Why is this chair conformation the most stable and which leaving group is best?

![enter image description here](https://i.stack.imgur.com/bnWO5.jpg)


I'm confused about how this chair conformation was constructed and why that particular Cl leaves. Why is one of the Cl's drawn axial while the others are drawn equatorial (it seems like the others could have been drawn axial as well)? And, to reiterate, why is it that Cl that's favored to leave over the other two?




In cyclohexane, the equatorial position is [energetically favored](http://en.wikipedia.org/wiki/Cyclohexane_conformation#Chair_conformation) over the axial position. So the lowest energy conformer is the one where the most substituents are in equatorial position. (Or rather: Where you minimize the energy according to the [A Value](http://en.wikipedia.org/wiki/A_value)).


The axial Cl is favored as leaving group because of the elimination reaction mechanism. Normally, we see $\beta\_{1,2}$ elimination in an anti-fashion. Which is what is proposed by your image.


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The chair conformation is the most stable due to the following reasons:
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* The carbons in cyclohexane have sp3 hybridization and hence they tend to have an angle of 109.5 degrees. The chair conformation allows it to have the closest possible angle which would be 110.9 degrees.Due to this the strain reduces and results in a lowered energy. The two chair conformations have the lowest total energy, and are therefore the most stable.
* Since cyclohexane's planar form results into bond angle being 120 degrees
(which means a higher deviation from 109.5). Hence the energy of the planar form is higher and it always has the tendency to convert to chair form.


Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.


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