July 13, 2020, 12:48:25 PM
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Topic: 1,2,3 trimethyl cyclopropane stereochemistry, maximum # of stereoisomers  (Read 8706 times)

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Offline spirochete

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Please stick with me on this convoluted post. I was helping a student solve this problem today and was temporarily stumped but I believe I figured it out. I'm trying to reconcile the formula for the maximum # of stereoisomers a compound can have with 1,2,3 trimethyl cyclopropane.

The formula for the maximum number of stereoisomers a compound can have is commonly said to be "2n - the number of meso compounds" where n is the number of stereocenters.

The definition of stereocener is "an atom where interchanging two groups give rise to stereoisomerism"

In 1,2,3 cyclopropane interchanging 2 groups on any of the carbons gives rise to stereoisomerism: there are two possible isomers, (cis,cis,trans) and (cis, cis, cis)), both of which are achiral.

But if the molecule has 3 stereocenters, acccording to the formula it should have 8 possible stereoisomers? This can't possibly be correct, even subtracting potential meso compounds. But there aren't any meso compounds, because none of the carbons are even chiral centers to begin with.

My interpretation here is that "n" in the formula really should be chiralcenters, not stereocenters. I believe some sources, including wikipedia, erroneously use chiral center and stereocenter synonymously when in fact they have different meanings. A chiral center can potentially give rise to chirality, where as a stereocenter gives rise to stereoisomerism.  So every chiral center is a stereocenter, but not vice versa. Is this correct?  Or is there a hole in my logic somewhere?


Offline Dan

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So every chiral center is a stereocenter, but not vice versa. Is this correct?  Or is there a hole in my logic somewhere?

Yes, this is also my understanding - stereogenic centres also include sp2 centres that give rise to E/Z stereoisomerism, for example.

The formula for the maximum number of stereoisomers a compound can have is commonly said to be "2n - the number of meso compounds" where n is the number of stereocenters.

I have never seen a formula that I found generally useful for working out the number of sterecentres. I used to just draw them all and eliminate degenerates to begin with, and with practice I did not have to draw out all 2n.

Using this formula has never really clicked with me - consider the pentitols (D-arabinitol, L-arabinitol, ribitol and xylitol). 3 Stereogenic centres, and we know there are two meso compounds (ribitol and xylitol). The formula gives 6 isomers, the reality is 4. My guess is that the problem is that arabinitol only has 2 stereogenic centres (the central -OH is non-stereogenic) which you would need to factor in somehow.

As soon as the molecule is cyclic you get many more degeneracies, epecially if all the substituents are the same, due to the symmetry of the ring.

I suppose the most general and reliable formula you can have is:

2n - degenerate structures

But how this is of any practical use eludes me - you still need to draw them all out and decide which are degenerate.

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My interpretation here is that "n" in the formula really should be chiralcenters, not stereocenters. I believe some sources, including wikipedia, erroneously use chiral center and stereocenter synonymously when in fact they have different meanings. A chiral center can potentially give rise to chirality, where as a stereocenter gives rise to stereoisomerism.

This doesn't help though - use this modified form and given that you argue 1,2,3-trimethylcyclopropane has 0 chiral centres and 0 meso compounds, it should have 0 stereoisomers, which is false.

My opinion is that looking for a magic formula is a waste of time.
My research: Google Scholar and Researchgate

Offline spirochete

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Please remind me: the central carbon in D-Ribitol has the S configuration because R chiral centers have higher priority than S chiral centers, correct?

Offline Dan

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Not quite, it is s - note the lower case, it is a pseudoasymmetric centre, not an asymmetric centre.
My research: Google Scholar and Researchgate

Offline TwistedConf

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I suppose the most general and reliable formula you can have is:

2n - degenerate structures

But how this is of any practical use eludes me - you still need to draw them all out and decide which are degenerate.

This 2^n stuff is usually introduced when referring to compounds containing sp3 chiral carbons... and then it works quite nicely (and you really don't need to draw them all out to envision if there are any meso isomers possible). Throwing in an E/Z double bond also works to add to the value of n.

These cis/trans diastereomers (or something even easier like the dimethyl cyclobutanes/propanes) without chiral centers are something a student would see first in order to appreciate the idea of stereoisomers long before they have to deal with a non-superimposable mirror image. There really is no formula in these cases and you just have to think about it and work out the possibilities. So, as you mentioned....  no magic.

Offline spirochete

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This doesn't help though - use this modified form and given that you argue 1,2,3-trimethylcyclopropane has 0 chiral centres and 0 meso compounds, it should have 0 stereoisomers, which is false.

My opinion is that looking for a magic formula is a waste of time.

I guess I should have added that if there are no actual chiral centers then the formula simply doesn't apply...but regardless your examples with the sugar alcohols showed an example where even that formula breaks down. 

Unfortunately a significant number of organic professors will continue to use stereocenter and chiral center synonymously. I'm pretty sure even my graduate school adviser didn't even know the difference.

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