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5.7: Meso Compounds

  • Page ID
    31426
  • Objectives

    After completing this section, you should be able to

    1. determine whether or not a compound containing two chiral carbon atoms will have a meso form, given its Kekulé, condensed or shorthand structure, or its IUPAC name.
    2. draw wedge-and-broken-line structures for the enantiomers and meso form of a compound such as tartaric acid, given its IUPAC name, or its Kekulé, condensed or shorthand structure.
    3. make a general comparison of the physical properties of the enantiomers, meso form and racemic mixture of a compound such as tartaric acid.
    Key Terms

    Make certain that you can define, and use in context, the key term below.

    • meso compound
    Study Notes

    You may be confused by the two sets of structures showing “rotations.” Of course in each case the two structures shown are identical, they represent the same molecule looked at from two different perspectives. In the first case, there is a 120° rotation around the single carbon-carbon bond. In the second, the whole molecule is rotated 180° top to bottom.

    Introduction

    A meso compound is an achiral compound that has chiral centers. A meso compound contains an internal plane of symmetry which makes it superimposable on its mirror image and is optically inactive although it contains two or more stereocenters. Remember, an internal plane of symmetry was shown to make a molecule achiral in Section 5.2.

    In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of reflected stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, resulting the molecule being optically inactive.

    Identification

    A meso compound must have:

    1. Two or more stereocenters.
    2. An internal plane of symmetry, or internal mirror, that lies in the compound.
    3. Stereochemistry that cancels out. This means reflected stereocenter should have the same substituents and be inverted. For instance, in a meso compound with two stereocenters one should be R and the other S.

    The compounds 2,3-dichlorobutane contains two chiral carbons and therefore would be expected to provide 22 = 4 different stereoisomers. These stereoisomers should be made up of two pairs of enantiomers.

    Meso 1.svg

    After drawing out all the possible stereoisomers of 2,3-dichlorobutane, the pair on the right in the figure below are mirror images. Also, they are non-superimposable because they have distinctly different conformation (R,R & S,S). This makes the pair enantiomers of each other. However, the pair on the left represent a meso compound, they both are identical despite being mirror images.

    Meso 2.png

    Upon further investigation, the meso compound has an internal plane of symmetry which is not present in the pair of enantiomers. The plane of symmetry in the meso compound comes about because there are two chiral carbons present, both chiral carbons are identically substituted (Cl, H, CH3), and one chiral carbon is R and the other is S. Despite being represented as mirror images, both structures represent the same compound. This is best proven by making molecular models of both representations and then superimposing them. Overall, 2,3-dichlorobutane only has three possible stereosiomers, the pair of enantiomers and the meso compound.

    Meso 3.svg

    When looking for an internal plane of symmetry, it is important to remember that sigma bonds (single bonds) can rotate. Just because the immediate representation of a molecule does not have a plane of symmetry does not mean that one cannot be obtained through rotation. Often the substituents attached to a stereocenter need to be rotated to recognize the internal plane of symmetry. As the stereocenter is rotated, its configuration does not change. Building a molecular model when considering a possible meso compound is an invaluable tool because it allows for easy rotation of chiral carbons. An example of how rotation of a chiral carbon can reveal an internal plane of symmetry is shown below.

    Rotation 1.svg

    Example \(\PageIndex{1}\)

    Below are the two mirror images of (meso)-2,3-Butanediol. Because it is a meso compound, the two structures are identical. Show that both mirror images can be obtained by simply rotating the three-dimensional structure provided belwo.

    Meso Example 1.svg

    Example \(\PageIndex{2}\)

    1 has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; 1 has chiral centers. Thus, 1 is a meso compound.

    Example \(\PageIndex{3}\)

    This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.

    Other Examples of Meso Compounds

    Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cycloalkanes. Although two chiral carbons must be present, meso compounds can have many more. Notice that in every case a plane of symmetry is present.

    Meso3 (1).bmp

    In general, a disubstituted cycloalkane is meso if the two substituents are the same and they are in a cis conformation. Trans disubstituted cycloalkanes are not meso regardless if the two substituents are the same.

    Cyclic.bmp

    Optical Activity Analysis of a Meso Compound

    When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+) because a meso compound is achiral (optically inactive). Investigations of isomeric tartaric acid (2,3-dihydroxybutanedioic acid), carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Tartaric acid, has two chiral but only three stereoisomers. Two of these stereoisomers are enantiomers and the third is an achiral a meso compound. Some physical properties of these stereoisomers of tartaric acid are given in the table below. Notice that the enantiomers have the same amount of optical rotation but in different directons. Meso-tartaric acid produces no optical rotation because it is achiral and not optically active. Meso-tartaric acid is actually a diastereomer of both (-) and (+)-tartaric acid, which gives it a distinctly different melting point.

    Tartaric Acid 1.png

    (+)-tartaric acid: [α]D = +13º m.p. 172 ºC
    (–)-tartaric acid: [α]D = –13º m.p. 172 ºC
    meso-tartaric acid: [α]D = 0º m.p. 140 ºC
    Exercise \(\PageIndex{1}\)

    1) Determine which of the following molecules are meso.

    Mesopracticeprob.bmp

    2) Explain why 2,3-dibromobutane has the possibility of being a meso compound while 2,3-dibromopentane does not.

    3) Observe the following compound and determine if it is a meso compound. If so indicate the plane of symmetry. Red = oxygen. Remember sigma bonds are able to rotate.

    Answer

    1) A C, D, E are meso compounds.

    2) One of the requirements of a meso compound is that the reflected chiral carbons have the same substituents. The compound 2,3-dibromobutane, fulfills this requirement (Br, H, CH3) and can possibly be a meso compound if the two chiral carbons have the appropriate configuration (R & S). The substituents of the two chiral carbons in 2,3-dibromopentane do not have the same substituents (Br, H, CH3 vs. Br, H, CH2CH3). This 2,3-dibromopentane cannot form a meso compound regardless of the configurations of its chiral carbons.

    Answer 2.svg

    3) The compound is meso.

    Answer 3.svg

    Exercise \(\PageIndex{1}\)

    Which of the following are meso compounds?

    5.8.1.svg

    Answer

    a) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) between the C’s (and it has stereochemistry of S & R).

    5.8.1a solution.svg

    b) This is not a meso compound. No matter how you rotate the C-C bond, you do not see a plane of symmetry (and its stereochemistry is S & S)

    c) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) that can be seen when you rotate the C-C bond (and it has stereochemistry of S & R).

    5.8.1c solution.svg

    Exercise \(\PageIndex{2}\)

    Which of the following are meso compounds?

    5.8.2.svg

    Answer

    a) This is not a meso compound. There is no plane of symmetry and has stereochemistry of S & S.

    5.8.2a solution.svg

    b) This is a meso compound. There is an internal plane of symmetry (dashed line shown in red) and it has stereochemistry of R & S.

    5.8.2b solution.svg

    c) This is not a meso compound (even though it has planes of symmetry). The plane of symmetry shown in red makes it so that both chiral centers have symmetrical groups (the ring) and thus the compound is not chiral (so it can’t be a meso compound).

    5.8.2c solution.svg

    Exercise \(\PageIndex{3}\)

    Which of the following are meso compounds?

    5.8.3.svg

    Answer

    a) This is a meso compound. If you rotate between the C-C bond, you can see that it has a mirror plane between the C’s (shown in red on the structure to the right). Notice how rotating a C-C bond doesn’t change the stereochemistry of the molecule (S & R).

    5.8.3a solution.svg

    b) This is a meso compound. You can see the plane of symmetry in red and the compound has stereochemistry of S & R.

    5.8.3b solution.svg

    c) This is not a meso compound. There is no plane of symmetry and it has stereochemistry of S & S.

    5.8.3c solution.svg

    Exercise \(\PageIndex{4}\)

    Determine (and draw) if any of the forms of 3,4-dichlorohexane are a meso compound.

    Answer

    Looking at the 4 different possibilities below, i & ii are equivalent structures (with R & S stereochemistry) so it is a meso compound. iii & iv are not meso compounds but are enantiomers to each other.

    5.8.4 solution.svg

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