Ch. 6 Textbook
6.1 Erythrose and threose
The simplest sugar glyceraldehyde, an aldotriose, is a very good example of the compounds with one asymmetric carbon atoms (▶go to S2.2 Eclipsed and Staggered FormsChapter 5.5, 21 and 22). Sugars with four carbon atoms are also a good example of the compounds with two asymmetric carbon atoms. Sugars which have one more HCOH unit in the middle of the molecule as compared with glyceraldehydes are aldotetroses 1. These have two asymmetric carbon atoms C2 and C3. For each asymmetric carbon atom, there are two possible configurations, R or S, regardless of the configuration of other asymmetric carbon atoms. This is valid whatever is the number of asymmetric carbon atoms.
A Fischer projection for compounds with two or more asymmetric carbon atoms will be drawn in the following way.1) Carbon atoms are arranged vertically. As previously mentioned, the most highly oxidized carbon atom occupies the top position (C1). 2) Horizontal ligands are above the plane of paper while vertical ligands are below the plane of paper. 3) As a matter of fact, the carbon chain does not form a straight line. So not all carbon atoms can be on the plane of the paper at the same time. When you consider the configuration of a given carbon atom, you have to suppose that carbon is on the plane although other carbon atoms may be out of the plane.
For reference, the name of each stereoisomer is given. Note that two names are used to differentiate the stereochemistry of ligand. Erythrose has the same two ligand on the same side of the Fischer projection (2, 3) while threose has on the opposite side (4, 5).
This nomenclature is extended to all compounds with two asymmetric carbon atoms. Isomers having the same stereochemistry as that of erythroseis given prefix "erythro-", while isomers having the same stereochemistry as that of threose is given prefix "threo-"
▶go to S6.1 Erythro form, threo form
6.2 Tartaric acid; meso form
Tartaric acid 9 is formally the oxidized product of aldotetrose 1. The two termini of 1, -CHO and -CH2OH, are both -COOH in 9. Like 1, 9 has two asymmetric carbon atoms C2 and C3, and may have 22 = 4 stereoisomers. To confirm whether this prediction is correct, we will draw stereoisomers 10～13 in analogy with 2～5.
There is some difference between the stereochemistry of tartaric acid 9 and that of aldotetrose 1 in that 9 has an identical set of ligands at C2 and C3. Will this affect the stereochemistry of 9?
Though 10 and 11 seem to have different stereochemistry, you will find that in fact 10 and 11 are the same molecule by careful examination of molecular model. Another important point is there is a center of symmetry in the center of 10 (and 11). This means that the upper half and the lower half of the molecule are identical. Hence 10 and 11 is "intramolecular" racemic compound and hence optically inactive. Optically inactive compounds with asymmetric carbon atom(s) are referred to as meso form.
This means that tartaric acid has only three stereoisomer, 12 and 13, the enantiomeric pair , and optically inactive (meso) 10 (= 11). You must remember that the number of stereoisomers will be less than 2ｎ when there is a symmetry in the molecule. (+)-Tartaric acid is widely distributed in nature, particularly in fruits as acid, and calcium or magnesium salts. Since c acid is obtained as the by-product of the wine production, it was known from olden times. Hence it is quite understandable that a great development of organic stereochemistry was made by the study of Pasteur on tartaric acid. When Pasteur initiated his study, (+)-tartaric acid and racemic acid (in modern terminology, (土)-tartaric acid: the word racemate cam from the name of this compound) were known, but the stereochemical relation between the two compounds was not understood at that time. Pasteur could successfully resolved (土)-tartaric acid by physical method (in fact with his own hands, magnifier and tweezers), establishing the relation among stereoisomers.
6.3 Configuration and conformation
Let us try to convert the Fischer projection of (-)-erythrose 2 and (+)-erythrose 3 into the corresponding Newman projection. If we strictly copy the Fischer projections, the obtained Newman projections 14 and 15 are necessarily in eclipsed conformation. Molecules are not likely in the eclipsed form, but in the more stable staggered conformation. It must be noted that more than one staggered conformation are possible.
There may be several factors which determine the most favorable conformation of the molecule. One is certainly the steric repulsion between ligands. It is highly likely that the molecule will be most stable when the largest ligand bonded to C2 and the largest ligand bonded to C3 are antiperiplanar. It is possible that an attractive interaction might occur between two ligands regardless of the size. This possibility will not be considered here for the sake of simplicity.
▶go to S6.3 Configuration and conformation
6.4 Compounds with three or more asymmetric carbon atoms
What we have learned on aldotetrose and tartaric acid can be extended to compounds with five carbon atoms (three asymmetric carbon atoms) or to even more complex compounds. When the given compounds have three asymmetric carbon atoms, each asymmetric carbon atom can be either R or S configuration, and the number of isomers is 2x2x2=23= 8 in maximum. If the molecule has symmetry, this number will be reduced as is the case with tartaric acid.
6.5 D, L nomenclature of sugars
Sugars are generally classified into two families, D-series and L-series. If the configuration of the asymmetric carbon atom furthest from the carbonyl carbon atom is similar to that of D-glyceraldehyde, the compound will belong to the D-series. If similar to that of L-glyceraldehyde, the compound will belong to the L-series. This method has some practical value since most of naturally occurring sugars belong to D-series.
▶go to S6.4 D, L-series (sugar)