In Part 1, we identified the diagrams below as representing enantiomers. Now we should find out how to name these two enantiomers so we don’t get them confused.
They’re both called amino(chloro)methanol. But the one on the right is known as (S)-amino(chloro)methanol, while the one on the left is (R)-amino(chloro)methanol. Each stereocenter can be labeled as R or S according to a system known as the Cahn-Ingold-Prelog rules.
We start by identifying each of the substituent atoms connected to the chiral center. The higher the atomic number of the atom, the higher the priority of the substituent group. Thus chlorine, with a an atomic number of 17, will have the highest priority of the four substituents and get labeled #1. Oxygen’s atomic number is 8. so it will have the second highest priority and be labeled #2, and nitrogen’s is 7 so it will be #3. Since hydrogen’s atomic number is 1, it will always get the lowest priority and be labeled #4. Now, we need to look at the substituent with the dashed bond. Remember, this means it’s positioned behind the solid wedge substituent.
If the dashed bond goes to substituent #4, simply draw a curved path with an arrow indicating the direction (clockwise or counter-clockwise) from substituent #1, to #2, and then to #3. Ignore substituent #4. If the path follows a counter-clockwise direction, as is the case with the diagram on the right, the enantiomer can be given the (S) designation. You may also see (S) enantiomers referred to as ‘left-handed.’ If the path instead follows a clockwise direction, it can be given the (R) designation and will sometimes be called ‘right-handed.’
If substituent #4 is in front, you will still identify the direction of the path following substituents #1 to #2 to #3. Remember, the dashed line indicates that the substituent is positioned behind the chiral center, while the dash lined means the substituent is facing the viewer. It’s counter-clockwise in the right diagram just like in the left diagram. But because the dashed line is not connected to substituent #4, you will invert the direction. Since the path was counter-clockwise in the left diagram, it will be given the designation of a clockwise path: (R).
Why is this? Well, you can draw the same enantiomer in different ways. Try manipulating the 3D models and checking out the images below. The green chlorine atom is positioned in front of the chiral center when the dark, wedged line is attached to it in the diagram. But notice how the positions of the nitrogen atom (blue) and oxygen atom (red) vary in each enantiomer even when the chlorine atom is in the same position.
Now let’s try assigning (R) or (S) based on the 2-ethyl-2-methylpentanoic acid diagram below. We’re not going to look at the longest chain like we would when deciding its IUPAC name. Instead, we’re going to look at the chiral center (blue star) and treat every attachment like a unique group.
There are four different substituents attached to the chiral center here: methyl, ethyl, propyl, and carboxyl groups. We need to assign each group a priority, but in this case, the first atoms in each of the substituents (the one bonded to the chiral center) is carbon. So let’s start by drawing a C next to each.
Since we can’t assign priorities just by looking at the first atom in the substituent group, we’ll look at the second group of atoms in line and identify the atom in each group with the highest atomic number.
Let’s start with the carboxyl group. Moving outward from the first substituent atom, we see that the carbon is bonded to two oxygens. Oxygen is the only option, so we’ll add O to our previous C. Now looking at the methyl group, we see our first substituent, moving down the line, is bonded to 3 hydrogens. Thus hydrogen is our only choice, and we can add an H to our C. Starting from the first substituent carbon in our ethyl group and moving outward, there is one C and two Hs. This is the case for the propyl group as well. Carbon has a higher atomic number than hydrogen, so carbon wins out and we’ll add it to our ethyl and propyl chains.
Now we have three different types of chains: C-O, C-H, and C-C. Looking at the second atoms in each chain, we can see that O should get priority over C, and C should get priority over H. So we know that the carboxyl group will have first priority and the methyl group will have fourth. But we’re still left to decide whether the ethyl or propyl group will get second priority. So let’s continue down the line for these two groups.
The second atom in the ethyl chain, moving outward, is bonded to no other carbons and three hydrogens. So the chain will end as C-C-H. In the case of the propyl group, the second atom in line, moving outward, is connected to another carbon and two hydrogens. Since carbon has a greater atomic number than H, the chain will become C-C-C, and it could continue, if necessary, to become C-C-C-H. But judging just from those third atoms in the chain, we can see that the propyl group takes priority over the ethyl group, and our final priority order is: 1. carboxyl, 2. propyl, 3. ethyl, and 4. methyl.
Lastly, will our diagram represent (R)-2-ethyl-2-methylpentanoic acid or (S)-2-ethyl-2-methylpentanoic acid? Try and guess yourself, and when you’re ready click the button below to reveal.