Meso vs. Identical compounds

[SGStudent]

Hi all, I just wanted to ask if somebody can help me clarify meso compounds vs identical compounds. This is one problem from bootcamp: I understand that it is identical since it has mirror-image and mirror-plane of symmetry and it has enantiomers, if assigned R and S. Left compound would have (top-bottom) S,R and right compound would have (top-bottom) R,S.
So it says that one way to identify identical compounds from meso-compounds is identical have enantiomers and meso-compounds has no enantiomers. Therefore if just looking at one compound lets say the left compound and ignoring the right compound, this compound will be meso-compound with mirror-plane of symmetry correct?
Next, if I flipped the right compound, with EtO pointing at the right and comparing this to the original left compound with the OEt pointing at the right, the compounds will be meso-compound, because when I assign R and S (top-bottom), it will be: S,R and S,R. It will have no enantiomers and have mirror of symmetry and superimposable. I am I correct? I am really trying to determine meso from identical compounds. Thanks.

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[gobinoob]

I think you are generally right, however you may be confusing some concepts. When we use the phrase "identical compounds", we refer to two images that are, in fact, the same molecule. The term "meso compound" however, tends to refer to just one molecule. For example, the two images above are not "meso compounds of one another". Rather, we should say that these two images are flipped representations of the same molecule (ie identical) and that this one molecule is meso.

The fastest method of determining if a compound is meso is to use these two checkboxes:

1. Two (or more) actual stereocenters (a "stereocenter" with 2 or more of the same substituent is not a stereocenter)
2. But achiral due to a plane of symmetry and "opposite" stereocenters

Opposite stereocenters, in this case, means that for each R on our molecule, there is a symmetrically placed S that "counters" it. Look at just the left image; it has two stereocenters, so #1 is fulfilled. Next, the two stereocenters are placed symmetrically around a central horizontal line and we find that the stereocenters are S, R. Since these two stereocenters, symetrically placed and opposite in stereochemistry, will counter and cancel one another out, we have "racemization". This fulfills #2 so the molecule is indeed meso!

Note that a meso compound, due to its inherent line of symmetry, has no enantiomers (it is "achiral"); its mirror image ("enantiomer") will be identical to the original compound. In this question, once we have identified that the left compound is meso, we know that the right one, which is its mirror image, must be the same molecule! Or, we could just realize (if we are able to see Fischer projections in our head) that the right image is just rotated 180' compared to the left one). Lastly, note that meso compounds, due to their "racemization", do not rotate plane polarized light (they are "optically inactive"). The R stereocenter will rotate light a certain amount in one direction, but the S stereocenter will rotate it the same amount in the opposite direction. Hope this helps!

 

[KyoPhan]

A lot of your terminology is slightly off as stated. Gobi has a great explanation to solve these types of problems. The only thing I would add to that two checkbox method is to check if the 3 attachments are the same for both centers.

I also use the idea that the S counters the R, making it unable to rotate the light; however, terminology wise, is it correct to say there is racemization for a meso compound? Whenever I deal with racemic questions, they usually only have 1 chiral center. It's not important to solving the problems, but I was just curious.

 

[gobinoob]

A lot of your terminology is slightly off as stated. Gobi has a great explanation to solve these types of problems. The only thing I would add to that two checkbox method is to check if the 3 attachments are the same for both centers.

I also use the idea that the S counters the R, making it unable to rotate the light; however, terminology wise, is it correct to say there is racemization for a meso compound? Whenever I deal with racemic questions, they usually only have 1 chiral center. It's not important to solving the problems, but I was just curious.

Oh yeah! I realize racemic refers to a 50:50 mixture of enantiomers and is perhaps not appropriate here! I just use that term with mesos because it helps me remember that there's equal and opposite amounts of both R and S. I will edit the original to make it more accurate! As for "making sure the attachments are not the same", that is the point of #1, as if a "stereocenter" had 2 or more identical attachments, it wouldn't be a stereocenter! I will edit this as well to make it clearer! Thanks!!

 

[NavyDDS1990]

I'm sorry but what does EtO stand for in questions like this. I was always confused

 

[SGStudent]

I think you are generally right, however you may be confusing some concepts. When we use the phrase "identical compounds", we refer to two images that are, in fact, the same molecule. The term "meso compound" however, tends to refer to just one molecule. For example, the two images above are not "meso compounds of one another". Rather, we should say that these two images are flipped representations of the same molecule (ie identical) and that this one molecule is meso.

The fastest method of determining if a compound is meso is to use these two checkboxes:

1. Two (or more) actual stereocenters (a "stereocenter" with 2 or more of the same substituent is not a stereocenter)
2. But achiral due to a line of symmetry and "opposite" stereocenters

Opposite stereocenters, in this case, means that for each R on our molecule, there is a symmetrically placed S that "counters" it. Look at just the left image; it has two stereocenters, so #1 is fulfilled. Next, the two stereocenters are placed symmetrically around a central horizontal line and we find that the stereocenters are S, R. Since these two stereocenters, symetrically placed and opposite in stereochemistry, will counter and cancel one another out, we have racemization. This fulfills #2 so the molecule is indeed meso!

Note that a meso compound, due to its inherent line of symmetry, has no enantiomers (it is "achiral"); its mirror image ("enantiomer") will be identical to the original compound. In this question, once we have identified that the left compound is meso, we know that the right one, which is its mirror image, must be the same molecule! Or, we could just realize (if we are able to see Fischer projections in our head) that the right image is just rotated 180' compared to the left one). Lastly, note that meso compounds, due to their racemization, do not rotate "plane polarized light" (ie they are "optically inactive").The R stereocenter will rotate light a certain amount in one direction, but the S stereocenter will rotate it the same amount in the opposite direction. Hope this helps!

Thanks for the reply and the clarification, this makes more sense. Yes I did realize that a meso-compound refers to only one compound, hence I said if I look at the single compound to the left and finding an internal mirror of symmetry this is meso-compound. The additional information is very helpful.

 

[KyoPhan]

EtO just means an ethyl attached to an oxygen.
-EtO is same as -OEt. So -OCH2CH3.

 

[SGStudent]

I'm sorry but what does EtO stand for in questions like this. I was always confused

I believe its Ethylene oxide C2H4O