Tartaric acid, HOOC–CH(OH)–CH(OH)–COOH, has two stereocenters, C2 and C3. The 2ⁿ rule from Stereocenters allows it up to 2² = 4 stereoisomers, but tartaric acid has only three. Two of the four R/S combinations turn out to be the same molecule, and that molecule is achiral. This section shows how that happens, how to spot it, and why it matters.
The exception to 2ⁿ
Start by writing out the four combinations: (2R,3R), (2S,3S), (2R,3S) and (2S,3R). The figure draws the first three in one layout. The C2–C3 bond runs across the page, with C3 on the left and C2 on the right. Both COOH groups point down, in the plane of the page, and the two OH groups sit above the bond on wedges or hashes.
(2R,3R) and (2S,3S) are mirror images that cannot be superimposed, so they are a pair of enantiomers. Both are chiral, and both rotate plane-polarized light: the specific rotation, [α], is +12° for (2R,3R) and −12° for (2S,3S).
(2R,3S) is different. Its two OH groups are both on wedges, and a mirror plane cuts its C2–C3 bond in half. Reflect the left half through that plane and you get the right half, with OH landing on OH, H on H and COOH on COOH. A molecule that contains a mirror plane is its own mirror image, so (2R,3S)-tartaric acid is achiral.
(2S,3R) is not a fourth compound. Number the chain from the other end and C2 becomes C3, so (2S,3R) is just another name for the (2R,3S) molecule.
That leaves three stereoisomers: (2R,3R), (2S,3S) and the achiral one, called meso-tartaric acid. Its [α] is 0. The chiral pair melts at about 170 °C, and the meso form melts lower.
A meso compound has two or more stereocenters and is still achiral, because internal symmetry makes it identical to its own mirror image. The symmetry is usually a mirror plane, the plane of symmetry from Chirality: a plane that cuts the molecule into two halves that are mirror images of each other. A meso compound has no enantiomer, and its specific rotation is zero.
So the 2ⁿ rule gives a maximum, not a count. With two stereocenters, as in tartaric acid, the meso form covers two R/S combinations, so it lowers the count from four to three.
Meso-tartaric acid also shows that chirality belongs to the whole molecule, not to single atoms. Both of its stereocenters are real, since each carbon carries four different groups. The two centers have opposite configurations, and because the two halves of the molecule are identical, the handedness of one half cancels the handedness of the other.
How to test for meso
Meso-tartaric acid is often drawn without its mirror plane in view. Draw it as a zigzag, the usual skeletal shape, and no plane cuts the drawing into mirror-image halves. It is still the same molecule, only in a different conformation. As Chirality showed with butane, the search for a plane covers every conformation a molecule reaches by turning about its single bonds. So before you look for a plane, turn the molecule until its two halves line up. This lined-up shape is the eclipsed conformation from Newman projections.
The zigzag is still achiral. It has a center of symmetry at the middle of the C2–C3 bond: every atom has an identical twin straight through that point, the same distance out. Chirality covers this second kind of symmetry. A mirror plane is simply easier to see, and the lined-up conformation, with matching groups side by side, is the one that shows it.
The R/S descriptors give a quick first check. In a meso compound with two stereocenters, such as tartaric acid, one center is R and the other S. The reverse is not always true: opposite descriptors mean meso only when the two halves of the molecule carry the same groups.
2,3-Dibromobutane and 2-bromo-3-chlorobutane show the difference. Both have stereocenters at C2 and C3, and the figure draws the (2R,3S) form of each in the same way. In 2,3-dibromobutane, CH₃–CHBr–CHBr–CH₃, both halves are CH₃–CHBr–, so the dashed line is a mirror plane and (2R,3S) is meso. In 2-bromo-3-chlorobutane, CH₃–CHBr–CHCl–CH₃, one half carries Br and the other Cl. No reflection can turn one half into the other, so the molecule is chiral, and (2S,3R) is its enantiomer, a different compound. All four of its stereoisomers exist, and none of them is meso.
A second drawing makes the plane easier still. The next section, Fischer projections, stands the lined-up conformation on end, with C2 above C3, and flattens it onto the page, so each stereocenter becomes a plain cross.
Only one rule is needed to read the cross for now: horizontal bonds come toward you, and vertical bonds go away. Because the drawing keeps the lined-up conformation, the mirror plane becomes a horizontal line across the middle of the cross.
Meso compounds in rings
Rings give some of the clearest meso compounds. Draw cis-1,2-dimethylcyclohexane as a flat hexagon, with both methyls on wedges, so both sit on the top face of the ring. It has two stereocenters, C1 (R) and C2 (S), and a mirror plane through the middle of the C1–C2 bond and the middle of the bond across the ring from it. The plane swaps C1 with C2 and one methyl with the other, so the compound is meso.
The trans isomer has no such plane. The same plane would reflect the C1 methyl, on a wedge, onto the H on a wedge at C2, because the C2 methyl sits on a hash. trans-1,2-Dimethylcyclohexane is chiral, and it exists as a pair of enantiomers, (1R,2R) and (1S,2S).
The real ring is a chair, and neither chair of the cis compound has a mirror plane. In one chair the C1 methyl is axial and the C2 methyl is equatorial. A ring flip swaps those roles, and both methyls stay on the top face, as Ring flips showed.
Each chair on its own is chiral, and the two chairs are mirror images of each other. The ring flips between them about 10⁵ times a second at room temperature, so any sample holds equal amounts of both at every moment. The compound is achiral.
This extends the rule from Chirality. A molecule is achiral if any conformation it can reach has a mirror plane or a center of symmetry, and also if it switches quickly between a chiral conformation and that conformation's mirror image. The flat hexagon stands for that average, which is why the plane shows in the hexagon and in neither chair.
Relationships among the tartaric acid stereoisomers
(2R,3R) and (2S,3S) are enantiomers of each other. Meso-tartaric acid is a diastereomer of both: it is a stereoisomer of each, but the mirror image of neither, since its own mirror image is itself.
Because it is a diastereomer, it has different physical properties from the chiral pair: a different melting point, a different solubility and a different density. Diastereomers explains why.
Why meso compounds matter in reactions
Some reactions give a meso product and nothing else, and recognizing this saves you from looking for optical activity that cannot be there. Bromine, Br₂, adds to a C=C double bond with anti stereochemistry (see Diastereomers): the two bromine atoms bond to opposite faces of the flat alkene. Alkenes & Alkynes shows how. Here only the result matters, and on but-2-ene that one fact decides the product.
- trans-But-2-ene + Br₂ gives meso-2,3-dibromobutane, a single achiral product.
- cis-But-2-ene + Br₂ gives (2R,3R)- and (2S,3S)-2,3-dibromobutane in exactly equal amounts.
The figure follows each product as it forms. The first bromine bonds to the front face of the alkene and the second to the back face, and every other group stays where the alkene had it. A half turn of C3 about the C2–C3 bond then brings both bromines to the front. From the trans alkene, the two methyls end up side by side, and a mirror plane runs between them. From the cis alkene, one methyl points up and the other down, and there is no plane.
Students often swap these two results, so learn the general pattern rather than the two examples. The pattern applies when a symmetric alkene, one that reads the same from either end, gains two identical groups. Syn addition, the opposite of anti, puts both new groups on the same face.
| Addition | cis alkene | trans alkene |
|---|---|---|
| anti (opposite faces) | chiral pair, racemic | meso |
| syn (same face) | meso | chiral pair, racemic |
You can rebuild the table from any one entry. Changing the alkene geometry flips the answer, and changing the addition mode flips it again, so changing both leaves it the same.
One entry can be cross-checked once you reach Alkenes & Alkynes. Syn dihydroxylation adds two OH groups to the same face, and it turns cis-but-2-ene into meso-butane-2,3-diol. That is the syn, cis entry. Flip the addition mode and you get the anti, cis entry, the chiral pair, which is what bromine gives.
What carries forward
Meso compounds are the standard trap in questions that ask you to count stereoisomers, and they come up whenever a symmetric alkene undergoes an addition with defined stereochemistry in Alkenes & Alkynes. They also reinforce the chapter's central claim: chirality belongs to the molecule, not to its atoms, and the only way to settle it is to look for symmetry in the whole structure.