Stereochemistry: Stereocenter Nomenclature: The R/S System

Discussion: As we have seen before organic compounds must be provided with unambiguous names that clearly delineate all structural features of the molecule. The same is true for naming stereocenters as well. Because a stereocenter can only exist in two absolute configurations, IUPAC nomenclature is easily modified to name stereocenters. To name a stereocenter, we assign priority to the groups attached to the stereocenter, then apply a label based upon the direction in which priorities decrease. Priorities are determined based on atomic number. The system was devised by Robert. S. Cahn (Royal Institute of Chemistry, London), Christopher K. Ingold (University College, London), and Vladimir Prelog (Swiss Federal Institute of Technology, Zurich) in the 1950's, and is thus called the Cahn-Ingold-Prelog convention. (Vladimir Prelog was awarded the 1975 Nobel Prize in chemistry for his work on organic stereochemistry.)

The procedure for using the Cahn-Ingold-Prelog convention for a given stereocenter is illustrated using one enantiomer of 2-chlorobutane shown below. It is extremely useful to use molecular models for this process.


Step 1. All groups attached to the stereocenter are assigned a priority. Identify the atoms attached to the stereocenter, and arrange these atoms in order of increasing atomic number. (A periodic table is located inside the back cover of your text.)

If the atoms have the same priority, then move one atom further out from the stereocenter until you find a difference. In this case, the two carbon atoms have the same priority, so we move further out. The methyl carbon is attached to three hydrogens, whereas the ethyl carbon is attached to two hydrogens and a carbon. Select the atoms that are different on these groups (hydrogen of the methyl vs. carbon of the ethyl), and assign their priorities. Hydrogen is lower atomic number and therefore lower priority than carbon, so C-H has less priority than C-C.

Thus the order of priorities is: Cl (highest) > C-C > C-H > H.

Step 2. Change your view of the molecule so that you are looking along the bond between the stereocenter and the lowest priority group, with the lowest priority group facing away from you. If you are using molecular models, use the bond between the stereocenter and the lowest priority group as a handle to hold the molecule. From this perspective, the molecule looks like this:

Step 3. Note the direction in which the priorities decrease. In the example, the priorities decrease in a clockwise manner, as shown by the curved arrow. If the priorities decrease in a clockwise manner, we label the stereocenter as R (Latin: rectus, right). If the priorities decrease in a counterclockwise manner, we label the stereocenter as S (Latin: sinister, left). (A useful mnemonic: At the top of the molecule, draw a curved arrow showing the direction in which priorities decrease. If that arrow points to the right (clockwise), then the stereocenter is R.) To provide the full name for the molecule, R or S is added in parenthesis in front of the name. Thus, this molecule is (R)-2-chlorobutane. If there is more than one stereocenter, each R or S is numbered with its position on the molecular skeleton. Example: (2R, 3S)-2,3-butanediol.

Example 1: Label the stereocenters in the following molecule as R or S.

Solution 1:

Step 1: Assign priorities to the groups attached to the stereocenter. The four atoms attached to the stereocenter are iodine and three carbons. Iodine has atomic number 53, and is therefore higher priority than carbon (atomic number 12). Now we need to differentiate between the three carbons by examining the attached atoms. Double bonds are treated as two single bonds to the element on the other end of the bond, as shown below. (Triple bonds are treated as three attachments). Thus we treat the carbon of the alkene as if it is attached to two carbons (for the double bond) and a hydrogen. The carbon of the carboxylic acid is treated as if it was attached to three oxygens (two from the carbonyl group, and one from the C-OH bond). The carbon with three oxygens gets priority over a carbon with two carbons or over a carbon with one carbon and two hydrogens (the methylene group). The relative priority of the last two groups is determined as before, by looking for a difference. Each carbon is attached to one other carbon and one hydrogen, so these cancel out. This leaves a carbon attached to the alkene carbon, and a hydrogen attached to the methylene carbon.

The alkene carbon gets priority over the methylene carbon, and our final order of priorities is: iodine (highest) > CO2H > C=C > CH2.

Viewing this with the lowest priority group in the rear:

The priorities decrease in a counterclockwise manner, so the stereocenter has the S configuration.

Exercises: Label each stereocenter as R or S.

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