2
$\begingroup$

Consider diequatorial trans-1,2-dimethylcyclohexane and diaxial trans-1,2-dimethylcyclohexane as shown in the figure below [1, p. 178]. In both compounds, the ring is free of angle strain.

trans-1,2-Dimethylcyclohexane has a conformation in which both methyl groups are equatorial and one in which both methyl groups are axial. As we would expect, the conformation with both methyl groups equatorial is the more stable one.

The two conformers of trans-1,2-dimethylcyclohexane with interconversion arrows labelled 'ring flip'

It can be clearly seen from the figure that in the diaxial, the methyl groups are much farther away than they are in the diequatorial. Hence, the diaxial conformer should be more stable due to less torsional strain or less repulsive dispersion forces. But my book states the opposite. Why?

Reference

  1. Solomons, T. W. G.; Fryhle, C. B.; Snyder, S. A. Organic Chemistry, 11th ed.; Wiley: Hoboken, NJ, 2013. ISBN 978-1-118-13357-6.
$\endgroup$
2
  • 3
    $\begingroup$ This might be of help. And the word is "equatorial", just as in equator. $\endgroup$
    – user55119
    Jun 3 at 14:09
  • $\begingroup$ @user55119 I think this is a common mistake. For e.g. the source of this answer wrote equitorial instead of equatorial. $\endgroup$ Jun 4 at 9:28

3 Answers 3

13
$\begingroup$

The key concept is 1,3-diaxial interaction. See the figure 1 below using methylcyclohexane as an example.

Depiction of 1,3-diaxial strain in methylcyclohexane
Figure 1: Equatorial and axial conformations of methylcyclohexane with 1,3-diaxial strain highlighted.

If a methyl group is in axial position of a cyclohexane ring, the angle strain-free chair conformation means that it ends up rather close to the axial hydrogen atom two carbons along the ring. This steric strain is depicted in the axial case with the inverted brackets. In trans-1,2-dimethylcyclohexane, the di-axial conformation would introduce this strain into the equation twice, destabilising the molecule.

Now you might want to argue that including a second methyl group into the depictions below would cause other types of strain to appear and you would be correct. In the di-axial conformation the two methyl groups are anti-to one another while the di-equatorial configuration has them gauche. However, realise that the cyclohexane ring also exists. Each methyl group in the di-equatorial conformation is gauche with respect to the other methyl but anti with respect to the cyclohexane ring. In the di-axial conformation, each methyl group is anti with respect to the other but gauche with respect to the cyclohexane ring. (See figure 2 below.)

Conformations of trans-1,2-dimethylcyclohexane with 1,3-diaxial strain highlighted in the axial confiormation for both methyl groups
Figure 2: di-equatorial and di-axial conformations of trans-1,2-dimethylcyclohexane with 1,3-diaxial strain highlighted for both methyl groups.

Comparing it in that way shows that the strain between the methyl groups and the ring remains more or less the same in both cases. Probably having the two methyl groups anti would be slightly preferred as the ring is conformationally less flexible and would thus induce marginally less strain. However, the clear driving force here is 1,3-diaxial strain which only occurs in the diaxial conformation.

$\endgroup$
9
  • $\begingroup$ Thanks, I understand that here the steric repulsion between 1-3 positions was the cause. But, I can't understand this point "" Each methyl group in the di-equatorial configuration is gauche with respect to the other methyl but anti with respect to the cyclohexane ring"" . How? $\endgroup$ Jun 5 at 3:41
  • $\begingroup$ Shiva: The diequatorial configuration [sic; conformation] is anti with respect to the cyclohexane ring. Consider methylcyclohexane in the equatorial chair conformation. The dihedral angles Me-C1-C2-C3 and Me-C1-C6-C5 are both 180 degrees. This "interaction" is taken as 0 Kcal/mol. To see how heats of formation play a role in this type of analysis, look here. $\endgroup$
    – user55119
    Jun 6 at 1:28
  • $\begingroup$ @user55119 But please reconsider the figure of di-equatorial 1,2-dimethyl cyclohexane . How can both methyl groups in them are anti? Clearly their dihedral angle is less than 90° and 180° is greater than 90° . How are they anti with respect to ring? $\endgroup$ Jun 6 at 16:43
  • $\begingroup$ Shiva: There are two types of interaction. Substituents with the cyclohexane ring and substituents with each other. In the diequatoral-1,2-dimethylcyclohexane the interactions of the first kind are taken as non-existent because of the anti relationship of each methyl group with the ring. However, there is a gauche interaction of the two methyl groups with each other (~0.9 Kcal/mol). In the diaxial conformation there are two gauche interactions of each methyl group with the ring (total: 4 x 0.9 Kcal/mol = 3.6 Kcal/mol). continued $\endgroup$
    – user55119
    Jun 6 at 17:23
  • $\begingroup$ ... See this link besides the one in the Comment I made in your original post. $\endgroup$
    – user55119
    Jun 6 at 17:24
-2
$\begingroup$

Basically, groups attached at the axial position are under some strain. There is angle strain between 2 axial groups and some repulsion between them.

The more stable form usually contains bulkier ps at equatorial position, which can relieve some strain in them and the lighter groups at axial position. It is a fact than equatorial positions are more stabilising, due to molecules being oriented in an ‘equator like’ manner.

You may want refer to:

https://www.researchgate.net/post/Why_is_a_molecule_most_stable_when_its_heavier_substituent_is_at_the_equatorial_position

$\endgroup$
-4
$\begingroup$

Generally , equatorial conformers are more stable due to less torsional strain . However in this case , where the ring is di-subsituted at vicinal carbons , the repulsion between electron clouds is a dominating factor than the repulsion between the methyl groups . The repulsion between electron clouds is more in axial position ( though their bonds seem somewhat to be antiparallel ) than it is between electron clouds of bonded pair of equatorial one .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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