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I had this question asked in my exam, where I was supposed to find if this pair is a diastereomer: enter image description here

Now, I have learnt this rule where: Making odd number of swaps in a molecule means they both are enantiomers, and making even number of swaps means they are identical.

So according to this rule, I made even no. of swaps in the molecule to the right, which means it is identical to the inital molecule that was on the right. Now this newly formed molecule superimposes the molecule to the left perfectly. Hence, in my opinion the two molecules in the picture above should be identical. But the answer given in my answer key is that they are diastereomers, and not identical.

Is there any flaw in my justification?

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    $\begingroup$ The trick you are using is not always correct. Follow the standard definitions rather than some rules which cease to be followed outside of some specific cases. $\endgroup$ May 27 '21 at 3:07
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    $\begingroup$ These are optically active optical isomers which are not identical and thus diastereomers. $\endgroup$ May 27 '21 at 3:09
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    $\begingroup$ Out of curiosity, which are some of these specific cases where this trick fails to follow and why? Are these cases just arbitrary exceptions, or is there any logic as to why this trick doesn't apply? $\endgroup$
    – Aashita
    May 27 '21 at 3:30
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    $\begingroup$ You should be asking where does this trick apply because I don't see many cases there this is absolutely correct. $\endgroup$ May 27 '21 at 3:33
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    $\begingroup$ @Aashita I forgot to mention the fact that it will be much more fun if you first prove this by yourself(there are many methods to do that, maybe like 3 or 4)and then use it(maybe if you have exams...). NOTE: I would not call this a trick, it is just a inference. $\endgroup$
    – Rishi
    May 27 '21 at 4:43
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While this rule is certainly not infallible, it is often used in organic compounds, especially for Fischer projections.

  1. The odd/even rule implies that no change of configuration occurs if an even number of swaps are made at a single chiral center and change of configuration occurs if an odd number of swaps are made at a single chiral center.

  2. Here, you have made 3 swaps at both ends which means you changed the configuration of 2 out of 4 chiral centers. Hence, you can make the inference that 2 chiral centers (C-2 and C-3) are the same, while another 2 (C-1 and C-4) are different. These are hence diastereomers.

  3. I am assuming that you thought you made 3 + 3 = 6 swaps, so the compound should be identical. But the odd/even swap rule is only applicable to a single chiral center, hence the error.

  4. Finally as pointed out already, this rule doesn't apply anywhere where the central atom doesn't have a tetrahedral geometry. As counterexamples, coordination complexes often have non-tetrahedral geometries:

    • octahedral: $\ce{[Co(NH3)6]^3+}$
    • square planar: $\ce{[PtCl4]^2-}$
    • dodecahedral: $\ce{[Ce(NO3)6]^2-}$

    Note that these specific examples are optically inactive but by changing the ligands for different ones, it is possible to create optically active versions with the given geometries.

    However, since organic chemistry mostly involves tetrahedral carbon as the chiral center, the rules certainly do apply there.

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  • $\begingroup$ Hello, please refrain from using unnecessary abbreviations like "no."; it's preferable to type the word "number" instead. Please also take a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange? In particular, chemical formulae should be typeset as $\ce{H2O}$, not just $H_2O$. $\endgroup$
    – orthocresol
    May 27 '21 at 14:25
  • $\begingroup$ @Ex_Machina Thanks a lot for your answer, it helped me. I was going wrong at the part: even number of swaps are made at a single chiral center only. $\endgroup$
    – Aashita
    May 27 '21 at 17:09

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