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I got this question wrong on a test and I don't know why:

Question: Which conformation is more stable?

enter image description here

Answer: the chair on the left is more stable.

I thought that the ring conformation has zero ring strain so I'm confused.

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    $\begingroup$ The conformation on the left is the one where all the bond angles can assume the natural tetrahedral angle, so... $\endgroup$ – user95 Jul 10 '12 at 0:05
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    $\begingroup$ Last not least the plane thing is not a conformer! This plane thing would "sit" on top of a local maximum of the conformational hypersurface, thus it does not "exist". $\endgroup$ – Georg Jul 11 '12 at 12:16
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    $\begingroup$ Yeah, the other main conformation of cyclohexane is boat, not flat. I recommend getting a molecular modelling kit and building them; That was how my orgo prof taught us, and it really lets you 'see' the bond angels and why it exists as a chair and boat. $\endgroup$ – Canageek Jul 11 '12 at 16:03
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By symmetry, the cyclohexane on the right requires each carbon atom to have a $\ce{C-C-C}$ angle of 120 degrees. From VSEPR theory, this is not optimal for tetrahedral carbon. A regular tetrahedron has angles between vertices of approximately 109.5 degrees and the equilibrium bond angles of a secondary carbon won't be too discrepant from this.

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Pointing out the stable conformations of cyclohexane was a major part of Barton's Nobel Prize. His original review is here (paywalled, but you can see the first page).

The topic is discussed by Henry Rzepa on his blog starting with a little history:

Like benzene, its fully saturated version cyclohexane represents an icon of organic chemistry. By 1890, the structure of planar benzene was pretty much understood, but organic chemistry was still struggling somewhat to fully embrace three rather than two dimensions. A grand-old-man of organic chemistry at the time, Adolf von Baeyer, believed that cyclohexane too was flat, and what he said went. So when a young upstart named Hermann Sachse suggested it was not flat, and furthermore could exist in two forms, which we now call chair and boat, no-one believed him. His was a trigonometric proof, deriving from the tetrahedral angle of 109.47 at carbon, and producing what he termed strainless rings. henry rzepa blog

Some related calculations are available here which show the relative energy levels of the different conformers.

Some of these results are easy to visualise with simple molecular modelling kit, but you need space-filling models to really appreciate some of the subtle detail (eg why a twist-boat conformation is more stable than a boat).

It is also worth noting that, at least at room temperature, the conformers will interconvert rapidly as there is enough thermal energy to overcome the barriers between them.

And, there is even a Youtube visualisation showing some of this.

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If you look at a Newman projection of "flat" cyclohexane, you'll see that along each C-C bond, the hydrogens and carbons are eclipsed. This leads to considerable torsional strain (imagine twisting a spring - it naturally wants to spring back) favoring relief of these unfavorable interactions. By contrast, in the chair form of cyclohexane, the Newman reveals that all of the hydrogens and carbons are staggered as you look along each C-C bond, which is a considerably more stable conformation.

The difference between the two forms is about 102 kJ/mol. For more references see here http://www.ch.ic.ac.uk/local/organic/tutorial/cyclohexane/index.html

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  • $\begingroup$ Bond dihedral effects are something I overlooked and on reflection (with reference to the energies of eclipsed/gauche conformations of the central bond of n-butane) 102 kJ/mol sounds about right for 6 bonds. How much water does my answer hold, so to speak? $\endgroup$ – Richard Terrett Jul 11 '12 at 12:41
  • $\begingroup$ I don't have a precise answer for you, but the ring strain effect (120 degrees) would be minimal when compared with torsional strain effects. $\endgroup$ – james_ash Jul 11 '12 at 22:22

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