Here is pentane-2,3,4-triol:

enter image description here

At first glance, without deciding the configurations at each chiral carbon, we can clearly see that this molecule has a plane of symmetry, perpendicular to the plane of paper and passing through the $\ce{-OH}$ and $\ce{-H}$ on the 3rd carbon atom.

However, if we then sit down to compute the R/S configuration at carbons 2 and 4, we find them to be R and S respectively. Thus, going by the definition of a pseudo-chiral carbon, given in Gold Book as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

we would infer that the 3rd carbon is indeed pseudo-chiral. This is confirmed by ChemSketch which generates the name of the above compound as "(2R,3s,4S)-pentane-2,3,4-triol".

Thus, we have observed that using the plane of symmetry to check chirality, we end up declaring the 3rd carbon to be achiral, rendering the molecule optically inactive. However, if we use the Gold Book's definition, we end up declaring the 3rd carbon to be pseudo-chiral, rendering the molecule optically active.

Another fact confirming that the molecule is optically active is that there exist two isomers, of two opposite configurations of the central carbon atom - (2R,3s,4S)-pentane-2,3,4-triol and (2R,3r,4S)-pentane-2,3,4-triol. Here are their structures:

enter image description here

I have looked at several questions on pseudochirality here, yet I have been unable to resolve this conundrum. I believe I am using a wrong fact or a misguided inference, but whatever is it, I am unable to grasp it, and thus need help resolving the conundrum.

  • 1
    $\begingroup$ Yeah, the middle carbon becomes pseudo chiral but that doesn't have an impact on the compound being meso. $\endgroup$ – Avnish Kabaj Mar 1 '18 at 3:53

The Gold Book definition for a meso compound is

the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

So, a meso compound has chiral subsets. Here we would have three such subsets:

  1. Leftmost carbon (2R)
  2. The centermost (pseudochiral 3S)
  3. The rightmost (4S) carbon

Even though the centermost carbon is pseudo chiral, the plane of symmetry still exists. The compound remains meso, achiral.

Nomenclature is just for naming a compound.

The following two molecules are diastereomers, not enantiomers.

enter image description here

  • $\begingroup$ Can you tell me about the small r and s .... How to use them when we have pseudosymmetric carbon $\endgroup$ – Scáthach Sep 19 '18 at 9:43

For the examples I've looked at, a small (r) or (s) is used for the pseudochiral carbon instead of (R) or (S). For example, see this section in Wikipedia. That means that this is not a real stereocenter. But we do need to assign something because there is intrinsic stereochemistry because it has relative stereochemistry with the two enantiomeric substituents, and changing the center produces diastereomers. As you noted, the reason it's a pseudocenter is because it cannot be a full stereogenic center by symmetry.

  • $\begingroup$ Thanks! Do you mind expanding your answer a bit? As it stands, "because there is intrinsic stereochemistry because it has relative stereochemistry with the two enantiomeric substituents, and changing the center produces diastereomers." makes very less sense me. Perhaps, splitting it into 3-4 sentences might be useful? $\endgroup$ – Gaurang Tandon Mar 1 '18 at 4:02

(2R,3s,4S)-Pentan-2,3,4-triol is a meso compound owing to the plane of symmetry. C2 and C4 are both chirotopic and stereogenic. Thus the R/S designation. C3 is ACHIROTOPIC and stereogenic (pseudoasymmetric). Hence, the lower case "s" (C3-OH>C2-R>C4-S>H, which are the CIP priorities). Inversion of C3 leads to a new meso diastereomer (2R,3r,4S) that has the same chirotopicity and stereogenicity at the three centers as did the original triol. To invoke enantiomers one needs the pair (2R,4R) and (2S,4S). C3 in each of these enantiomers has no designation (non-stereogenic and chirotopic) because inversion returns the same enantiomer. For a discussion of stereogenicity and chirotopicity in lieu of "pseudoasymmetry", see Mislow, K. and Siegel, J., J. Am. Chem. Soc., 1984, 106, 3319.

  • $\begingroup$ Can you contextually explain what you mean by "achirotropic"? The goldbook definition is a bit vague - goldbook.iupac.org/html/C/C01065.html Also what does "OH>R>S>H" mean? $\endgroup$ – Gaurang Tandon Mar 8 '18 at 3:14
  • $\begingroup$ @Gaurang Tandon: Sorry for the late reply. Your Goldbook link seems clear to me. Inversion of C3 returns the same structure, i.e., achirotopic. As to the priorities, O(H) is highest, H the lowest. What about the 2 enantiotopic groups, C2 and C4? Follow CIP rules. R precedes S. Thus, O(H)>R>S>H. $\endgroup$ – user55119 Apr 13 '18 at 15:41
  • $\begingroup$ When do we decide the compound has plane of symmetry... By analysjng it's 3D structure before making the isomers? $\endgroup$ – Scáthach Sep 19 '18 at 9:44
  • $\begingroup$ Take a look here: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… $\endgroup$ – user55119 Sep 19 '18 at 15:25

Pseudoasymmetric centres are stereogenic but achiritopic. Now, the 3rd carbon is pseudoasymmetric centre. It has stereogenicity only when the two groups are different in configuration. Note that I have used the word stereogenicity, and not chirality, because whatever the configuration of the two $\ce{CH(OH)CH3}$ group may be 3 rd carbon does not induce optical activity or chirality in the molecule. That is why it is a pseudo yet asymmetric centre (In the sense that it has 4 different groups).

So the configuration you have given is obviously meso, but as a whole if you ignore the stereochemistry in the structure then for this compound you have 4 isomers: 2 optically active and two meso (these two meso compound arising from the 3rd carbon being stereogenic)

  • $\begingroup$ Thanks for your answer! "Note that I have used the word stereogenicity, and not chirality, because whatever the configuration of the two CH(OH)CH3 group may be it does not induce optical activity or chirality" That was exactly my question actually, as I was unable to justify it through a plane of symmetry. Still, +1 for describing it through words. Though I don't get your second part. How have you counted four isomers? $\endgroup$ – Gaurang Tandon Mar 7 '18 at 16:51

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