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In a particular question in one of my reference book, I was asked if the given compound (meso-butane-2,3-diol) was a meso compound or not. In it’s solution part, to show that the compound was actually meso, they did the following. enter image description here

As I was not convinced about rotation of C2, I searched the web where I found this which says

While rotating the bonds about any atom (chiral) the stereochemistry of the molecule does not changes.

This is what I am confused. I mean if we rotate meso-butane-2,3-diol in the given way such that we have a conformer other than eclipsed or staggered, wouldn’t it be optically active, without any plane or centre of symmetry, which means rotation does change stereochemistry. Where am I going wrong?

Klyen Dave
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  • Duplicate of: https://chemistry.stackexchange.com/questions/48995/is-2r-3s-butane-2-3-diol-chiral/87415#87415 Any conformations of the meso cmpd that do not have a plane or center of symmetry constitute a racemic pair. – user55119 Nov 02 '19 at 19:02
  • @user55119 I agree about the racemic pair. What I was asking is that different conformers of meso compounds are either optically active or inactive, so rotation of groups about C2 and C3 results in different compounds (with different stereochemistry), unlike mentioned in the question. – Klyen Dave Nov 03 '19 at 02:01
  • Of the three staggered conformations about the C2-C3 bond, one has a center of symmetry and two are a racemic pair. If you could "freeze" the trio, say on Pluto, and put each one in a polarimeter, each enantiomer would have equal and opposite rotations. The center of symmetry one would have no rotation. On Earth, the trio is in rapid equilibrium and gives no rotation. Drawing the eclipsed conformation with a plane of symmetry is merely a test for a meso compound and a minor contributor. See my link above. – user55119 Nov 03 '19 at 02:16
  • @user55119 I think I am getting it. So all conformers would have same configurations (R,S), which in turn result in same stereochemistry, but since in eclipsed form, the POS is easily visible, they are rotating it as shown, correct? – Klyen Dave Nov 03 '19 at 02:24
  • Rotation doesn't affect R,S configuration. Make 3 molecular models and see for yourself! – user55119 Nov 03 '19 at 02:40

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The method shown in OP's question to find a meso-compound is the way most people used. However, I usually used the Cahn-Ingold-Prelog (CIP) assignments of chiral centers for this purpose. It is a fact that the mirror image of (R)-chiral center is corresponding (S)-chiral center. For example, bromochlorofluoromethane is a chiral compound and the morror image of its (R)-stereoisomer is its (S)-stereoisomer as depicted in the diagram below:

Mirror Images

When (S)-isomer is rotated $180^\circ$ vertically, the resultant structure would not be superimposed on (R)-isomer, as shown in the diagram. This is to justified the above mentioned fact.

Now, let's assign the compound in hands with CIP system. I'd call it (2 R, 3 S)-butane-2,3-diol (it could also be (2 S, 3 R)-butane-2,3-diol, based on the numbering). Either way, carbon #2 has three different groups attached to it, which are identical to those on carbon #2. Based on the given assignments on carbon #2 and #3, if you'd put a mirror between these carbons, you must see the mirror image of one part (carbon #2) on other side (carbon #3) as depicted in the diagram. Therefore, the compound has plane of symmetry (the mirror) and hence, it is the meso-isomer.

To conclude this fact, I included two enanthiomers of this compound ((2 R, 3 R)-butane-2,3-diol and (2 S, 3 S)-butane-2,3-diol), none of which gives mirror images when a mirror placed in between two chiral carbons (see the diagram).

Mathew Mahindaratne
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  • The staggered conformation of the meso-isomer you have shown does not have a (mirror) plane of symmetry but rather a center of symmetry, which accounts for this conformations lack of optical activity. See the links in the comments above. – user55119 Nov 03 '19 at 00:45
  • @user55119: I disagree. If you rotate carbon #2 $180^\circ$, you'd see that plane of symmetry. – Mathew Mahindaratne Nov 03 '19 at 01:02
  • I believe you misunderstood my point. The molecule is achiral and meso. The staggered conformation you have shown has a center of symmetry in the center of the C2-C3 bond. The other two staggered conformations form a racemate. Yes, the rotation by 180 degrees gives the eclipsed conformation with a plane of symmetry, no rotation and low concentration. The eclipsed conformation is a test for a meso cmpd. The other 2 eclipsed conformations are also a racemate. Your diagram should show "center of symmetry". – user55119 Nov 03 '19 at 02:03
  • I just want to confirm one thing-that all conformers of a meso compound are meso compounds? – An_Elephant Jun 16 '22 at 15:17