Meso-compounds

PROBLEMS For each pair compounds below, determine whether the pair are enantiomers or diastereomers. 

The analogy goes like this when you have a lot of stereocenlers in a compound, there will be many stereoisomers (brothers and sisters). But, they will be paired up into sets of enantiomers (twins). Any one molecule will have many, many diastereomers (brothers and sisters), but it will have only one enantiomer (its mirror image twin). For example, consider the following compound  

This compound has five stereocenters, so it will have many diastereomers (compounds where only some of the wedges have been inverted). There are many, many possible compounds that fit that description, so this compound will have many brothers and sisters. But this compound will only have one twin—only one enantiomer (there is only one mirror image of the compound above)  

It is possible for a compound to be its own mirror image. In such a case, the compound will not have a twin. It will be all by itself, and the total number of stereoisomers will be an odd number, rather than an even number. That one lonely compound is called a meso compound. If you try to draw the enantiomer (using either of the methods we have seen), you will find that your drawing will be the same compound as what you started with. 

A meso compound has stereocenters, but the compound also has symmetry that allows it to be the mirror image of itself. Consider cis-1,2-dimethylcyclohexane as an example. This molecule has a plane of symmetry cutting the molecule in half. Everything on the left side of the plane is mirrored by everything on the right side  

Now for a rather unexpected twist. We have seen that if there are n chiral centres there should be 2 configurational isomers, and we have considered each of these for n = 2 (e.g. ephedrine, pseudoephedrine). It transpires that if the groups around chiral centres are the same, then the number of stereoisomers is less than 2 . Thus, when n = 2, there are only three stereoisomers, not four. As one of the simplest examples, let us consider in detail tartaric acid, a component of grape juice and many other fruits. This fits the requirement, since each of the two chiral centres has the same substituents. 

We can easily draw the four predicted isomers, as we did for the ephedrine-pseudoephedrine group, and two of these represent the enantiomeric pair of (—)-tartaric acid and (+)-tartaric acid. Now let us consider the other pair of isomers, and we shall see the consequences of 

We can see why a compound with chiral centres should end up optically inactive by looking again at the eclipsed conformer. The molecule itself has a plane of symmetry, and because of this symmetry the optical activity conferred by one chiral centre is equal and opposite to that conferred by the other and, therefore, is cancelled out. It has the characteristics of a racemic mixture, but as an intramolecular phenomenon. A meso compound is defined as one that has chiral centres but is itself achiral. Note that numbering is a problem in tartaric acid because of the symmetry, and that positions 2 and 3 depend on which carboxyl is numbered as C-1. It can be seen that (2R,3S) could easily have been (3R,2S) if we had numbered from the other end, a warning sign that there is something unusual about this isomer. 

The same stereochemical principles apply to both acyclic and cyclic compounds. With simple cyclic compounds that have little or no conformational mobility, it can even be easier to foUow what is going on. Let us first look at cyclopropane-1,2-dicarboxylic acid. These compounds were considered in Section 3.4.3 as examples of geometric isomers, and cis and trans isomers were recognized. 

This is essentially the same as the tartaric acid example, without the conformational complication. Thus, there are two chiral centres, and the groups around each centre are the same. Again, we get only three stereoisomers rather than four, since the cis compound is an optically inactive meso compound. There is a plane of symmetry in this molecule, and it is easy to see that one chiral centre is mirrored by the other, so that we lose optical activity. 

I were given the first compound only, you could have drawn the enantiomer by using 

Let s look at one more example of a compound with more than one chirality center, the tartaric acid used by Pasteur. The four stereoisomers can be drawn as follows  

Tile mirror-image 2R,3R and 2S,3S structures are not identical and therefore represent a pair of enantiomers. A close look, however, shows that the 2R,3S and 2S.3R structures are identical, as can be seen by rotating one structure 180°. 

Some physical properties of the three stereoisomers are listed in Table 9.3. The (+)- and (-j-tartaric acids have identical melting points, solubilities, and densities but differ in the sign of their rotation of plane-polarized light. The meso isomer, by contrast, is diastereomeric with the (+) and (-) forms. As such, it has no mirror-image relationship to (+)- and (-)-tartaric acids, is a different compound altogether, and has different physical properties. 

Distinguishing Chiral Compounds from Meso Compounds 

Does ris-1,2-dimethylcyclobutane have any chirality centers Is it chiral  

Does c/,s-l,2-diniethvlcvclobutane have anv cliiraliiv centers Is it chiral  

Whereas 2,3-dibromopentane has two stereogenic centers and the maximum of four stereoisomers, 2 -dibromobutane has two stereogenic centers but fewer than the maximum number of stereoisomers. 

With two stereogenic centers, the maximum number of stereoisomers = 4. 

To find and draw all the stereoisomers of 2,3-dibromobutane, follow the same stepwise procedure oudined in Section 5.7. Arbitrarily add the H, Br, and CH3 groups to the stereogenic centers, forming one stereoisomer A, and then draw its mirror image B. A and B are nonsuperimposable mirror images—enantiomers. 

To find the other two stereoisomers (if they exist), switch the position of two groups on one stereo-genic center of one enantiomer only. In this case, switching the positions of H and Br on one stereo-genic center of A forms C, which is different from both A and B and is thus a new stereoisomer. 

However, the mirror image of C, labeled D, is superimposable on C, so C and D are identical. Thus, C is achiral, even though it has two stereogenic centers. C is a meso compound. 

A symmetry plane through the C2-C3 bond of meso-tartark acid makes the molecule achiral. 

To see whether a chirality center is present, look for a carbon atom bonded to four different groups. To see whether the molecule is chiral, look for the absence of a symmetry plane. Not all molecules with chirality centers are chiral—meso compounds are an exception. 

There are only three stereoisomers of 2,3-dibromobutane because two of the four structures are identical. The diastereomer on the right is achiral, having a mirror plane of symmetry. The asymmetric carbon atoms have identical substituents, and the one with (R) configuration reflects into the other having (5) configuration. It seems almost as though the molecule were a racemic mixture within itself. 

Compounds that are achiral even though they have asymmetric carbon atoms are called meso compounds. The (2R,3S) isomer of 2,3-dibromobutane is a meso compound most meso compounds have this kind of symmetric structure, with two similar halves of the molecule having opposite configurations. In speaking of the two diastereomers of 2,3-dibromobutane, the symmetric one is called the meso diastereomer, and the chiral one is called the ( ) diastereomer, since one enantiomer is ( + ) and the other is ( —). 

MESO COMPOUND An achiral compound that has chirality centers. 

The term meso (Greek, middle ) was used to describe an achiral member of a set of diastereomers, some of which are chiral. The optically inactive isomer seemed to be in the middle between the dextrorotatory and levorotatory isomers. The definition just given ( an achiral compound with chirality centers ) is more easily applied, especially when you remember that chirality centers are usually asymmetric carbon atoms. 

We have already seen other meso compounds, although we have not yet called them that. For example, the cis isomer of 1,2-dichlorocyclopentane has two asymmetric carbon atoms, yet it is achiral. Thus it is a meso compound. cw-l,2,-Dibromocy-clohexane is not symmetric in its chair conformation, but it consists of equal amounts of two enantiomeric chair conformations in a rapid equilibrium. We are justified in looking at the molecule in its symmetric flat conformation to show that it is achiral and meso. For acyclic compounds, the Fischer projection helps to show the symmetry of meso compounds. 

There are only two types of stereoisomers Enantiomers are stereoisomers that are mirror images. Diastereomers are stereoisomers that are not mirror images. 

Label the two stereogenic centers in each compound and draw all possible stereoisomers (a) CH3CH2CH(CI)CH(0H)CH2CH3 (b) CH3CH(Br)CH2CH(C0CH3. 

One of the following molecules (a)-(d) is D-erythrose 4-phosphate, an intermediate in the Calvin photosynthetic cycle hy which plants incorporate CO2 into carbohydrates. If D-erythrose 4-phosphate has R stereochemistry at both chirality centers, which of the structures is it Which of the remaining three structures is the enantiomer of D-erythrose 4-phosphate, and which are diastereomers  

Assign R,S configuration to each chirality center in the following molecular model of the amino acid isoleucine  

How many chirality centers does morphine have How many stereoisomers of morphine are possible in principle  

We saw that the molecule 2-bromo-3-chlorobutane contains two distinct stereocenters, each with a different halogen substituent. How many stereoisomers are to be expected if both centers are identically substituted  

Two identically substituted stereocenters give rise to only three stereoisomers 

The first pair of stereoisomers, with R and S,S configurations, is clearly recognizable as a pair of enantiomers. However, a close look at the second pair reveals that (5,f ) and mirror image R,S) are superimposable and therefore identical. Thus, the S,R diastereomer of 2,3-dibromobutane is achiral and not optically active, even though it contains two stereocenters. The identity of the two structures can be readily confirmed by using molecular models. 

Brominating racemic 2-bromobutane at C3 is stereochemically equivalent to pairing two identical pairs ( racemates ) of shoes only three distinct combinations. 

Meso diastereomers can exist in molecules with more than two stereocenters. Examples are 2,3,4-tribromopentane and 2,3,4,5-tetrabromohexane. 

---

The same kind of spontaneous racemization oc curs for any as 1 2 disubstituted cyclohexane in which both substituents are the same Because such compounds are chiral it is incorrect to speak of them as meso compounds which are achiral by definition Rapid chair-chair interconversion however converts them to a 1 1 mixture of enantiomers and this mix ture IS optically inactive... 

A second type of optically inactive chiral compounds, meso compounds, will be discussed in the next section. 

Multiple Chiral Centers. The number of stereoisomers increases rapidly with an increase in the number of chiral centers in a molecule. A molecule possessing two chiral atoms should have four optical isomers, that is, four structures consisting of two pairs of enantiomers. However, if a compound has two chiral centers but both centers have the same four substituents attached, the total number of isomers is three rather than four. One isomer of such a compound is not chiral because it is identical with its mirror image it has an internal mirror plane. This is an example of a diaster-eomer. The achiral structure is denoted as a meso compound. Diastereomers have different physical and chemical properties from the optically active enantiomers. Recognition of a plane of symmetry is usually the easiest way to detect a meso compound. The stereoisomers of tartaric acid are examples of compounds with multiple chiral centers (see Fig. 1.14), and one of its isomers is a meso compound. 

When the asymmetric carbon atoms in a chiral compound are part of a ring, the isomerism is more complex than in acyclic compounds. A cyclic compound which has two different asymmetric carbons with different sets of substituent groups attached has a total of 2 = 4 optical isomers an enantiometric pair of cis isomers and an enantiometric pair of trans isomers. However, when the two asymmetric centers have the same set of substituent groups attached, the cis isomer is a meso compound and only the trans isomer is chiral. (See Fig. 1.15.)... 

Propylene glycol, dipropylene glycol, and tripropylene glycol all have several isomeric forms. Propylene glycol has one asymmetric carbon and thus there are two enantiomers (R)-I,2-propanediol and (3)-1,2-propanediol. 1,3-Propanediol is a stmctural isomer. Dipropylene glycol exists in three stmctural forms and since each stmctural isomer has two asymmetric carbons there are four possible stereochemical isomers per stmcture or a total of twelve isomers. These twelve consist of four enantiomer pairs and two meso- compounds. Tripropylene glycol has four stmctural isomers and each stmctural isomer has... 

The situation is different if the substrate is a prochiral or meso compound. Since these molecules have a center or plane of symmetry the binding of pro-S or pro-R forms is equivalent. The chirahty appears only as a result of the transformation. Hence, at least theoretically, the compound can be converted to one enantiomer quantitatively. 

Meso stereoisomer (Section 7.11) An achiral molecule that has chirality centers. The most common kind of meso compound is a molecule with two chirality centers and a plane of symmetry. 

Enantioselective desymmetrization of achiral or meso compounds with formation of enantiomerically enriched products, among them heterocycles 99JCS(P1)1765. 

Distinguishing Chiral Compounds from Meso Compounds... 

A look at the structure of cis-1,2-dimethylcyclobutane shows that both methyl-bearing ring carbons (Cl and C2) are chirality centers. Overall, though, the compound is achiral because there is a symmetry plane bisecting the ring between Cl and C2. Thus, the molecule is a meso compound. 

Which of the following structures represent meso compounds ... 

Problem 9.17 Does the following structure represent a meso compound If so, indicate the symmetry plane. 

Meso compounds contain chirality centers but are achiral overall because they have a plane of symmetry. Racemic mixtures, or racemates, are 50 50 mixtures of (+) and (-) enantiomers. Racemic mixtures and individual diastereomers differ in their physical properties, such as solubility, melting point, and boiling point. 

Which, if any, of the following structures represent meso compounds (Blue -N, 5relloW green - Cl.)... 

Draw the structure of a meso compound that has five carbons and three chirality centers. 

D-Allaric acid has a symmetry plane and is a meso compound, but o-glucaric acid is chiral. 

The substituted carbons are stereogenic carbons. This means that there are not only two isomers. In the most general case, where W, X, Y, and Z are all different, there are four isomers since neither the cis nor the trans isomer is superimposable on its mirror image. This is true regardless of ring size or which carbons are involved, except that in rings of even-numbered size when W, X, Y, and Z are at opposite comers, no chirality is present, (e.g., 68). In this case, the substituted carbons are not chiral carbons. Note also that a plane of symmetry exists in such compounds. When W = Y and X=Z, the cis isomer is always superimposable on its mirror image, and hence is a meso compound, while the trans isomer consists of a dl pair, except in... 

Rings with more than two differently substituted carbons can be dealt with on similar principles. In some cases, it is not easy to tell the number of isomers by inspection. The best method for the student is to count the number n of differently substituted carbons (these will usually be asymmetric, but not always, e.g., in 68) and then to draw 2" structures, crossing out those that can be superimposed on others (usually the easiest method is to look for a plane of symmetry). By this means, it can be determined that for 1,2,3-cyclohexanetriol there are two meso compounds and a dl pair and for 1,2,3,4,5,6-hexachlorocyclohexane there are seven meso compounds and a dl pair. The drawing of these structures is left as an exercise for the student. 

Of course, the trans isomer will give the opposite results the threo pair if the addition is syn and the erythro pair if it is anti. The threo and erythro isomers have different physical properties. In the special case where Y=W (as in the addition of Br2), the erythro pair is a meso compound. In addition to triple-bond compounds of the type ACsCA, syn addition results in a cis alkene and anti addition in a trans alkene. By the definition given on page 166 addition to triple bonds cannot be stereospecific, though it can be, and often is, stereoselective. 

In open-chain compounds, the molecule can usually adopt that conformation in which H and X are anti periplanar. However, in cyclic systems this is not always the case. There are nine stereoisomers of 1,2,3,4,5,6-hexachlorocy-clohexane seven meso forms and a dl pair (see p. 161). Four of the meso compounds and the dl pair (all that were then known) were subjected to elimination of HCl. Only one of these (1) has no Cl trans to an H. Of the other isomers, the fastest elimination rate was about three times as fast as the... 

Of the two former processes shown in Scheme 5.2, the kinetic resolution of race-mates has found a much greater number of applications than the desymmetrization of prochiral or meso compounds. This is due to the fact that racemic substrates are much more common than prochiral ones. However, kinetic resolution suffers from a number of drawbacks, the main being the following ... 

Consider, for example, a hydrolase-based desymmetrization of a pseudo-meso compound 20 (Scheme 5.18) [91]. 

Scheme 5.18 Enzymatic hydrolysis of a pseudo-meso compound 20.

If a molecule has an internal plane of symmetry, then it is a meso compound. If yon try to draw the enantiomer (nsing either one of the two methods we saw), you will... 

---