Cfrgtkky

I was asked to draw all possible stereoisomers of inositol (1,2,3,4,5,6-cyclohexanehexol). To obtain the answer I had to assume that all six carbons of the molecule are asymmetric, which (bearing in mind possible molecular symmetry) results on the following nine stereoisomers:^[1]

The problem is that I fail to see why these carbons can be considered chiral centers. Given:^[2]

The most common cause of chirality in an organic molecule, although not the only one, is the presence of a carbon atom bonded to four different groups [...]. These carbons are now named chiral centers [...]

The chirality of the molecule is not the reason of my doubts, I can see that the only chiral isomers are D- and L- chiro-inositol. Since all carbons in the ring possess a hydroxyl group, can they not be considered to be bonded to equivalent substituents, making them symmetric? Is the possibility of the -OH groups to be above or below the plane of the ring what makes them asymmetric?

1. From Wikipedia
   


2. McMurry, John. Organic Chemistry. CENGAGE Learning, 2008. 7th edition.

When we count different substituents to determine chiral centers, we have to include the relative locations of atoms as well as stoichiometry.

Let's look at the top carbon atom in the D-chiro form. If you go clockwise around the ring you encounter five CHOH groups, and if you go counterclockwise you get the same five CHOH groups. The substituent atomic formulas look the same. But:

• in the clockwise direction you see the hydroxyl group attached to the first carbon from the top is on the same side of the ring as the hydroxyl group on the top carbon itself. Whereas:




• in the counterclockwise direction you see the hydroxyl group attached to the first carbon from the top is on the opposite side of the ring as the hydroxyl group on the top carbon itself.

That difference is enough to count as two different substituents even though the formulas in terms of atoms are the same. So you have a chiral center, and the CIP convention even identifies the orientation of the top carbon as R. Similarly any of the other ring carbons can be a chiral center.

Contrast this with the scyllo form, for instance, where the substitution on any carbon is truly fully symmetric including stereochemistry. So no chiral centers there. Most of the other forms listed above do have chiral centers by the method above, but end up being meso compounds. so only the forms labeled "chiro" actually have a net molecular chirality.

Consider 2,3-butanediol.

This compound has 2 stereocenters. The meso version of this compound is achiral. However, there are still two stereo centers. As drawn, the left one is (R) and the right one is (S). It is precisely because these two stereocenters are identical but opposite in configuration that leads the molecule to be meso.

A carbon center is asymmetric if it has 4 different substituents regardless of whether or not the whole molecule is chiral. The ring substitution in inositol is such that the centers around the ring are not identical (including when you look at stereochemistry--for example, (R) has higher priority than (S)). This of course isn't enough to help you easily make stereochemical assignments for all of the centers of inositol, but hopefully, this allows you to understand why the centers are still chiral centers.

As an aside, there are a few centers that cannot be chiral because they lie on a mirror plane. Interestingly though, changing the configuration of the center produces a different molecule. These centers are called pseudochiral, and we label them with lower case r and s.