Path to take when labeling R/S

Question

I'm trying to find the R/S for the carbon marked with the asterisk. As I branched out to from the asterisk carbon to the red carbons, the methyl group has the 3rd priority (with the hidden hydrogen 4th). Here comes the confusion. In order to find the first point of difference for the two carbons, for one of the carbon, I can simply move outwards to the next green carbon (C, C, H). But for the other carbon, there are two paths I can take, one which gives (C, H, H) and the other (C, C, H). Now, which path should I take?

Assuming I take the path that gives me the (C, C, H) carbon. This gives me a tie. Again, I have two paths to take (indicated by the green arrow). Which path should I take? If I choose to go 'down', the two arrow clash at a single carbon.

I have consulted numerous textbooks but none of them seem to even surface this as an issue. In general, if there are multiply paths, which path should one take?

Extension of question: In a scenario that there are two carbon, C1 and C2 that 'ties', and it so happens that each of the two carbons has 'two paths' to take: C1 with path 1 and path 2, and C2 with path 3 and path 4. What happens if the priority of taking the different paths is not clear. Eg. Priority when taking path 1 is higher than the path taken by path 3, but path 2 has a lower priority then when taking path 4?

Answer 1

The compound you are presenting is a natural product, compactin. The numbers of the carbon atoms are arbitrary. The digraph of C₉ shows how the configuration is determined. Sphere 2 contains the atoms C₈, C₁₀, C₁₄ and H. Clearly, hydrogen has the lowest priority. Sphere 3 has the locants C₁₀{15,5,H}, C₈{7,(7),H} and C₁₄{H,H,H}. In a one-to-one comparison of the locants, C₁₀ and C₈ have priority over C₁₄ because C>H. So C₁₄ has the next lowest priority. C₍₇₎, which is a duplicate atom of atomic number 6, is attached to three atoms in sphere 4 of atomic number zero. Zero is as low as you can go. Therefore, the priorities are C₁₀>C₈>C₁₄>H. C9 has the (S)-configuration.

Answer 2

You always have to compare all paths at the same level of depth.

I will use the numbering in user55119’s answer.

The first level of depth gives us $\ce{C,C,C,H}$, we can assign H-9 4th priority.

The second level of depth for the three C atoms gives us:

• $\ce{C,C,H}$ for carbon 10;
• $\ce{C,(C),H}$ for carbon 8; and
• $\ce{H,H,H}$ for carbon 14.

Thus, carbon 14 gets 3rd priority.

The third level of depth gives us:

• carbon 10
  • $\ce{C,C,C}$ for carbon 5
  • $\ce{C,H,H}$ for carbon 15
• carbon 8
  • $\ce{C, (C), H}$ for carbon 7
  • nothing for ghost carbon (7)

The first point of difference is carbon 5 versus carbon 7; 5 is attached to three carbons while 7 is attached to two carbons and a hydrogen. Thus, $5>7$, thus $10>8$. This is reinforced by 15 being a real atom with bonds while (7) is a ghost atom not attached to anything. Thus, even if carbon 7 is methylated 10 still wins over 8.