In our discussion of VSEPR theory, we saw how chemists used Coulomb's Law to rationalize molecular shapes. What we did not consider was the experimental evidence that allowed chemists to deduce those shapes in the first place. Here we will consider two lines of evidence that were central to the conclusion that a carbon atom is tetrahedral when it connected to four other atoms. The first line involves a series of experiments in which one or more hydrogen atoms attached to a carbon were replaced with halogen atoms. The second approach involved the measurement of the optical activity of these compounds.
We have already seen that information about the polarity of a molecule offers insight into its molecular structure. For example, the fact that methane has a dipole moment of zero means that the structure of methane is not pyramidal. This information does not, however, differentiate between a square planar and a tetrahedral geometry. More information is needed to establish one specific geometry.
An important line of additional evidence came from the determination of the number of isomers of derivatives of methane. As summarized in Equation 1, irradiation of a sample of methane and dichlorine produces a mixture of compounds in which 1, 2, 3, and 4 of the hydrogen atoms are replaced by chlorine atoms. Only one compound with the molecular formula CH3Cl has ever been isolated. In other words, no one has ever obtained an isomer of CH3Cl . The same is true for the other compounds shown in Equation 1.
If the chloromethane is separated from the other components of the mixture and treated in a similar fashion, but with dibromine instead of dichlorine, another series of halogenated compounds is obtained. The reaction is summarized in Equation 2. No one has ever obtained an isomer of any of the products shown in Reaction 2 either.
Exercise 1 If you assume that carbon is square planar, which of the products shown in Equation 2 could exist as isomers? CH2BrCl CHBr2Cl CBr3Cl
Exercise 2 If you assume that carbon is tetrahedral, which of the products shown in Equation 2 could exist as isomers? (Make two molecular models of each compound and compare them.) CH2BrCl CHBr2Cl CBr3Cl
If bromochloromethane is converted into bromochlorofluoromethane, as shown in Equation 3, two products are formed.
Exercise 3 If you assume that carbon is square planar, how many isomers of CHBrClF are possible? one two three
Figure 1 presents interactive models of the two isomers of CHBrClF.
Whenever a tetrahedral atom is connected to four different atoms it is chiral. Figure 2 presents the structures of some biologically important compounds that contain chiral carbon atoms.
Chiral Molecules of Biological Interest
In our discussion of host-guest chemistry, we encountered the lock and key metaphor to describe the snug fit of a guest into a host molecule. How well a guest fits into its host depends upon the 3-dimensional shapes of both partners as witnessed by the dramatic difference in odor between caraway oil and spearmint. Presumably these enantiomers have different aromas because they fit into different olfactory receptor sites.
Exercise 4 How many chiral carbon atoms are there in cholesterol? How many are there in b-(D)-glucopyranose?
One way to determine if a carbon atom is chiral is to identify the four groups attached to it. If they are different, the carbon is chiral. Another way is to look for at least two of the same group attached to the carbon. If you see that, then the carbon is not chiral. For example, the carbon is not chiral in any of the following molecules: CH2Cl2, CH2ClF, CH3CHCl2, CH3OH, CBrCl3, etc. The reason for this should be apparent from Figure 3 in which you can see that whenever at least two of the four groups attached to a carbon are identical, the molecule will contain in internal plane of symmetry. Since chirality arises from the asymmetry of a molecule, compounds containing internal symmetry planes are not chiral.