One stereocenter

Stereocenters in a ring which can be severed by a disconnective transform, but which are not part of the retron, can be eliminated prior to disconnection if they are clearable. Removal of such stereocenters may convert a non-strategic bond into a strategic one. Stereocenters should also be cleared if that sets up the retron for a disconnective transform. Such strategic stereocenter eliminations commonly involve transforms which remove a 3 ° or 4 ° stereocenter to generate C=C (endo- or exocyclic), C=0 or C=N. Elimination of two clearable vicinal stereocenters to generate C=C retrosynthetically is strategically indicated whether or not that leads to a disconnectable retron. 

EXERCISE 7.1 In the compound below, there is one stereocenter. Find it. 

In the previous problems, you knew that you were looking for just one stereocenter. Hopefully, you started to realize some tricks that make it faster to find the stereocenter (for example, ignore CH2 groups). So, now, we will move on to examples where you don t know how many stereocenters there are. There could be five stereocenters or there could be none. 

When we learned how to name compounds, we said that we would skip over the naming of stereocenters until we learned how to determine configuration. Now that we know how to determine whether a stereocenter is R or S, we can now see how to include this in the name of a compound. It is actually quite simple. If there is only one stereocenter, then you simply place either (R) or (S) at the beginning of the name. For example, 2-butanol has one stereocenter, and can either be (I )-2-bu-tanol or (S)-2-butanol, depending on the configuration of the stereocenter. If more than one stereocenter is present, then you must also use numbers to identify the location of each stereocenter. Consider the following example ... 

Fischer projections can also be used for compounds with just one stereocenter, as above, but they are usually used to show compounds with multiple stereocenters. You will utilize Fischer projection heavily when you learn about carbohydrates at the end of the course. 

SnI and Sn2 reactions produce almost the same products. In both reactions, a leaving group is replaced by a nucleophile. The difference in products between SnI and Sn2 reactions arises when the leaving group is attached to a stereocenter. In this situation, the Sn2 mechanism will invert the stereocenter, while the SnI mechanism will produce a racemic mixture. That s the main difference—the configuration of one stereocenter. It seems like a lot of work to go through to determine the configuration of one stereocenter (which matters only some of the time). 

Similarly, you may encounter situations where only one stereocenter is being formed. For example,... 

In cases like this, the stereochemistry is still irrelevant (as long as the compound does not possess any other stereocenters). Why With only one stereocenter, there will only be two possible products (not four). These two products will represent a pair of enantiomers (one will be R and the other will be S). You will get both of these products whether the reaction proceeds through a syn addition or through an anti addition. If the reaction is a syn addition, the OH group can come from above the plane or from below the plane of the double bond, giving both possible products. Similarly, if the reaction is an anti addition, the OH group can come from above the plane or from below the plane of the donble bond, giving both possible products. Either way, we get the two possible products. 

Answer We begin by looking at the regiochemistry. We are not told what the re-giochemistry is, becanse it is irrelevant (we are adding two gronps of the same kind OH and OH). Next, we look at the stereochemistry. The reaction is a syn addition. But in this case, we are only creating one stereocenter ... 

In any reaction, the mechanism should explain not only the regiochemistry, but the stereochemistry as well. In this particular reaction (addition of H—X across alkenes), the stereochemistry is generally not relevant. Recall from the previous section that we need to consider stereochemistry (syn vs. anti) only in cases where the reaction generates two new stereocenters. If only one stereocenter is formed, then we expect a pair of enantiomers (racemic mixture), regardless of whether the reaction was anti or syn. You will probably not see an example where two new stereocenters are formed, because the stereochemical outcome in such a case is complex and is beyond the scope of our conversation. 

We do not need to think about stereochemistry here, because the product does not contain two stereocenters (in fact, it doesn t even have one stereocenter). Therefore, stereochemistry is irrelevant in this example. As we have said, the stereochemistry of this reaction (H—addition) will generally not be relevant in the problems that you will encounter. 

Until now we have focnsed on the regiochemistry of this reaction. We did not explore the stereochemistry, because it is beyond the scope of the course. In situations where two stereocenters are formed, the results are dependent on the starting alkene and on the temperature. Therefore, we will only present problems where no stereocenters are formed, or where only one stereocenter is formed. 

Answer HBr indicates that we will be adding H and Br across the double bond. The presence of peroxides indicates that the regiochemistry will be anti-Markovnikov. To determine whether stereochemistry is relevant in this particular case, we need to look at whether we are creating two new stereocenters. When we place the Br on the less substituted carbon (and the H on the more substituted carbon), we will only be creating one new stereocenter. With only one stereocenter, there are not four possible stereoisomers but just two possible products (a pair of enantiomers). And we will get this pair of enantiomers regardless of whether the reaction was syn or anti ... 

Answer If we compare the starting material and product, we see that we must add H and OH. We look at the regiochemistry, and we see that OH is ending up at the more substituted carbon—so we need a Markovnikov addition. Then, we look at the stereochemistry and we see that we are not creating two stereocenters in this reaction (in fact, we are not even creating one stereocenter). Therefore, the stereochemistry of the reaction will be irrelevant. So we need to choose reagents that will give a Markovnikov addition of H and OH. We can accomplish this with an acid-catalyzed hydration ... 

IB Naming Compounds with More Than One Stereocenter... 

Diastereomers are nonenantiomeric isomers that result when more than one stereocenter is present in a molecule. The distinction between diastereomers and enantiomers is not always clear but, in general, enantiomers have mirror images, whereas diastereomers are not mirror images of one another. As such diastereomers have different physical properties such as boiling and melting points, solubilities, etc. 

We have seen that when more than one stereocenter is present in a molecule, both enantiomers and diastereomers are possible. Distinguishing between enantiomers requires the relative configurations of each stereogenic center to be specified. However, to distinguish diastereomers, only the relative spatial orientation of groups needs to be specified. For example, aldotetroses have two stereocenters and the four stereoisomers are shown below ... 

Both coniine and nicotine have one stereocenter, and two possible stereoisomers (one pair of enantiomers). 

Let us consider an addition reaction to one of the cited C=X double bonds during which at least one stereocenter is produced. The topicity of the faces of the C=X double bond in question allows one to predict whether such an addition can, in principle, take place stereo-selectively. This is because depending on the topicity of the faces of the reacting C=X double bond, the transition states that result from the reaction with reagents of one kind or another (see below) from one or the other face are enantiotopic or diastereotopic. 

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