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cis-trans (Geometric) Isomers of Alkenes

Key Concepts

  • Carbon atoms can freely rotate around the C-C bond in straight chain alkanes1.

  • Cabon atoms cannot freely rotate around the C=C in alkenes.
        The double bond, C=C, is said to be rigid.

  • If each carbon atom in the C=C has two different groups attached to it, then different spatial arrangements of the atoms around the rigid C=C bond are possible.

  • The different spatial arrangements of the atoms result in cis-isomers and trans-isomers2:

    cis
    Latin for "on the same side"
    trans
    Latin for "across from"
    X X
    \ /
    C=C
    / \
    H H
    X H
    \ /
    C=C
    / \
    H X
    Both X's on the same side of C=C X's on different sides of C=C

  • A cis-isomer will have different a boiling point, melting point and solubility compared to the trans-isomer.

  • Cis-trans isomers used to be called geometric isomers3.

Identifying cis-trans Isomers

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.

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.

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 180o in space makes the first 2 structures identical.   Rotating the molecule 180o in space makes the second 2 structures identical.

rotating this molecule 180oC 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.

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-isomertrans-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:

    (i) 1,2-dibromoethene has rigidity: a C=C

    (ii) 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)

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)
      Note: commas are used between numbers and numbers (1,2)
      Note: hyphens are used between numbers and letters (2-)

  2. 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)

Example : 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

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

  3. Add the prefix cis- because both CH3 groups are on the same side of the double bond:
    cis-2-butene (or cis-but-2-ene)

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

  2. Name the alkene: name the longest carbon chain
        2 carbon atoms in the longest chain = eth
        double bond = en
        no other functional groups present = e (ethene)

  3. Name the alkene: identify and name halogens
        Cl atoms = chloro
        2 x Cl atoms = dichloro
        1 Cl is bonded to carbon 1, the other is bonded to carbon 2: 1,2-dichloro

  4. Name the alkene: halogen atom part of the name is attached as a prefix to the name of the longest carbon chain:
        1,2-dichloroethene

  5. Add the prefix trans- to the name because the chlorine atoms are on different sides of the double bond:
        trans-1,2-dichloroethene


What would you like to do now?

1This 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.

2When 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.

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

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