Kaplan FL #5R Orgo Question

[tuhtles]

I was between choices A and C. I ended picking C, but A was the right answer.
So here are my reasonings for each answer choice:

A would make sense if you consider that the equatorial positions of compound B makes it more stable since it reduces the 1,3-diaxial interactions.
C would make sense if you consider that the staggered anti confirmation is more stable than the gauche conformation.

So how do gauge which steric strains would be more overwhelming without given some sorta of energy chart? I went with C because I thought hydrogens are relatively small compared to the greater steric issues of the gauche conformation since chlorine and the methyl groups are way larger and closer (granted there's 2 hydrogens 1,3 diaxial interaction per bulky group) .. lol I think I'm thinking way into this. I get that equatorial positioning will tend to be more stable, but things can more complicated when you take other factors into account or think about it in a different way like how deuterium is more stable in axial position.

Actually.. I vaguely remember learning about energy differences of these factors and just maybe the math turned out that 1,3-diaxial is more unstable.. but questions like these make me over think ugh. If the methyl group and chlorine weren't right next to each other, I would have picked A in a heart beat. no brainer there

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[ashtonjam]

The diequatorial configuration is definitely favored. I've been told that halogens and methyl groups are very similar in terms of steric hindrance, so this link may be of help to you: http://research.cm.utexas.edu/nbauld/teach/cyclohex.html. See the very bottom for 1,2-dimethylcyclohexane.

The 1,2-disubstitution pattern is very much like the 1,4 pattern, in that the two groups can only be equatorial if they are trans, so the trans isomer is more stable than the cis. The diaxial trans conformer has 3.6 kcal of steric strain, of course and is much less favored than the diequatorial conformer...

The key difference between the 1,2 and 1,4 patterns, is that in the diequatorial 1,2 conformation, the methyl groups are gauche as in gauche butane (remember that gauche essentially implies a 60 degree dihedral angle ). Consequently, this diequatorial conformation no longer is strain free, as was the 1,4-trans diequatorial conformation, where the methyls are very far apart. The 1,2-diequatorial isomer is 0.9 kcal/mol less stable than the 1,4-diequatorial isomer, because of this gauche butane-like interaction (recall that the gauche isomer in butane is destabilized by exactly this amount).

Basically, you're right about both effects existing. It's safe to say that having the two biggest groups (methyl and -Cl) diequatorial wins over the anti configuration. Note that the quoted bit above confirms that the gauche orientation of equatorial 1,2-dimethylcyclohexane DOES destablize it in comparison to equatorial 1,4-dimethylcyclohexane. The equilibrium will be shifted towards the diequatorial for 1,2, but the Keq for 1,4 will be larger because of the missing gauche interaction that shoves it back the other way.

 

[tuhtles]

Ah, I see. This makes sense. Thank you!!