in this organic chemistry tutorial we're going to be talking about the absolute configuration of chirality centers also known as r and s configurations here are the three things you'll learn by the end of this video you're going to learn what the absolute configuration is and why you need it you're going to learn the three steps to assigning r or s to a chirality center and finally you're going to learn what to do if your fourth priority group is not pointing away from you and if you don't know what this means you'll learn by the time we get there taking notes is going to help all of this stick in your brain so grab a paper and pen or whatever you like to take notes with and if you'd like a copy of the slides that i'm using just click the link below and while you're at the website make sure you check out all of the other organic chemistry resources that i've got on there when a molecule has a chirality center which is a tetrahedral atom with four different groups connected to it we can describe the exact arrangement of those atoms around the chirality center using what's called the absolute configuration the two possible labels for the absolute configuration are the r configuration and the s configuration those are the names for the two possible mirror images that a chirality center can have this system is useful because it helps us to distinguish different stereoisomers and give them different names so for example the chirality center on the left hand molecule has the r configuration and the chirality center on the right hand molecule has the s configuration these labels get incorporated into the iupac names for the molecules so the one on the left is called r2 chlorobutane and the one on the right is s to chlorobutane the r and s configurations are assigned using a set of steps which we're going to talk about right now the first step is to assign priorities to the four substituents around the chirality center and we do this using the khan ingold prelog rules the four substituents are ranked from high priority number one to low priority number four based on their atomic numbers atoms with higher atomic numbers have higher priority if you're comparing two isotopes of the same atom the one with the higher atomic mass is going to be the one with the higher priority in this molecule the substituents are a methyl group a chlorine a hydrogen and a fluorine atom the atom with the highest atomic number connected directly to the chirality center is the chlorine atom atomic number 17 so it's gonna get priority number one next is fluorine it gets second priority the carbon of the methyl group is third it has atomic number six and finally number four is the hydrogen atom which only has an atomic number of one so that's our lowest or fourth priority atom sometimes all four groups on the chirality center are different but two of the directly connected atoms are the same so in this example the chirality center has a hydrogen and a fluorine as well as a methyl and an ethyl group fluorine has the highest atomic number of those directly connected atoms and so it's priority number one and hydrogen has the lowest priority number four but how are we going to decide between the two carbon atoms when we have two of the same atom we're going to write out a little list of all the connected atoms for the two that we are comparing between and we're going to make sure that list is in order from highest atomic number to lowest for this methyl group carbon atom the connected atoms are h h and h hydrogen hydrogen hydrogen and for the ethyl group carbon we have c h and h one carbon and the other two things that it's connected to our hydrogens then we're going to go down both lists until we find the first point of difference in this case it's just the first atom so carbon ranks higher than hydrogen therefore the ethyl group carbon ranks higher than the methyl group carbon and so we're going to give ethyl priority 2 and the methyl priority 3 respectively if you apply the rule from the previous slide and you still don't have a winner you just repeat the process one bond away so let me show you an example in this ring there's one chirality center the four substituents are hydrogen a fluorine and the two ring carbon atoms and because the ring is asymmetrical those carbons are considered different groups hydrogen is going to rank number four and fluorine is going to rank number one just like the previous examples the two carbon atoms that we're comparing are both connected to a carbon hydrogen and hydrogen c h and h so our two lists are the same that means we have to move down one more bond now we're comparing these two carbons and this is a situation that we haven't run into before we have a double bond when you have a double bond you consider it as two single bonds to the same atom so for example this carbon we would consider it bonded to a carbon a carbon and a hydrogen and this other carbon we would consider it to be bonded to a carbon a carbon and then another carbon so this carbon that's connected to three carbons is the winner and because it came from this side of the ring we would make this carbon priority two and then this one priority three once you have all your substituents ranked from one to four it's time for the next step the second step in assigning the r or s configuration is to view the molecule with the fourth priority group which is usually a hydrogen atom pointing away from you usually this means that you are looking at the molecule so that the hydrogen or the fourth priority group is on a dashed bond that means it's pointing away from you the viewer and into the page we want to be able to look at our one two and three priority groups as if they make a steering wheel and the bond between the chirality center carbon and the fourth priority group is like the steering column pointing away from us so when you're checking out your molecule for its orientation look for that steering wheel with number four in the back for now we're going to move on to the next step and we're only going to use examples where the fourth priority group is on a dashed bond but later i'm gonna give you a ton of strategies to use for cases where you don't have the fourth priority group pointing away from you with the fourth priority group pointing away from us we can then draw a circle tracing through groups one through three on this molecule i would start at the number one and then draw my circle through two and three and here on the enantiomer the mirror image of that first molecule we would also start at one and then the circle goes the other way through two and three notice how those circles were traced in opposite directions this is our differentiator this is the key difference between these two enantiomers if the circle that you traced is clockwise the configuration is r and if the circle that you traced is counterclockwise the configuration is s an easy way to remember this is that from the top clockwise goes to the right so r for right and for the other configuration the s from the top it goes to the left and the latin word for left starts with s that's why it's s and not l so that one's a little more complicated but if you can remember r for right from the top then you will remember how to assign the configuration let's try putting all of these steps together for an example here's a molecule and this molecule has a chirality center at the top there's a hydrogen and implicit hydrogen on this carbon which means that there are four different groups attached to that carbon a methyl hydrogen and then two unequal sides of a ring so first let's rank those groups we knew that hydrogen will be the lowest it'll be fourth and then we have to compare the three other carbon atoms in order to rank them the top carbon on the methyl group is connected to h h and h the carbon on the bottom right is connected to a carbon and then two hydrogens and the carbon on the bottom left is connected to another carbon by a double bond so we count that as two carbons and one hydrogen c c and h so being connected to two carbon atoms gives that bottom left carbon a higher priority so it's going to be number one the bottom right carbon will be number two and then the top methyl carbon will be number three now that we have our priorities let's check that the fourth priority group is pointing away from us and yes it is the hydrogen is on a dashed bond so now we're gonna begin at priority number one and trace a circle through two and three and we ended up with a counterclockwise circle so that means the configuration of the chirality center is s so that example was great because our fourth priority group our hydrogen was already pointing away from us on that dashed bond but on an exam your professor might not be so nice you need to know what to do if you come across a molecule where the fourth priority group isn't drawn on the dashed bond that hydrogen or that fourth priority might be pointing towards you or it might be in the plane of the page and you need some strategies to help you solve those problems really quickly and be able to assign rns no matter which way the molecule is drawn so here's a few of those strategies first and i'd only recommend this if your molecule is very simple you can imagine rotating your molecule and redrawing it to put the fourth priority group in the back for example we could draw an axis right through the chirality center like this and then we would do a 180 degree rotation so that would switch the sides of the methyl and ethyl groups and it would bring the hydrogen to the back and move the oh to the front then we could rank the other groups oh would be one ethyl would be two and methyl3 so this chirality center would have the s configuration another possibility is using the double swap trick because swapping two groups on a chirality center generates the enantiomer doing a second swap will regenerate the original molecule and you can use this to your advantage to get the fourth priority group pointing away from you on a dashed bond in this example we're going to swap the hydrogen and the fluorine and also swap the ethyl and the chlorine so these two are drawn in different perspectives but they are the same molecule and now to assign that configuration chlorine would be number one fluorine two and ethyl three and so this chirality center has the r configuration this is my favorite method to use when you have a hydrogen or a fourth priority group that is in the plane of the page that means it's not on a dash or a wedge it's just on a flat bond in line with the page itself just like on this molecule below so here i'm going to imagine what i would see if i was standing within the page looking straight down the bond from the chirality center to the hydrogen which is pointing away from me so if i'm standing there in the page i would see a chlorine atom down into my right i would see a fluorine down into my left so chlorine down to my right flooring down to my left and then straight up above my head i would see that ethyl group so i'm going to imagine those three groups as if they're on my steering wheel i've got that steering column between the chirality center and hydrogen pointing away from me so if i traced that circle from group 1 to group 2 to group three in the view that i am standing in that circle would be clockwise and so this chirality center would have the r configuration finally when the hydrogen or the fourth priority group is pointing to the front so on a wedged bond you can do two things you can either swap the circle or you can swap the rule i'm going to teach both and just use the one that you are more comfortable with if the hydrogen is on the front so on a wedged bond pointing towards you and only if this is the case you can draw the circle from three to two to one instead of one to two to three and then apply the rs label as usual r for clockwise s for counterclockwise i like to do it this way because that keeps the r to the right intact so in this case doing the circle from three to two to one would give us the r configuration we could instead draw our circle as usual from one to two to three and then remember to swap the rule so then by swapping the rule we would get s for clockwise and r for counterclockwise only when the hydrogen is pointing towards you of course if we do this for this same molecule we do the swap the rule method we would also get r so here's our first practice problem this is asking us to assign the absolute or s configuration to each chirality center this molecule has two chirality centers this one up on top has an implicit hydrogen we'll just draw it in on the dashed bond but we're going gonna look at the other carbon first so we know that hydrogen is gonna be priority four because it's the lowest atomic number and bromine will be number one and then comparing between the two carbons this one's attached to three hydrogens and this one is attached to three carbons two from the double bond and one from the single bond so that carbon is priority two and the methyl will be priority three and now to view our molecule we're going to put ourselves in the page so i'm going to draw myself here and i'm going to draw what i'd see if i'm looking down the bond between the chirality center carbon and the fourth priority hydrogen atom just like this so i would see to the top sort of the rest of the molecule we'll use r down into my right of bromine and down into my left the methyl group so i'll add those priorities on so i can draw my circle and then the circle is going counterclockwise so this is an s configuration on the other chirality center hydrogen will rank four nitrogen will rank one this carbon is attached to ch and h and this one to c c and h c c and h is better than c h so the double bonded carbon will get priority two and this one over here will get priority three we've already got a hydrogen in the back so our circle can be drawn it's counterclockwise so this is also an s configuration okay here's practice problem number two we're going to assign r and s to each chirality center in this molecule this molecule also has two chirality centers here and here both of those chirality centers have an implicit hydrogen so let's draw those in on this one that hydrogen is on a wedge so it's pointing out towards us and on the other one on a dash so that means it's pointing away from us we'll start with the left-hand carbon here o-h is going to rank one and hydrogen is going to be the fourth priority and then this carbon here is connected to ch and h and this one is to h so the ch and h is better it's 2 and hh is 3. hydrogen's pointing towards us so i'm going to draw my circle the opposite direction from three to two to one and so i get clockwise which is r because i used to swap the circle method instead i could have used the swap the rule method so drawn my circle normally one two three and then a counter class counterclockwise circle is s but we change it to r for swap the rule so in both cases you get r so the other carbon has fourth priority on hydrogen this carbon is connected to o o and c this carbon connected to ch and h and that one connected to c l h and h so chlorine is the highest atomic number of the first connected atoms comparing with oxygen and carbon so that one gets first that one gets second and this one gets third and hydrogen's pointing away already so we can just make our regular circle one two three counterclockwise so this one's s if you've made it all the way to the end it means you're putting in time and effort for your organic chemistry studying so good for you i have a feeling you're going to do well and if you want to help even more people find this video i would love it if you could give it a like and if you like this style of tutorial feel free to check out my website i've got lots more resources on there including more r and s practice problems see you in the next one happy studying