Chemistry
Meso Compounds
Meso compounds are molecules with chiral centers that possess an internal plane of symmetry, resulting in overall achirality. This unique property makes meso compounds optically inactive despite having chiral elements. They are often found in organic chemistry, and their internal symmetry allows for distinct stereochemical properties compared to other chiral compounds.
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5 Key excerpts on "Meso Compounds"
eBook - PDF- T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
Meso Compounds are not optically active. Another test for molecular chirality is to construct a model (or write the structure) of the molecule and then test whether or not the model (or structure) is superposable on its mirror image. If it is, the molecule is achiral. If it is not, the molecule is chiral. We have already carried out this test with structure C and found that it is achiral. We can also demonstrate that C is achiral in another way. Figure 5.15 shows that structure C has an internal plane of symmetry (Section 5.6). The following two problems relate to compounds A–D in the preceding paragraphs. Last, we check that none of these formulas is identical to another by testing the superposability of each one with the others. We should not expect any to be identical because none of the formulas has an inter- nal plane of symmetry. The case would have been different for 2,4-dibromopentane, however, in which case there would have been one meso stereoisomer (a type of stereoisomer that we shall study in the next section). Br H Br H FIGURE 5.15 The plane of symmetry of meso-2,3- dibromobutane. This plane divides the molecule into halves that are mirror images of each other. PRACTICE PROBLEM 5.20 Which of the following would be optically active? (a) A pure sample of A (b) A pure sample of B (c) A pure sample of C (d) An equimolar mixture of A and B (continued) 5.12 Molecules with More than One Chiral Center 227 PRACTICE PROBLEM 5.21 The following are formulas for three compounds, written in noneclipsed conformations. In each instance tell which compound (A, B, or C on the previous page) each formula represents. Br H Br H (c) Br H H Br (b) H Br H Br (a) Which of the compounds shown below (X, Y, or Z) is a meso compound? Strategy and Answer In each molecule, rotating the groups joined by the C 2 C 3 bond by 180° brings the two methyl groups into comparable position. In the case of compound Z, a plane of symmetry results, and therefore, Z is a meso compound.
eBook - PDF- T. W. Graham Solomons, Craig B. Fryhle, Scott A. Snyder(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
When we write the new structure C (see below) and its mirror image D, however, the situation is different. The two structures are superposable. This means that C and D do not represent a pair of enantiomers. Formulas C and D represent identical orientations of the same compound: D C Br Br Br H H H H Br The molecule represented by structure C (or D) is not chiral even though it contains two chirality centers. • If a molecule has an internal plane of symmetry it is achiral. • A meso compound is an achiral molecule that contains chirality centers and has an internal plane of symmetry. Meso Compounds are not optically active. Another test for molecular chirality is to construct a model (or write the structure) of the molecule and then test whether or not the model (or structure) is superposable on its mirror image. If it is, the molecule is achiral. If it is not, the molecule is chiral. We have already carried out this test with structure C and found that it is achiral. We can also demonstrate that C is achiral in another way. Figure 5.15 shows that structure C has an internal plane of symmetry (Section 5.6). The following two problems relate to compounds A–D in the preceding paragraphs. Last, we check that none of these formulas is identical to another by testing the superposability of each one with the oth- ers. We should not expect any to be identical because none of the formulas has an internal plane of symmetry. The case would have been different for 2,4-dibromopentane, however, in which case there would have been one meso stereoisomer (a type of stereoisomer that we shall study in the next section). Br H Br H FIGURE 5.15 The plane of symmetry of meso-2,3- dibromobutane. This plane divides the molecule into halves that are mirror images of each other. PRACTICE PROBLEM 5.20 Which of the following would be optically active? (a) A pure sample of A (b) A pure sample of B (c) A pure sample of C (d) An equimolar mixture of A and B
- Vasyl Andrushko, Natalia Andrushko, Vasyl Andrushko, Natalia Andrushko(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
In such Meso Compounds (or meso isomers), the mir- ror images are superposable, i.e., identical. Despite the fact that Meso Compounds have stereogenic centers, they are not optically active members of a set of ster- eoisomers. This is because the rotation of plane polarized light by one stereogenic center within the molecule is equal and opposite to the rotation caused by the other (mirror image) stereogenic center, and the two effects cancel. The famous example is tartaric acid with two stereogenic centers (at C2 and C3) (Figure 1.20). The two stereogenic centers have identi- cal substituents, and one of the possible stereoisomers of this compound has an internal mirror plane such that C2 is one mirror image of C3. As noted, when two stereogenic centers are equivalent, the number of stereoisomers is less than the maximum of 2 n , but in fact, it is n þ 1. In the case of two stereogenic centers (n ¼ 2), there are 3 stereoisomers (e.g., L-(R,R)-tartaric, D-(S,S)-tartaric, and meso-tartaric acid). 1.1.3.8. Chirality without Stereocenters Not only does a plane of symmetry, which divides a molecule into two exactly identical halves, guarantee achirality, but there are also other elements of symmetry that will make a molecule with stereogenic centers optically inactive. For example, a center of symmetry, a point at which all the straight lines joining identical points in the molecule cross each other, will be responsible that such molecules do not show an enantio- merism and therefore are not optically active (Figure 1.21).
- David R. Klein(Author)
- 2017(Publication Date)
- Wiley(Publisher)
(a) (b) (c) (d) (e) 5.59. (a) The compound in part (a) of the previous problem has an internal plane of symmetry, and is therefore a meso compound: (b) The structures shown in parts (b) and (c) of the previous problem are enantiomers. An equal mixture of these two compounds is a racemic mixture, which will be optically inactive. (c) Yes, this mixture is expected to be optically active, because the structures shown in parts (d) and (e) of the previous problem are not enantiomers. They are diastereomers, which are not expected to exhibit equal and opposite rotations. 5.60. As we saw in problem 5.58, it is helpful to use a numbering system when converting one type of drawing into another. When drawing each substituent in the Fischer projection, you must decide whether it is on the right or left side of the Fischer projection. For each chiral center, make sure that the configuration is the same as the configuration in the bond-line drawing. If necessary, assign the configuration of each chiral center in both the Fischer projection and the bond-line drawing to ensure that you drew the configuration correctly. With enough practice, you may begin to notice some trends (rules of thumb) that will allow you to draw the configurations more quickly. (a) H OH CH 2 OH HO H O OH HO OH O OH OH 1 2 3 4 1 2 3 4 CHAPTER 5 151 (b) (c) 5.61. As shown below, there are only two stereoisomers (the cis isomer and the trans isomer). cis trans With two chiral centers, we might expect four possible stereoisomers, but two stereoisomers are Meso Compounds, as shown above, so these are the only two isomers. 5.62. With three chiral centers, we would expect eight stereoisomers (2 3 = 8), labeled 1–8. However, structures 1 and 2 represent one compound (a meso compound), while structures 3 and 4 also represent one compound (a meso compound).
eBook - PDF- Allan Blackman, Steven E. Bottle, Siegbert Schmid, Mauro Mocerino, Uta Wille(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
Achieving stereochemical control over synthetic reactions, however, is one of the greatest challenges facing chemists today. To master such challenging methodology, we must first learn about the key functional groups, their properties and how they react; this will be covered in the following chapters. CHAPTER 17 Chirality 901 SUMMARY 17.1 Define stereoisomers and recognise stereocentres. Isomers are molecules with the same molecular formula but different structures. Constitutional isomers are isomers with different sequences of atom connectivity. Stereoisomers have the same order of attachment of atoms, but a different three- dimensional orientation of their atoms in space. A chiral object is one that is not superimposable on its mirror image. Enan- tiomers are stereoisomers that are chiral. An achiral object is the same as its mirror image. Diastereomers are stereoisomers that are not superimposable and are not mirror images. 17.2 Recognise enantiomers as non-equivalent mirror images. Most biological molecules are enantiomeric. Enantiomers have the same physical and chemical properties but each of a pair reacts differently with other chiral molecules. A plane of sym- metry is an imaginary plane passing through an object, dividing it such that one half is the reflection of the other half. Objects with planes of symmetry are achiral. A stereocentre is the part of a molecule that can be assembled in two different ways to generate stereoisomers. The most common type of stereo- centre among organic compounds is a tetrahedral carbon atom with four different groups bonded to it. Inorganic molecules can also have stereocentres and, therefore, enantiomers and diastereomers. 17.3 Apply the R,S system to name enantiomers. The configuration at a stereocentre can be specified by the R,S system.
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