-1
$\begingroup$

How can identify which one is cis and which one is trans . If I consider it as a planar molecule (though it is not) I understand that if the substituents are in same side it is cis else trans but this one confused meenter image description here

$\endgroup$
6
  • 3
    $\begingroup$ At each carbon there is one substituent pointing in the general direction of up, and one pointing in the general direction of down, no? $\endgroup$ – orthocresol Sep 1 '20 at 9:55
  • $\begingroup$ @orthocresol shouldn't it be cis if botha re down and trans gor the other $\endgroup$ – Saniya Sep 1 '20 at 9:59
  • 1
    $\begingroup$ Well, that's pretty much it, really. What exactly are you confused about? $\endgroup$ – orthocresol Sep 1 '20 at 10:03
  • $\begingroup$ @orthocresol in 1st both the substitients lie in perpendicular plane I.e in different planes so it should be trans whereas in 2nd both are in same plane isn't it? $\endgroup$ – Saniya Sep 1 '20 at 10:09
  • 3
    $\begingroup$ Don't overthink it, there's no need for planes. Even in the chair, non-planar conformation that you've attached, you can already see that for every carbon, there is one substituent pointing upwards and one pointing downwards. Try building a model for yourself, if the 2D depictions on paper aren't sufficient. $\endgroup$ – orthocresol Sep 1 '20 at 10:11
1
$\begingroup$

Question: How to identify cis- and trans-forms of cyclohexane?

This is the title and main question raised in the OP's text. So I think that should be addressed before explaining the example. Following image shows the chair conformation of cyclohexane (in the box):

Chair Conformation of cyclic compounds

All $\color{red}{\text{red}}$ $\ce{C-H}$ bonds are in axial orientation (ax) and $\color{black}{\text{black}}$ $\ce{C-H}$ bonds are in equatorial orientation (eq). If you concentrate only on axial positions, you'd see 1,2-ax,ax-bonds are always trans (opposite direction). Therefore you can conclude followings:

  1. Compounds with 1,2-ax,ax-substitutions are always trans.
  2. Compounds with 1,2-eq,eq-substitutions are always trans as well.
  3. Compounds with 1,2-ax,eq- or 1,2-eq,ax-bonds are always cis.

Similarly, since 1,3-ax,ax-bonds are always cis (same direction):

  1. Compounds with 1,3-ax,ax-substitutions are always cis.
  2. Compounds with 1,3-eq,eq-substitutions are always cis as well.
  3. Compounds with 1,3-ax,eq- or 1,2-eq,ax-bonds are always trans.

Furthermore, since 1,4-ax,ax-bonds are always trans (opposite direction):

  1. Compounds with 1,4-ax,ax-substitutions are always trans.
  2. Compounds with 1,4-eq,eq-substitutions are always trans as well.
  3. Compounds with 1,4-ax,eq- or 1,2-eq,ax-bonds are always cis.

For example, I have put structures of $\alpha$- ($\bf{1}$) and $\beta$-hexoses ($\bf{2}$) in the same image. Just look at orientations of 2,3-dihydroxy-groups in chair conformation of $\bf{1}$. They are in 1,2-ax,eq-orientation, thus are in cis-orientation (see point (3) above). You can confirm that by looking at planer version of the molecule.

In addition, if you look at the orientations of 2,4-dihydroxy-groups in the same chair conformation, you'd see they are in 1,3-ax,eq-orientation as well. Thus are in trans-orientation (see point (6) above). You can again confirm that by looking at planer version of the molecule.

$\endgroup$
2
$\begingroup$

According to Master Organic Chemistry-Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes

$5$. Geometric Isomerism: “cis-” And “trans-” Isomerism In Cycloalkane Rings

[...]For the case where the two groups are on the same side of the ring, we refer to it as ‘cis‘ (from the Latin, meaning, “same side of”.) For the case where the two groups are on the opposite side of the ring, we refer to it as “trans” (meaning “opposite side of'”).[...]

These are the possible structures of 1,4-dichlorocyclohexane (I've taken $\ce{-Cl}$ instead of $\ce{-CH3}$ to better highlight the difference using color as well)

Now, if you notice in the first structure, both chlorines are "above" the ring which means that they are the cis-isomer of 1,4-dichlorocyclohexane. (cis-1,4-dichlorocyclohexane)

In the second structure, one is "above" the ring and the other is "below" the ring. This means that they are 'trans' to each other and so this structure would be the trans-isomer of 1,4-dichlorocyclohexane. (trans-1,4-dichlorocyclohexane)

$\endgroup$
2
$\begingroup$

When reading the formulae, you need to translate these into 3D. Imagine to see the cyclohexanes from the side for either the (cis)-1,4-substitution,

enter image description here

or the (trans)-1,4-substitution (here, intentionally the less stable conformer is chosen to highlight the difference),

enter image description here

From the side, the six carbon atoms of the cyclohexane are like an equator. The substitutents are either on the same side of this (marked by the red line), or not.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.