Substituted Cyclohexanes – Equatorial vs Axial

by James

in Conformations, Organic Chemistry 1

Just to bring you up to speed, here’s what we mentioned in the last post. At the bottom, I’ll also correct a little fib I made.

1. Cyclohexane, at room temperature, undergoes a conformational interconversion known as a “chair flip”. In this chair flip, all axial groups become equatorial, and all equatorial groups become axial.[but all “up” groups remain up, and all “down” groups remain down].

2. While the two chair forms of cyclohexane itself are completely indistinguishable, this is not true in most cases. For example, in 1-methylcyclohexane, one chair conformer has an axial methyl group, and in the other the methyl group is equatorial. These are conformational isomers, or simply, “conformers”.

3. At room temperature, these two conformations are in rapid equilibrium with each other. There is an activation barrier of about 10 kcal/mol for this interconversion, since the high-energy “half-chair” conformer is an intermediate in this process. Trying to observe both conformations of 1-methylcyclohexane at room temperature with a device for taking “molecular snapshots” (an NMR spectrometer is what we use – more precise details on this in future posts) results in a blurred picture. Like an old camera trying to take pictures of spokes on a moving bicycle wheel, the “shutter speed” is too slow, and the result is that the images blend together to give an average. Using this device, it’s simply not possible to see both cyclohexane conformers of 1-methylcyclohexane at room temperature.

4. At very low temperatures (about 80 degrees above absolute zero) equilibrium between the two chair forms stops, because there isn’t enough thermal energy available to ascend the activation barrier of 10 kcal/mol. Now,  when we try to take “molecular snapshots” of 1-methylcyclohexane, we do indeed see the two conformations separately.

Now, the correction to the fib. In the last post, we assumed that these two conformations would be equal in energy, and therefore we would see a 50:50 mixture of the two conformations.

Is this true?

There’s only one way to find out. Do the experiment with 1-methylcyclohexane.

When we do this, here’s what we find. Instead of being equal, the ratio of “equatorial methyl” to “axial methyl” conformers is about** 95:5  favouring the conformation where the methyl group is equatorial.


Very interesting! This must mean that the equatorial conformation is of lower energy than the “axial” conformation.

Why might that be?

Let’s look at the Newman projection of the chair. Imagine looking along the C-1 to C-2 bond (which is coplanar with the C-4 to C-5 bond). Here’s what you’d see.


Note that in the conformation where methyl is axial,  there is a gauche interaction between the axial methyl group and C-3. This is absent in the conformation where methyl is equatorial.  This gauche interaction is an example of [van der waals strain] is what makes the axial conformer higher in energy.

There is actually a second gauche interaction if you look along C-1 to C-6 . This gauche interaction is with C-5.
A simple way to keep track is to think of it as the methyl group interacting with the other ‘axial’ hydrogens, at C-3 and C-5. These are called “diaxial interactions” since they are steric interactions between axial substituents.

steric interactions

Bottom line: in two unequal conformations of a cyclohexane ring, the conformation where steric interactions are minimized will be favoured. **

Now here’s a neat consequence of this knowledge. Since this ratio of conformers (95:5) represents a system at equilibrium, we can actually use it to calculate the difference in energy of these two conformers using the following equation:

equation 2

For a 50:50 mixture (K = 1) the energy difference ΔG would be zero.

For methylcyclohexane at room temperature (298 K) the 95:5 ratio of equatorial to axial conformers translates to an energy difference of 1.70 kcal/mol. 

In other words, the equatorial conformer is more stable by 1.70 kcal/mol.

Since there are two gauche interactions, and the strain energy is 1.70 kcal/mol, it’s easy to calculate the value of each interaction: 0.85 kcal/mol .

Now this opens up all kinds of questions. If a methyl group (CH3) leads to an energy difference of 1.70 kcal/mol, then what effect would an ethyl group (CH2CH3) have? Or a Cl? Or OH ? Or tert-butyl ?

We can use the same approach to measure all of these numbers. More about that in the next post.


* This is a bit of a cheat. In the equation  ΔG = –RT ln K  , the value of K is related to T, so the equilibrium ratio at –80 °C will be a bit different than the value at room temperature [quiz time – do you think it will be higher than 95:5 or lower than 95:5 ?]   However, one can then solve for ΔG and use this number to calculate what K is at room temperature.

** Enterprising students might ask what happens if the axial hydrogens on C-3 and C-5 are removed. Would this change the equilibrium? Absolutely!



In the molecule above, the CH2 groups at C-3 and C-5 have been replaced by oxygen. Since there are no longer any significant diaxial interactions between the methyl group and substitutents on the ring, there is no significant energy difference between the equatorial and axial conformations of this molecule.

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