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I can see that these compounds are identical as we have basically just done a 180 degree rotation of the molecules.

But if we go by the concepts of chirality, the carbons 2 and 4 are chiral centers and since they have inverted configurations at both chiral centers, they should be enantiomers. Or is it just that I cannot assume the hydrogens not shown to be out of plane?

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    $\begingroup$ Have you heard of meso compounds? $\endgroup$
    – Mithoron
    Apr 5 '19 at 18:58
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    $\begingroup$ No, you just said that they're identical. Why would you then say that they are enantiomers? The concepts of chirality don't say that. They have to be nonsuperimoposable mirror images, and they are clearly superimposable. $\endgroup$
    – Zhe
    Apr 5 '19 at 18:59
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    $\begingroup$ An enantiomer is a non-superimposable mirror image. The definition does not say anything about configurations. True, having opposite configurations at all chiral centres often makes two things enantiomers, but not always. $\endgroup$
    – orthocresol
    Apr 5 '19 at 19:05
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    $\begingroup$ The basic question: Yes the molecules are identical. Optically activity None, this is a meso compound. The two chiral centers cancel each other out so the molecule as a whole is not optically active. $\endgroup$
    – MaxW
    Apr 5 '19 at 19:13
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    $\begingroup$ On a mildly related note, the general relativity books I'm reading keep saying don't use coordinates, just use coordinate-free descriptions because that will be true in every coordinate system. This is exactly like that. The definition of chirality is fully independent of identifying the configurations of centers. Does this molecule have an improper axis of rotation? If yes, it's not chiral, and you don't need to assign stereochemistry to any center. This is the equivalent coordinate-free language. $\endgroup$
    – Zhe
    Apr 5 '19 at 19:21

Sameer. The key to understanding this is that the two asymmetric carbons are indistinguishable because of the symmetry in the molecule. When you do your inversions on them, the molecule looks different, and should be an enantiomer, but after you do the 180 deg rotation, you have essentially switched them over (now the molecule looks the same as the first molecule, but only because you can’t differentiate the two asymmetric carbons).

Imagine that you switched one of the terminal methyl groups with an ethyl, then try the same trick. Because you have broken the symmetry of the molecule, the 180 deg rotation doesn’t get you back to the start. Hence, this meso isomerism relies on the mirror plane of symmetry (or a point of inversion) existing in addition to multiple asymmetric centers.


This molecule has a plane of symmetry. Imagine a vertical plane go through $\ce{C}$-3 atom of pentane chain. One side of that plane is the mirror image of other side. Hence, this is not optically active molecule by definition. It is called meso-isomer. If you look at (R,S)-configuration and number the chains as they are drown, first compound is (2R,4S)-2,4-dichloropentane and second compound is (2S,4R)-2,4-dichloropentane. But, since each have plane of symmetry, you may number second compound from opposite direction. Then it is (2R,4S)-2,4-dichloropentane. Therefore, they are identical.

The two enantiomers relevant to this meso-isomer are (2R,4R)-2,4-dichloropentane and (2S,4S)-2,4-dichloropentane.

To understand more on enantiomers and relevant meso-isomer, study halogenation of cis-/trans-2-butene or 1,2-diphenylethene (stilbene).


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