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Consider (R)-3-bromo-1,1-dimethylcyclohexane.

I wish to clarify how I assign R,S priorities in a cyclic compound. My book says to compare the atoms along the "path of higher priority." The problem is in a cyclic compound, if you go far along this path of higher priority without finding a point of difference, you may collide (picture below):

I'm assuming that because the path taken initially results in no difference, we can still assign the top path a higher priority than the lower path because the top path has two methyl groups. My book says that if the paths are identical then choose the one with more of the same atoms.

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If you "collide" without finding some difference, then the molecule is not chiral. In the example at hand, it appears that you have analyzed it correctly. First, you look at the 2 carbons adjacent to the chiral center. They are both $\ce{CH_{2}{'}s}$; no difference, therefore move out to the next carbon on each side of the ring. Here we find a $\ce{C-CH_{2}-C}$ on one side and a $\ce{C-CMe_{2}-C}$ on the other side. The latter has higher priority than the former. Stop at this point, no need to go further, and complete the "1-2-3-4" prioritization.

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  • $\begingroup$ Good idea to keep my wits about me; if we really do collide and find that the molecule is symmetrical then by definition there cannot be chirality! Thanks for breaking it down for me. It is very much appreciated, ron. $\endgroup$ – Dissenter Jun 4 '14 at 17:17
  • $\begingroup$ So to conclude in the R configuration of the above specified molecule, the bromine must be equatorial? $\endgroup$ – Dissenter Jun 4 '14 at 17:20
  • $\begingroup$ No, if you flip the ring you'll still have (R)-3-bromo-1,1-dimethylcyclohexane, but the bromine will now be axial. In other words, (R)-3-bromo-1,1-dimethylcyclohexane has 2 conformers, one with the bromine equatorial and one with the bromine axial. The mirror image compound (S)-3-bromo-1,1-dimethylcyclohexane also has 2 conformers, one with the bromine equatorial and one with the bromine axial. The 2 bromine-axial compounds are mirror images of each other as are the two conformers with the bromines equatorial. $\endgroup$ – ron Jun 4 '14 at 17:32
  • $\begingroup$ Good point. I'll draw the other conformer too. Shouldn't have forgot about conformational isomers. $\endgroup$ – Dissenter Jun 4 '14 at 17:39

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