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I am trying to figure out the R and S configuration of bridging carbons in bicyclic systems.

In the image, there are two bicyclic systems and depending on the position of the Br substituent, the R or S configuration of the bridging carbons change. How is the configuration actually determined? I don't understand how the prioritization works in this case, as two substituents are similar. Thus, the reason must be the adjacent stereocenter... But I am not able explain it properly.

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    $\begingroup$ Are you sure the bridgehead carbons are stereogenic? $\endgroup$
    – Zhe
    Commented Jul 17, 2020 at 16:51
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    $\begingroup$ @Zhe Yes, because the solution to my exercise marked the carbons with R and S tags and when drawing these structures in ChemDraw, the R and S configurations were assigned automatically, too... $\endgroup$
    – QuestionCookie
    Commented Jul 17, 2020 at 17:08
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    $\begingroup$ You should continue in all directions at once, not just along the cycle. $\endgroup$
    – Ivan Neretin
    Commented Jul 17, 2020 at 17:33
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    $\begingroup$ To be sure, the carbons of two bis-methylene bridges are heterotopic because of their in-space relationship to the substitution, but again, does that matter? If you say yes, then you would also need to assign a configuration to the isolated methylene unit. That has two hydrogen atoms, but they are diastereotopic. But you're not claiming that that is a stereogenic center. Therefore, something is contradictory in your reasoning. $\endgroup$
    – Zhe
    Commented Jul 17, 2020 at 18:26
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    $\begingroup$ So the assignment of Chemdraw and the solution to our exercise is kind of confusing. I totally agree with you, that the two loops are identical, hence, defining R/S configurations does not make sense. Can you explain why the solution and chemdraw assign R and S? Or what exactly does the R/S at bridging carbon mean? I mean there must be a reason why it is written like this, mustn't it? $\endgroup$
    – QuestionCookie
    Commented Jul 17, 2020 at 18:59

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Several days ago I offered a solution to the R/S assignment of the bridgehead carbons of (1R,2S,4R)-2-bromo-2-methylbicyclo[2.2.2]octane 1. During the middle of the night I realized that the method was faulty because the solution would not accommodate the (1S,2R,4S) enantiomer 3 but would rather predict (1R,2R,4R) which clearly is a non-existent diastereomer. The shortcoming lay in not applying, in a heirarchical sense, CIP rule 4b (RR/SS>RS/SR) over rule 5 (R>S).
Focusing on bromide 1, the configuration of C2 is "S" with Br>C1>C3>C9. Structure 2 is structure 1 from a different perspective to which we will refer during the ensuing discussion. To determine the configurations of the bridgehead carbons, C1 and C4, a digraph is the method of choice. In the "Digraph of C1" one starts at the non-duplicate atom C1 and traces the six possible paths back to duplicate atoms C(1) that are each attached to three phantom atoms of atomic number zero. The vertical chain has the top priority while the hydrogen (not shown) at C1 has the lowest priority. The two horizontal chains are mirror images as far as the atom connectivity is concerned and they must be assigned to the second and third priorities. In these chains C2 has been predetermined as having the S-configuration. Locate C4 in the left chain. It has the R-configuration because C3>C8>C5. Locate these atoms in structure 2 and convince yourself. [I use my hands. Point your right thumb from C4 to H and your fingers will point from C3 to C8 to C5]. In the right hand chain C4 has the S-configuration because C3>C5>C8. Check these priorities in structure 2.
Now the left hand chain is designated as RS and the right hand chain as SS. Since RR/SS>RS/SR, the order of carbon atoms attached to non-duplicate C1 is C2>C6>C7. Locate these atoms in structure 2 and convince yourself that C1 does indeed have the R-configuration! Look at "Digraph of C4". Now that you are versed in the method, the priorities around non-duplicate C4 are C3>C8>C5 because C3-chain>SS>RS>H. C4 also is of the R-configuration! <continued>


The digraphs below confirm the assignments to the enantiomer 3 (4), (1S,2R,4S)-2-bromo-2-methylbicyclo[2.2.2]octane. I'll leave the analysis to the reader.


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