[azmanam]

A molecule with one stereocenter has one pair of stereoisomers: a pair of enantiomers.



A molecule with two stereocenters has 6 pairs of stereoisomers: 4 pairs of diastereomers and 2 pairs of enantiomers.



Given that, how many stereopairs does a molecule with 3 stereocenters have (assume no meso)? How many pairs of diastereomers? How many pairs of enantiomers?

Can you come up with formulas for the number of total stereopairs, diastereomer pairs, and enantiomer pairs given a molecule with n stereocenters?

First one with the right answer is the coolest!

For more fun, don’t just reply when you have the right answer. Reply with some initial guesses, first approximations, and thought process. A good discussion is more fun than one person posting the right answer

[fledarmus]

How do you define a "pair" of diastereomers? Diastereomers are molecules which are stereoisomers but not enantiomers - they don't come in pairs.

[azmanam]

Systematically compare the relationship of any two stereoisomers. Are they enantiomers or are they diastereomers? Do this for all pairs (duos, diads, OtherGroupsOfTwo).

[yesway]

I agree with fledarmus. For n=2 the possible configurations are

RR    SS
RS    SR

which, in my opinion leads to two possible diastereoisomers. The diastereoisomeric relation includes two pairs of enantiomers, but I don't see how that would increase the total number of diastereoisomers. 

So in conclusion, I would expect 2^n stereoisomers to consist of 2^(n-1) diastereoisomers.

[Schrödinger]

Fischer projections and combinatorics? I got an answer that I've PM'd you.

[azmanam]

I agree with fledarmus. For n=2 the possible configurations are

RR    SS
RS    SR

which, in my opinion leads to two possible diastereoisomers. The diastereoisomeric relation includes two pairs of enantiomers, but I don't see how that would increase the total number of diastereoisomers. 

So in conclusion, I would expect 2^n stereoisomers to consist of 2^(n-1) diastereoisomers.

There are more than 4 possible pairings of any two molecules: RR w SS, RR w RS, RR w SR, etc... Some of those pairings will be enantiomers, some will be diastereomers. 2^n-1, then, is incorrect.

[yesway]

Sorry, I don't get it. 

I agree that there a four diastereoismeric relations for n=2 namely

RR <-> RS
SS <-> RS
RR <-> SR
SS <-> SR

but only two diastereoisomers - in the physical sense as two distinguishable compounds.
E.g. if I follow the diastereomeric relations from RR to RS or either SR it will not be distinguishable by achiral means whether I have SR or RS configuration. So, the pair  SR/RS will behave as one of two diastereoisomers.

[azmanam]

I think you're overthinking it.

I appreciate the chemical reactivity difference, that's not what I mean by 'pairings' here.

We agree that for 2 stereocenters, there are 4 total stereoisomers, yes? (RR, RS, SR, SS).

From those 4, choose 2 (randomly RR and RS). What is their relationship? They're diastereomers.
Choose another 2 (randomly RS and SR). What is there relationship? They're enantiomers.

From these 4 stereoisomers, compare all possible pairings of two molecules. There are 6 such pairings: (RR RS) (RR SR) (RR SS) (RS SR) (RS SS) (SR SS). 2 of those pairings are enantiomers (RR SS), (RS SR). 4 of those pairings are diastereomers (RR RS) (RR SR) (RS SS) (SR SS). That's all we're doing here.

So, for 3 stereocenters, how many total pairings are there? How many of those pairings are enantiomers? how many of those pairings are diastereomers?

If I'm still not explaining it well, I do apologize. Perhaps someone else can phrase it differently to get my point across?

[fledarmus]

Mathematically it can be done - for n stereocenters, there will be 2n possible stereoisomers (assuming no meso structures), and for any one of those 2n stereoisomers, there will be 1 enantiomer. Any other combination of that stereoisomer with the remaining 2n-2 stereoisomers will be (according to your definition) a "diastereomeric pair", and adding them all up and subtracting redundancies will give you the total.

The question is, why bother? There is no particular utility in grouping diastereomers as pairs unless there is something actually relating the two diastereomers of all the other possible stereoisomers. The number of enantiomeric pairs possible is important, the number of total stereoisomeric compounds is important, but the number of ways of combining two diastereomers and not getting an enantiomer doesn't seem to be a useful value.

I'm not trying to be flip or demeaning, it really is a serious question. Is there some aspect of possibly statistical or computational chemistry that I'm missing in respect to the question?

[azmanam]

Nope, there's no utilitarian purpose for this exercise at all. Just some Monday maths fun. That's why it's a riddle.

[yesway]

I think there is a difference between diastereomeric pair (I would rather call it relation) and a diastereomeric compound (relative to another). In a chromatography run with n=3 I will only isolate 4 separate compounds, although the number of diastereomeric relations is higher. I would call those four compounds diastereoisomers, as I think is common sense. For C5 aldehyde sugars, one commonly speaks of four diastereoisomers lyxose, arabinose, xylose, ribose. There are many other examples.

I'm not trying to argue here - trying to reconcile practical experience with the formal rigorosity (diastereomeric relations).