Determining R-S configuration with lowest priority groups in plane of paper

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)

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:

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?

• You seem to know how to prioritize groups. Follow the instructions here and believe in yourself. chemistry.stackexchange.com/questions/151831/… Commented Aug 29, 2021 at 16:00
• "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. Commented Sep 4, 2021 at 23:41

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.)

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.

• 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? Commented Sep 4, 2021 at 22:01
• 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. Commented Sep 4, 2021 at 22:46
• Change of mind. The Fischer projections of the hexahalide are linked in the ADDENDUM. Commented Sep 4, 2021 at 23:23
• 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. Commented Sep 5, 2021 at 8:34
• Fair enough. BTW, I should have said above, "Oh look! They are mirror images". Commented Sep 5, 2021 at 13:18

One way to assess the configuration without having yo swap or ritate functional groups us to use the concept of even and odd permutations.

Say you have ranked the substituents on a stereocenter and a group other than Group 4 is oriented to the bottom. Proceed as follows:

1. List the rankings of the three groups directed to you in clockwise order.

2. Attach the bottom group at the end of the list.

3. Count the pairwise interchages that you would need to get your permutation to the "standard" 1234. An even count, called an even permutation, corresponds to R and an odd permutarion corresponds to S.

Let us look at the left carbon in molecule A. Bromine ranks 1, chlorine 2, fluorine 3 and the other half of the molecule, attached through carbon, ranks 4. So 2 is directed to the bottom and the other groups in clockwise order are 134. Thus you render your permutation as 1342.

To get from 1342 to 1234 you then need to interchage the 2 with the 4 and then with the 3; if you try to interchage 2 directly with 3 first you have 3 and 4 in the wrong order. Thus you need two interchanges, an even number, and so the left carbon is configured R.

Can you work out the other centers with this method?