bond:

    trans-1,2-dichloroethene

Do you understand this?

Join AUS-e-TUTE!

Take the test now!


Footnotes:

(1) This is true for "straight-chain" alkanes, but not true for carbon atoms joined together in a ring. These aliphatic cyclic (or alicyclic) compounds can also exhibit cis-trans isomerism.

(2) When there is no clear distinction between the groups bonded to the C=C carbon atoms, then the E and Z designations, or absolute geometric configuration, are used.

(3) IUPAC Recommendations 1996 strongly discourage the use of the term geometric isomers and favour the term cis-trans isomers.