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4.6: Compounds with multiple chiral centers

  • Page ID
    291629
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    So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers.

    Molecule with two chiral centers.

    To avoid confusion, we will simply refer to the different stereoisomers by capital letters.

    Look first at compound A below. Both chiral centers in have the R configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the S configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers.

    Four molecules with two chiral centers. A and C on left side of mirror plane and their mirror images on right side. Red arrows for enantiomers and blue arrows for diastereomers. Molecule A has two R centers, B has two S centers, C and D have one S and one R. A and B are enantiomers and C and D are enantiomers. A and C, A and D, B and D, B and C are diastereomers.

    Now, look at compound C, in which the configuration is S at chiral center 1 and R at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other.

    Notice that compounds C and B also have a diastereomeric relationship, by the same definition.

    So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B.

    This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers:

    Stereoisomer shortcuts

    If all of the chiral centers are of opposite R/S configuration between two stereoisomers, they are enantiomers.

    If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers.

    (Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later)

    Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together.

    Molecule with two chiral centers; Chiral center 1 in red and chiral center 2 in blue. Possible configurations (chiral center 1 listed first and chiral center 2 listed second): R R, R S, S S, S R.

    We know, using the shortcut above, that the enantiomer of RR must be SS - both chiral centers are different. We also know that RS and SR are diastereomers of RR, because in each case one - but not both - chiral centers are different.

    Now, let's extend our analysis to a sugar molecule with three chiral centers. Going through all the possible combinations, we come up with eight total stereoisomers - four pairs of enantiomers.

    Molecule with three chiral centers (number 1 is red, number 2 is blue and number 3 is green.) Possible configurations are (center one listed first, center two second and center three last): R R R, R R S, R S R, R S S, S S S, S S R, S R S, S R R.

    Let's draw the RRR stereoisomer. Being careful to draw the wedge bonds correctly so that they match the RRR configurations, we get:

    Stereoisomer with three chiral centers (each chiral center has an O H group). All chiral centers are in R configuration. Center one and three have O H group on wedge, center two has O H group on dashes.

    Now, using the above drawing as our model, drawing any other stereoisomer is easy. If we want to draw the enantiomer of RRR, we don't need to try to visualize the mirror image, we just start with the RRR structure and invert the configuration at every chiral center to get SSS.

    S S S stereoisomer. O H on dashes for centers 1 and 3, O H on wedge for center 2.

    Try making models of RRR and SSS and confirm that they are in fact nonsuperimposable mirror images of each other.

    There are six diastereomers of RRR. To draw one of them, we just invert the configuration of at least one, but not all three, of the chiral centers. Let's invert the configuration at chiral center 1 and 2, but leave chiral center 3 unchanged. This gives us the SSR configuration.

    Diastereomers of a molecule with three chiral centers. Left: R R R configuration. Right: S S R configuration. Two chiral centers have been changed and one chiral center stays the same.

    One more definition at this point: diastereomers which differ at only a single chiral center are called epimers. For example, RRR and SRR are epimers:

    Diastereomers of a molecule with three chiral centers. Left: R R R configuration. Right: S R R configuration. One chiral center has been changed on two chiral centers stay the same so they are epimers.

    The RRR and SSR stereoisomers shown earlier are diastereomers but not epimers because they differ at two of the three chiral centers.

    Exercise 3.17

    a) Draw the structure of the enantiomer of the SRS stereoisomer of the sugar used in the previous example.

    b) List (using the XXX format, not drawing the structures) all of the epimers of SRS.

    c) List all of the stereoisomers that are diastereomers, but not epimers, of SRS.

    Solutions to exercises

    The epimer term is useful because in biochemical pathways, compounds with multiple chiral centers are isomerized at one specific center by enzymes known as epimerases. Two examples of epimerase-catalyzed reactions are below.

    Top: Ribulose-5-phopsphate with R R configuration in equilibrium with xylulose-5-hposphate (same molecule in S R configuration). Phosphopentose epimerase used as catalyst. Bottom: Epimers Glucose-UDP and galactose-UDP in equilibrium. Glucose-UDP epimerase as catalyst.

    We know that enantiomers have identical physical properties and equal but opposite magnitude specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to distinguish between them. In addition, the specific rotation values of diastereomers are unrelated – they could be the same sign or opposite signs, similar in magnitude or very dissimilar.

    Exercise 3.18

    The sugar below is one of the stereoisomers that we have been discussing.

    The only problem is, it is drawn with the carbon backbone in a different orientation from what we have seen. Determine the configuration at each chiral center to determine which stereoisomer it is.

    Exercise 3.19

    Draw the enantiomer of the xylulose-5-phosphate structure in the previous figure.

    Exercise 3.20

    The structure of the amino acid D-threonine, drawn without stereochemistry, is shown below. D-threonine has the (S) configuration at both of its chiral centers. Draw D-threonine, it's enantiomer, and its two diastereomers.

    Solutions to exercises

    Here is some more practice in identifying isomeric relationships. D-glucose is the monosaccharide that serves as the entrance point for the glycolysis pathway and as a building block for the carbohydrate biopolymers starch and cellulose. The 'D' in D-glucose stands for dextrarotatory and is part of the specialized nomenclature system for sugars, which we will not concern ourselves with here. The open-chain structure of the sugar is shown below.

    D-glucose molecule numbered one through six starting with the aldehyde carbon.

    Because D-glucose has four chiral centers, it can exist in a total of 24 = 16 different stereoisomeric forms: it has one enantiomer and 14 diastereomers.

    Now, let's compare the structures of the two sugars D-glucose and D-gulose, and try to determine their relationship.

    D-glucose molecule with hydroxy groups attached to carbons 3 and 4 on wedges and it's diastereomer, D-gulose with hydroxy groups on carbons 3 and 4 on dashes.

    The two structures have the same molecular formula and the same connectivity, therefore they must be stereoisomers. They each have four chiral centers, and the configuration is different at two of these centers (at carbons #3 and #4). They are diastereomers.

    Now, look at the structures of D-glucose and D-mannose.

    D-glucose compared to D-mannose. D-glucose has hydroxy group on carbon 2 on a wedge and D-mannose has hydroxy group on carbon 2 on dashes. They are epimers.

    Here, everything is the same except for the configuration of the chiral center at carbon #2. The two sugars differ at only one of the four chiral centers, so again they are diastereomers, and more specifically they are epimers.

    D-glucose and L-glucose are enantiomers, because they differ at all four chiral centers.

    Left: D-glucose; Hydroxy groups 1 through 3 wedged and 4 dashed. Right: L-glucose; hydroxy groups 1 through 3 dashed and 4 wedged. Enantiomers.

    D-glucose is the enantiomer commonly found in nature.

    D-glucose and D-fructose are not stereoisomers, because they have different bonding connectivity: glucose has an aldehyde group, while fructose has a ketone. The two sugars do, however, have the same molecular formula, so by definition they are constitutional isomers.

    Left: D-glucose; has aldehyde group. Right: D-fructose; has ketone group. Constitutional isomers.

    D-glucose and D-ribose are not isomers of any kind, because they have different molecular formulas.

    Left: D-glucose; six carbon chain. Right: D-ribose; five carbon chain. Not isomers.

    Exercise 3.21

    Identify the relationship between each pair of structures. Your choices are: not isomers, constitutional isomers, diastereomers but not epimers, epimers, enantiomers, or same molecule

    Exercise 3.22

    Identify the relationship between each pair of structures. Hint - figure out the configuration of each chiral center.

    Solutions to exercises

    Khan Academy video tutorial on stereoisomeric relationships


    This page titled 4.6: Compounds with multiple chiral centers is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Tim Soderberg.