8
$\begingroup$

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]

InositolIsomers

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.
$\endgroup$
  • 2
    $\begingroup$ Seven of the nine inositols are achiral. Any carbons lying in a plane of symmetry are in achiral environments (r and s). Those carbons not in a plane of symmetry are in chiral environments (R and S). I Googled "inositols chirality" and found this: ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/… $\endgroup$ – user55119 Feb 26 at 2:57
2
$\begingroup$

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.

$\endgroup$
9
$\begingroup$

Consider 2,3-butanediol.

2,3-butanediol structure

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.

$\endgroup$
2
$\begingroup$

I offer a different perspective to your question about the inositols and how the chirality of the "top" carbons in D-chiro-inositol 2 (C2; S-configuration) and scyllo-inositol 3 (C2; r-configuration) are determined.

Generic inositol 1 is numbered the same way as D-chiro-inositol and scyllo-inositol. The method for assigning the CIP descriptor for each inositol at C2 will be illustrated. Starting from C2 in structure 1, the counterclockwise path C2-->C3-->C4-->C5 is colored blue while the clockwise path C2-->C1-->C6-->C5 is in red.

Starting with D-chiro-inositol 2, temporary descriptors (Ro/So) at all carbons except C2 are assigned. At each of these carbons the hydroxyl group has the top priority while hydrogen has the lowest priority. C2 is taken as the next to highest priority group, no matter how far the carbon in question is from C2.1 The temporary assignments for D-chiro-inositol are illustrated in structure 2. To determine whether the blue or red path has the higher priority, step 1 is to compare the assignments of C3/C4 (So/Ro) with C1/C6 (So/Ro). No decision can be made at this point. In step 2, one compares C3/C5 (So/So) in blue with C1/C5 (So/Ro) in red. Because "like" descriptors (RR or SS) take precedent over "unlike" descriptors (RS or SR), whether they are temporary or not. The priorities for C2 are OH>SoSo>SoRo>H. C2 is of the S-configuration. Digraph 4 illustrates this point as well. Because D-chiro-inositol is chiral, each of the carbons is both stereogenic and chirotopic. Each carbon must be analyzed independently as was done for C2. Notice that the temporary assignments (Ro/So) have no relationship to the final assignments (R/S). continued

enter image description here

enter image description here

Scyllo-inositol 3 is achiral having three planes of symmetry passing through C1/C4, C2/C5 and C3/C6. All carbons are equivalent; assign one, you have assigned all six. After assigning temporary descriptors and proceeding all the way to step 3 (see rules next to structure 1), no decision can be made. C2 and all other carbons are pseudoasymmetric (stereogenic and achirotopic). The default position is to use the temporary assignments at C1 and C3 (see digraph 5). CIP Rule 5 states R>S (and Ro>So). The priorities for C2 are OH>Ro>So>H. C2 is of the r-configuration. Lower case r/s descriptors are reserved for pseudoasymmetric carbons.

The two inositols discussed here have descriptors that are either R/S or r/s. This need not be the case. For example, muco-inositol has four chirotopic (R/S) carbons and two achirotopic (r/s) carbons.

1) This method is a simplification of a more complex methodology.

$\endgroup$
  • $\begingroup$ OK, I'll bite. How is the orientation around the top carbon atom in the scyllo form determined when there is obviously a mirror plane going right through that carbon atom? (And one mirror plane or another through all the other carbon atoms too.) $\endgroup$ – Oscar Lanzi Mar 6 at 2:28
  • $\begingroup$ @Oscar Lanzi: Because of the plane of symmetry thru C2/C5, the red/blue paths are indistinguishable. C1 is Ro while C3 is So. Ro > So. Priorities OH>Ro>So>H. You can do the same exercise for anyone of the scyllo carbons because they are all equivalent. C2 in D-chiro is S and not R. See the colored picture. Don't fret. It took me a long time to understand this stuff. $\endgroup$ – user55119 Mar 6 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.