Structure and Bonding: Chirality, It's not just for Tetrahedral Carbon
Chiral centers are very typical in tetrahedral carbon compounds as you explored in More About Isomers. How would you define a chiral center?
View the structures of glyceraldehyde given below. Is there a chiral center in glyceraldehyde? If so, are the structures below isomers? Explain why or why not.
Can a molecule have more than one chiral center? Examine the chain structure of glucose below and determine the number of chiral centers. Remember a chiral carbon atom must have four different groups attached to it.
Glucose has four chiral centers. How many possible optically active isomers, which come in pairs, could glucose possibly have?
In theory, each chiral center could produce a pair of enantiomers. So the maximum possible number of optically active isomers, I, is given by the formula: I = 2n where n is the number of chiral centers in the molecule.
View the two pairs of images of 2,3-dibromobutane given below. How many chiral centers does 2,3-dibromobutane have? Are the pairs of images different? Measure the torsion angle between the two bromine atoms. Right click, select mouse click action, then select torsion. To measure a torsion angle you click on four adjacent atoms in succession. The torsion angle is the angle between the first and fourth atom while sighting down the length of the second and third. The angle is reported after clicking on the fourth atom. A good animation of a torsion angle can be found at Colby College's Chemistry website. You will need Shockwave installed to view it. What is the maximum number of possible optically active isomers? How many actual isomers are possible?
Torsion angle ______ Torsion angle ______
Torsion angle ______ Torsion angle ______
If other symmetry occurs in the molecule, the maximum number of possible optically active isomers will not be obtained. To explore symmetry in simple molecules- click here. 2,3-dibromobutane has a pair of optically active isomers (the bottom pair) plus a third isomer (top pair, mirror images are superimposable). Stereoisomers with two or more chiral centers that are superimposable are called mesoisomers.
Now, does a molecule need an atom with a chiral center to be optically active? View the two octahedral structures of Cr[H2N-(CH2)2-NH2]3. Are they isomers? If so, describe their relationship. It may help with viewing these two structures to turn off the display of the hydrogen atoms (right click, go to options, select display hydrogen).
These two structures are nonsuperimposable mirror images of each other; hence they are enantiomers. Does the central Cr atom have different groups around it?
The Cr atom is not like the C atom in the glyceraldehyde, where four different groups are bound to the carbon. When one chiral carbon occurs in a molecule, the molecule is asymmetric. The Cr compounds do not have an asymmetric atom at all, but they are nonsuperimposable (they are dissymmetric).
View the structure of methyl ethyl amine given below. Is their a chiral center in this molecule? If so, identify it and draw the mirror image of the molecule.
The nitrogen atom is a chiral center, it has four different groups on it - a hydrogen, a methyl, an ethyl, and a lone pair of electrons. Chiral centers can be found in N, P, and Si compounds that are derived from tetrahedral geometries such as tetrahedral and trigonal pyramidal molecules.
However, asymmetric amines are not isolatable as the two individual isomers, they always occur as a 50-50 mixture of both (a racemic mixture). View the animation of the ammonia molecule, the building block of amines, given below. Why do asymmetric amines occur as a racemic mixture?
This is called inversion of the trigonal pyramidal geometry. If you start with one isomer, the low-energy barrier for this inversion allows it to convert to the other quickly, and you get an equal mixture of the two isomers. A movie illustrating the inversion that occurs in amines can be found at Colby College's Chemistry website- click here. You will need Shockwave installed to view it. The trigonal planar transition state that forms then can flip either way to go back to the trigonal pyramidal geometry, and an equal mix of both enantiomers. If you examine the normal vibrational modes of ammonia, the wagging of the N-H bonds is a strong IR vibration - click here to see the motion.
Could a salt of an asymmetric 3o amine (see The Behavior of Amines) be optically active or exist as enantiomers? Explain why or why not. Use an example to support your position.
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