After completing this section, you should be able to
- identify a compound as being prochiral.
- identify the Re and Si faces of prochiral sp2 centre.
- identify atoms (or groups of atoms) as pro-R or pro-S on a prochiral sp3 centre.
Make certain that you can define, and use in context, the key terms below.
Prochiral substituents on tetrahedral carbons
When a tetrahedral carbon can be converted to a chiral center by changing only one of its attached groups, it is referred to as a ‘prochiral' center. The actual example shown below is the reduced form of a molecule called nicotinamide adenine dinucleotide (NADH), an important participant in many biochemical oxidation/reduction reactions (section 16.4A).
Note that if, in a thought experiment, we changed either one of the indicated hydrogens on NADH to a deuterium (the 2H isotope of hydrogen), the carbon would become a chiral center. Prochirality is an important concept in biological chemistry, because enzymes can distinguish between the two ‘identical’ groups bound to a prochiral carbon center due to the fact that they occupy different regions in three-dimensional space. For example in the following reaction, which is a key step in the oxidation of fatty acids, it is specifically HA and HD that are lost, while HB and HC remain in the resulting conjugated alkene.
The prochiral hydrogens on C2and C3 of the fatty acid can be designated according to a variation on the R/S system. For the sake of clarity, we'll look at a much simpler molecule (ethanol) to explain this system.
To name the two prochiral hydrogens on ethanol, we again need to engage in a thought experiment. If we, in our imagination, were to arbitrarily change HB to a deuterium, the molecule would now be chiral and the stereocenter would have the S configuration (D has a higher priority than H). For this reason, we can refer to HB as the pro-S hydrogen of ethanol, and label it HS. Conversely, if we change HA to D and leave HB as a hydrogen, the configuration of the molecule becomes R, so we can refer to HA as the pro-R hydrogen of ethanol, and label it HR.
Looking back at our fatty acid example, we see that it is specifically the pro-R hydrogens on carbons 2 and 3 that are lost in the reaction.
Prochiral hydrogens can be designated either enantiotopic or diastereotopic. If either HR or HS on ethanol were replaced by a deuterium, the two resulting molecules would be enantiomers (because there are no other stereocenters)
Thus, these two hydrogens are referred to as enantiotopic.
In glyceraldehyde-3-phosphate (GAP), however, we see something different:
If either HR or HS is replaced by a deuterium, the two resulting molecules will be diastereomers - thus, in this molecule, HR and HS are referred to as diastereotopic hydrogens. The importance of the distinction between enantiotopic and diastereotopic groups will become apparent when we learn about the analytical technique called nuclear magnetic resonance.
Two hydrogens on the same carbon of a substituted ring structure can be diastereotopic - we determine this by carrying out the same thought experiment as discussed above. In the example below, the diastereotopic hydrogens indicated are either on the same side or on the opposite side of the ring relative to the hydroxyl group.
Finally, hydrogens that are completely identical and can be designated neither enantiotopic nor diastereotopic are called homotopic. If a homotopic hydrogen is replaced by deuterium, a chiral center is not created. The three hydrogen atoms on the methyl (CH3) group of ethanol (and on any methyl group) are homotopic.
Even to an enzyme, all three of these hydrogens will look the same.
Groups other than hydrogens can be considered prochiral. The alcohol below has two prochiral methyl groups - methyl A is the pro-R methyl, and methyl B is pro-S. (How do we make these designations? Simple - just arbitrarily make methyl A higher priority than methyl B, and the compound now has the R configuration).
Citrate provides a more complex example. The central carbon is a prochiral center with two 'arms' that are identical except that one can be designated pro-R and the other pro-S.
In a reaction of the citric acid cycle (Krebs cycle), a water molecule is specifically lost on the pro-R arm (we will study this reaction in section 14.1B).
Notice also that it is specifically the pro-R hydrogen on the pro-R arm of citrate that is lost - one more layer of stereoselectivity!
Although an alkene carbon bonded to two identical groups is not considered a prochiral center, these two groups can be diastereotopic. Ha and Hb on the alkene below, for example, are diastereotopic: if we change one, and then the other, of these hydrogens to deuterium, the resulting compounds are E and Z diastereomers.
Carbonyl and imine carbons as prochiral centers
Trigonal planar, sp2-hybridized carbons are not, as we well know, chiral centers– but they are referred to as prochiral centers if they are bonded to three different substitutuents. As you might expect, we (and the enzymes that catalyze their reactions) can distinguish between the two planar ‘faces’ of a prochiral sp2 - hybridized group. These faces are designated by the terms re and si. To determine which is the re and which is the si face of a planar organic group, we simply use the same priority rankings that we are familiar with from the R/S system, and trace a circle: re is clockwise and si is counterclockwise.
Below, for example, we are looking down on the re face of the ketone group in pyruvate:
If we flipped the molecule over, we would be looking at the si face of the ketone group. Note that the carboxylate group does not have re and si faces, because two of the three substituents on that carbon are identical (when the two resonance forms of carboxylate are taken into account).
As we will see beginning in chapter 11, enzymes which catalyze reactions at carbonyl carbons act specifically from one side or the other.
Notice that the 'hydrogenation' reaction above is specific not only in terms of which face of the carbonyl group is affected, but also in terms of which of the two diastereotopic hydrogens on NADH is transferred (we will study this type of reaction in more detail in section 16.4).
State which of the following hydrogen atoms are pro-R or pro-S.
Identify which side is Re or Si
Left compound: Ha = pro-S and Hb = pro-R
Right compound: Ha = pro-R and Hb = pro-S
A – Re; B – Si; C – Re; D – Si