Q. 20

[newno558362]

Hey everyone, I've got a pretty basic question here.

I understand that an equatorial substituent is almost always preferred over axial one. I also understand the reasoning behind II and III in this question.

I am having a hard time figuring out why I B is correct. I figured that the two CH3 groups would interfere with each other, and thus I A seemed more attractive (even though I B has both substituent in equatorial positions)



It's not that hard, I just don't remember the "rules" for this.

Please help!

[newno558362]

Hmm, I suppose I understand now.

No matter what, the 1,2-dimethylcyclohexane still will have two CH3's right next to eachother...

so in order to reduce interference from the ring, both substituents adopt the equatorial position?


Anyways, please let me know if my reasoning is right, and any additional info is appreciated

[admin]

"so in order to reduce interference from the ring, both substituents adopt the equatorial position? "

Yes, it is generally referred to as 'steric hindrance' which comes down to the 'bulk effect' or that electron shells repel and therefore want to be maximally apart. To really understand why equatorial positions are further apart from each other than axial positions, 3 D modeling is helpful (of course, you could choose to just memorize the rule for the MCAT; there are exceptions but not relevant for the MCAT).

www.chem.ucalgary.ca/courses/351/Carey5th/Ch03/ch3-06.html

NB. You used the word "adopt" in your question which should be replaced by "prefer" to be more accurate since conformations are constantly changing at room temperature though some conformers may be dominant due to their stability.

[newno558362]

Ah yes, they "prefer" equatorial positions.

Thanks

[harman_bai5664]

What about the given changes in G? The free energy changes are all negative, doesn't that imply that the products in each of those reactions are more stable?

[morganzee6964]

[admin]

Yes you are correct however the question asks the comparisons of each molecule in each answer category or number such as for I is it a or b or for II, is it a or b and for III, is it a or b so the Gibbs free energy would only be useful if one were to compare I with II with III and this question does not ask for that.

So, the explanation from the exam is correct and it is the fact that the diastereomer with equatorial substituents are the most stable as you get the least repulsion or electronic hindrance with the respective configurations (equilatiral,equilateral).

[dnpgr16513]

I still don't understand why 1 would be B. There is steric hinderance with the CH3 groups being so close to each other. This makes the ring more unstable, correct? At least in A they're farther apart from each other, but I don't know which weighs more: steric hinderance or equatorial positions.

please explain.

thanks.

[pinoalejan6739]

Someone messed up on the Gibbs free energy of II because there is no way it can be negative and have the reactant be more stable than the products.

This is a bad question, why would the cis and trans be in equilibrium. What should be in equilibrium is the 2 different conformations when they undergo ring flipping. In either case if the gibbs is negative the products are more stable than the reactants.