R vs S Configuration (Absolute Configuration)

We said earlier that stereoisomers are molecules with the same connectivity of atoms that differ only in the spatial arrangement of the atoms. There are 2 specific ways a stereogenic carbon can arrange its groups spatially, which we call R vs S configuration (commonly called the absolute configuration of a molecule). To determine whether an atom is R vs configuration, we:

a. Rank each group coming off the stereogenic center from 1-4, 1 being the group with the highest molecular weight and 4 being the group with the least; however, you do not just add up the molecular weight of the entire group. You must “walk your way down the molecule.” This means that you start at the atoms closest to the stereogenic center, and you move (or “walk”) outward from the stereogenic center. Let’s look at an example to illustrate:

Here is an enantiomer for Adderall, the medication commonly given to treat ADHD:




To prioritize each group by molecular weight, we must start closest to the stereogenic center (the carbon atom), and work our way outward. So we will zoom in on just the groups directly attached to the carbon, which are labeled in red:


The first step in determining the R vs S configuration is assigning priority to the atoms of the chirality center.

It’s immediately clear that the –N will be priority 1 because it has the highest molecular weight, while the -H will be priority 4 because it has the lowest molecular weight. But what about the two alkyl chains? Both initially tie because they are both carbon atoms. So we then look at the atoms each carbon is attached to (colored blue).



The heaviest atom directly attached to the left carbon is another carbon atom, while the heaviest atom attached to the right carbon is a hydrogen atom. Because of this, the left carbon gets priority 2, and the carbon gets priority 3.




For the rest of the process, we will only call groups by their number (1-4).


Keep it Simple

How do we prioritize a double or triple bond when determining R vs S configuration? We treat it like so:





Prioritize the groups attached to the stereogenic center as 1-4 based on molecular weight.


We narrow our focus to just the atoms located on the stereogenic center:


It’s now clear that chlorine will be priority 1 and hydrogen will be priority 4. But the carbons both tie for 2 and 3 right now. So we now move outward from the stereogenic center and look at the atoms each carbon is directly attached to:



The highest molecular weight on both carbons then is another carbon, so we still have a tie. The second highest priority atom on the left carbon is another carbon, but on the right, the second highest priority atom is a hydrogen atom. Because of this, the carbon on the left gets 2, and the carbon on the right gets 3.



Keep it Simple

Deuterium is commonly used on questions that involve stereoisomers. It is an isotope of hydrogen, as it has 1 more neutron than a typical hydrogen atom. Because of this, it is heavier than hydrogen; therefore, deuterium will have a lower priority number (indicating its higher molecular weight) than hydrogen.

b. Arrange group 4 so it is going back into the page (so it’s on a wedge). This can be done by mentally rotating the molecule or by the double displacement trick (see below).


Some people are able to mentally rotate from one structure orientation to the other. This would probably be done by:


For those who can’t mentally rotate objects (the vast majority of the population), there is a solution! It’s called the double displacement rule. In this approach, you would first swap the positions of 2 group’s priority numbers. You must then swap the other two numbers. So if I swap the position of 3 and 4 (to put 4 onto the dash), then I must swap 1 and 2 to make sure the molecule is exactly the same. If I only swapped 3 and 4 without changing 1 and 2, the molecule would be different:




Note that although products obtained by the mental rotation and the double displacement above look different, they are indeed identical (check your model kit if you’d like to confirm).


c. Now, look at how groups 1, 2, and 3 are arranged. Note the direction (clockwise versus counterclockwise) you must take to count from 1 to 2 to 3.



Keep it Simple

As a time saving tip, if the 4th priority group is a wedge, then you can simply solve whether the molecule is in the R vs S  configuration (with the 4th group as the wedge), and flip the result you get (if you get R, then the answer’s S and vice versa).