I have come across a question in my textbook: enter image description here

The textbook gives this answer: enter image description here

enter image description here

I understand the explanation for the first compound. However, for the second structure, I am not understanding why the 2 methyl groups are diastereotopic. Is there not free rotation about the $\ce{C-C}$ bond, connecting the $\ce{CH(CH3)2}$ group to the ring such that the 2 methyl groups become equivalent? The answer key makes it seem as though there is no rotation and the methyl groups are fixed in place. I feel as if I am misunderstanding the question. Please help me clear up the confusion.

  • $\begingroup$ Substitution is frequently the easiest approach. Replace a hydrogen on each of those two methyl groups with a deuterium to create two new compounds? What is the relationship between those two compounds? Identical? Hydrogen atoms are homotopic. Enantiomers? Enantiopic. Diastereomers? Diastereotopic. $\endgroup$ – Zhe Nov 23 '18 at 19:04

The two methyl groups do not see the same average environment

The issue here is not whether the two methyl groups are freely rotating about the bond to the CH group: the issue is whether they see the same environment on average when the bond between the CH and the cyclohexyl ring rotates.

The easiest way to see that they don't is to build a model and play with it.

If you do you will see that their average environment can never be the same no matter how much rotation you do. Imagine rotating them so that the CH is facing the same way as the Ph with respect to the ring. The two methyls are different: one will be closer to the Ph than the other. Now rotate the CH-ring bond 180 degrees. The CH now points away to the opposite side of the ring to the Ph. But the environments of the methyl groups are not reversed: the other methyl is now closer to the Ph but is now on the opposite side of the ring so sees a different environment. So that rotation does not equalise the average environment for each methyl.

This is generally true in chiral compounds.

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  • $\begingroup$ Your explanation makes sense. I made a model and tried rotating the CH-ring bond and noticed that at no particular instance are the methyl groups the same distance away from the phenyl group AND the cyclohexane ring. I was wondering, however, if we were to remove the phenyl group, then would the methyl groups become equivalent? (I also noticed an instance of symmetry in my new structure without the phenyl. Is that at all relevant?) $\endgroup$ – James Bond Nov 23 '18 at 21:49
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    $\begingroup$ @JamesBond I think that, were there no phenyl, the methyl groups would be equivalent as the average environment in the ring would now be the same from any direction and, perhaps more importantly, the molecule would no longer be chiral. $\endgroup$ – matt_black Nov 23 '18 at 22:49

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