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I was studying R-S nomenclatures and determining configurations of molecules when I came across a certain problem, which had the 4th priority group on the plan of paper (not on a dash or wedge)

According to this: https://www.masterorganicchemistry.com/2016/10/20/introduction-to-assigning-r-and-s-the-cahn-ingold-prelog-rules/#five

All I need to do to determine the R-S config in these cases is to switch the group in the 4th place with the one in the back, thus "flipping its configuration", then determine the R-S for that molecule and the actual result is the opposite. I know why this works and it has worked for me in the previous problems that I have attempted. However, in this question:

enter image description here

For the left compound (looking at the carbon on the bottom), clearly, $\ce{Br}$ has priority $1$, $\ce{Cl}$ is $2$, $\ce{F}$ is $3$ and the other carbon attached to it is clearly $4$ in priority which is in the plane of paper. Swapping the groups in priority $4$ with that in the back i.e., $\ce{Cl}$, and going from $1,2,3$, I get S as the absolute configuration of the flipped molecule, so the actual configuration should be R. But the answer does say that S is indeed the configuration of both the carbons in the 1st compound and R is the configuration for both carbons on the right molecule.

Why does the swapping of the groups in back and lowest priority not work here?

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  • $\begingroup$ You seem to know how to prioritize groups. Follow the instructions here and believe in yourself. chemistry.stackexchange.com/questions/151831/… $\endgroup$
    – user55119
    Aug 29, 2021 at 16:00
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    $\begingroup$ "the 4th group on wedge, dash, or on the plane of paper" This is not the right question to ask. You don't need to, and should not, memorise a series of rules for each individual case. Instead, you rotate the molecule until it matches the description given in the rules (lowest priority group facing away). To do this, you either build a model which you can physically rotate, or learn to draw out the rotation on paper, or learn to visualise it in your head. $\endgroup$
    – orthocresol
    Sep 4, 2021 at 23:41

1 Answer 1

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Techie5879: If you have followed my suggestion in the Comment above, you probably have answered this question yourself. Now that you no longer "flip" structures, you should have the R/S assignments for A and B (vide infra). Because each carbon in A and B bears F, Cl and Br and both A and B have R and S configurations, they are identical meso compounds. I was not familiar with the term "homomer", so I looked it up.

Homomers : Homomers are the identical representations of the same compound. It means that when the same compound is represented in different ways, then all the representations are known as homomers of each other. These representations are superimposable on each other. (Italics added.)

Config. A and B

If you pickup A (mentally) and superimpose the S-carbons onto B, pairing halogens (F with F, etc.), then the R-carbons will superimpose halogens but they will not match halogen for halogen. In this context they are not homomers. That is to say, they are different conformations of the same compound. Because A and B are identical compounds, they cannot be enantiomers or diastereomers.

I would appreciate any constructive feedback as to whether or not my interpretation of "homomer" is correct or whether or not the term serves a meaningful purpose. Exclude any applications to polymers or biochemistry.

ADDENDUM: Per the OP's request, the Fischer projections of A, aka B, are provided here.

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  • $\begingroup$ Yes your interpretation of homomers is correct. Also, this is the same answer I am getting, but the answer in the book says they are enantiomers and apparently drawing the Fischer projection gives opposite configurations at the carbons? Could you also update your answer with what the Fischer projection of the two molecules would be? $\endgroup$
    – Techie5879
    Sep 4, 2021 at 22:01
  • $\begingroup$ The person who wrote the question was not necessarily the person who provided the answers. They may have simply placed a mirror plane between A and B and said, "Oh look! they are images. Therefore, they are enantiomers." Nonsense, they are identical meso compounds. I could draw Fischer projections but they are more suitable for carbon chains greater than two. $\endgroup$
    – user55119
    Sep 4, 2021 at 22:46
  • $\begingroup$ Change of mind. The Fischer projections of the hexahalide are linked in the ADDENDUM. $\endgroup$
    – user55119
    Sep 4, 2021 at 23:23
  • $\begingroup$ Okay, this matches with my conclusion then. I'll wait a bit for another answer, if I see any change of conclusion as to whether there are enantiomers or homomers, and if not I'll accept this answer. Hope thats fair. $\endgroup$
    – Techie5879
    Sep 5, 2021 at 8:34
  • $\begingroup$ Fair enough. BTW, I should have said above, "Oh look! They are mirror images". $\endgroup$
    – user55119
    Sep 5, 2021 at 13:18

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