# Is (2R,3S)-butane-2,3-diol chiral?

Is the attached compound a chiral or achiral overall? My lecturer said that this is an achiral compound because it has a superposable mirror image but I don't get that. So I tried to figure out the configuration of both stereogenic centers because what I understood is when you have an even number of chiral centers and if they have opposite configuration, say (R,S) or (S,R), they will cancel out each other so that specific rotation is zero and two chiral centers are overall achiral, which gives you a meso compound.

• – CoffeeIsLife Apr 5 '16 at 7:01
• I think a superposable image is mirror symmetry, ie you can put a mirror somewhere in the molecule and you would be able to describe the molecule with the molecule part being mirrored and the mirror image – CoffeeIsLife Apr 5 '16 at 7:03
• Your process in determining chirality is correct. However, you have to check if the compound is meso as the last step. If it is meso, then it is achiral. – CoffeeIsLife Apr 5 '16 at 7:04
• – user7951 Apr 5 '16 at 10:14
• rotate one of the ends by 180 degrees and you will see a plane of symmetry. – user223679 Apr 8 '16 at 16:58

The staggered, conformation of (2R,3S)-butane-2,3-diol (1a) that you have shown does not have mirror planes but rather a center of symmetry, which is the reason that this conformation does not rotate plane polarized light. There are two other stable, staggered conformations of the diol, namely 1b and 1c. They are a racemic pair, each of which is chiral. But like any racemate, it is optically inactive. Each of these staggered conformations has also been represented as Newman projections with the S-carbon in the front (1a', 1b' and 1c''). Rotation of 1c'' by 180o about a vertical axis passing through the center of the C2, C3 bond leads to structure 1c', which is the mirror image of structure 1b'. A time tested device for ascertaining whether or not a compound is meso is to pair like groups in an eclipsed conformation (1a''). While this technique serves its purpose and conformation 1a'' has a plane of symmetry, its population in solution is negligible as is the racemic pair of eclipsed conformations (not shown).

• At room temperature this is a free rotation and the staggered forms are transition states. They will still be significantly populated, even though they are not stable. – Martin - マーチン Dec 16 '17 at 11:20

The molecule you show has two mirror planes. To see this rotate the molecule around the central bond until the OH groups are pointing in the same direction (C-C single bonds usually rotate freely). In this configuration there is a mirror plane through the OH groups along the C-C bond and a mirror plane perpendicular to the C-C bond.

Build a model of the molecule and rotate it around that central bond if you can't visualise this in your head.

• The mirror plane including the $\ce{O-C-C-O}$ atoms would map carbon on hydrogen and vice-versa. The molecule can only have one plane of symmetry after the rotation you propose. – Jan Dec 15 '17 at 8:52

A molecule is chiral when it doesn't exhibit any kind of symmetry.

This particular molecule exhibits centre of inversion symmetry, i.e. it's symmetrical about a particular point.

In this question that point is the midpoint of the bond joining the two substituted carbons.

If you would like to know about the different types of symmetry, Wikipedia has a good page on the topic.