assigning r and s in this lesson we're going to talk about what are termed absolute configurations now these are a designation given to a chiral center now this is the second lesson in a whole chapter on isomers and stereochemistry in the first chapter part of what we learned was what a chiral center was and how to recognize them in a molecule and that they exist in two different forms that are mirror images of each other in this lesson we're going to learn that there's a system of rules called the khan ingold prelog system and they help us give a designation to those two different forms of a chiral center as either r or s so that we can distinguish them one from another now if this is your first time joining me my name is chad and welcome to chad's prep my goal here for the channel is to help make science both understandable and maybe even enjoyable now this is part of my new organic chemistry playlist i'll be releasing these lessons weekly throughout the 2020-21 school year so if you don't want to miss one subscribe to the channel click the bell notifications you'll be notified every time i post a lesson all right so let's get right into this here so absolute configurations assigning rns we'll look at some chiral centers here and as a reminder chiral centers have four different groups attached and so in this case we can see one right here this carbon right here is bonded to a chlorine it's also bonded to a hydrogen that's not drawn in that would be a dashed bond and that hydrogen is technically right behind that wedge bond the dash is always right behind the wedge so so again this chiral center right here this carbon is bonded to a chlorine a hydrogen then it's bonded to the carbon of a methyl group and the carbon of an ethyl group so those carbons are definitely different now the first part of your con angled prelog system says you're going to give the four things that chiral centers attach to a designation of or a priority we should say of one two three and four and you first start those priorities by uh according to atomic number here so out of hydrogen chlorine carbon and carbon chlorine's got the highest atomic number so we're gonna give him a designation of one hydrant's got the low atomic number we're gonna give him a designation of four and from there if you've got a tie like we do with these two carbons then you'll take a deeper look and see what are those carbons bonded to so in this case they both have a bond coming from the chiral center itself we don't recount that so but then you say what other three atoms are we bonded to well this one right here is one of the three hydrogens and so we just list those kind of hhh now the next one here it's bonded to a carbon right here and then it's bonded to these two hydrogens right here and so in this case it's a carbon and two hydrogens and you want to list those in descending order of atomic number so carbon first and then hydrogen and then hydrogen and so when comparing these two carbons right here i'm actually comparing these three to these three but you're not actually comparing them all at the same time you're comparing them until you find the first point where they're different and the first place where the different carbon beats hydrogen once again according to atomic number and so as a result we'd give this carbon right here a designation of two this one a designation of three and again the hydrogen being the low atomic number got the designation of four now from there you're then going to make a circle from one to two to three now some people say go one two to three to four but don't really worry about four just one to two to three is suffice it to say and in this case we're going to go around and we can see that this is clockwise now i don't like saying clockwise i like saying a right-handed turn like if you were turning a steering wheel this would be a right-handed turn and that's actually where r and s come from so r comes from the latin word rectus which means right so s comes from latin word sinister which means left and so a right-handed turn is going to go with r and a left-handed turn is going to go with s as we'll see here so but they had to give us just a little more than just that so because it depends on which side of the molecule you're looking at so notice if i make a i'm going to put my hands up do me a favor right on the other side of the camera put your hands up so i'm waiting for you come on yep there you go put your hands up now i'm going to make a right-handed turn so and if you're putting your hands up and matching my hands look and you should be making a left-handed turn but notice again i was making a right-handed turn and so they had to give you some perspective here they couldn't just tell you make a right or left-handed turn they had to tell you which side of the molecule you're looking on because in this case if i look at this molecule from this side it looks like a right hand turn but if i could crawl in behind the board and look at the molecule from the other side it would actually end up looking like a left-handed turn so going around this way instead and so what they said is that when you look at your turn whether it be left hand or right-handed your number four priority group has to be going away from you in the dashed position that's kind of the designation they gave and so this is indeed a right-handed turn and as long as your number four party is in the back going away from you then right-handed turn is going to mean r cool so what if your lowest priority group is not in the dashed position and that's we're going to deal with in this next example so in the in this next example chlorine is still going to be number one so this carbon's still going to be number two and this carbon is still going to be number three and once again that hydrogen that's not drawn in is going to be right there and that's going to be priority number four and so now we have a problem i make my circle from one to two to three and for me it still looks like a right-handed turn and it should look like a right hand turn to you the problem is we're looking at this molecule from the wrong side they said no no no your number four priority is not going away from you it's coming out towards you what you need to do is actually flip this molecule over so that that hydrogen is actually going away from you well we're not actually going to flip it over what we're going to do is just take advantage of the fact that what we just learned that if from this perspective which is exactly 180 degrees the wrong perspective it looks like a right-handed turn then if actually again i crawled in behind the board and was looking at it from the other side that hydrant would be going away from me and it would look like a left-handed turn and so the idea is this then if your number four priority rather than being a dash where it's supposed to be if it's a wedged bond it's just opposite of the way it looks so it looks like a right-handed turn from this side which is the wrong side from the correct side it would have looked like a left-handed turn and this corresponds to the s isomer instead okay so last possibility though and this is the worst one of them is what if your lowest priority group is not a dashed bond not a wedge bond but it's one of the two bonds that's in the plane of your paper in this case in the plane of the board so well that kind of sucks and i'm gonna warn you it's the worst case scenario and it's the one that students struggle with the most and there's a couple of different approaches for how to look at this now i'm gonna give you the approach that i look at it with and then i'm gonna recommend that you don't look at it that way just yet but the way i look at this is instead of looking at this head on so i go through assign my priorities one two three and four so same same notice it's the same groups we had before these other two examples and what i do is instead of looking at it head on i come off to the side here and look at it from the side with the hydrogen going away from me and so as a result the chlorine here on a wedge bond is out over here out in front of the board so number two is up in the plane of the board and then number three is back behind the board down below at that methyl group so one two three one two three one two three one two three and that's a left-handed turn for me with the h going away from me and that means this is going to be s so however students often struggle with the three-dimensionality of this early on in ocam and so as a result maybe this is not the best way now for students who have had a year of organic chemistry and coming back and reviewing some organic chemistry this is actually what i recommend they do so but for sophomores who are taking and experiencing organic chemistry for the first time this might not be the best approach so we're going to take a little bit different view of this now one of the things we can do is we could take the other bond that's in the plane and just kind of rotate that bond around until this hydrogen was either on the dash or the wedge and that's one way to approach and it's commonly taught that way but students screw that up all the time so i'm actually not going to teach that way either but you can do that as long as you're just rotating a bond it's the same molecule but what i'm going to do and recommend is something a little bit different now in the last lesson we learned that a chiral center is an example of a stereocenter if you take any two groups and you trade them places it you get the opposite version of that chiral center so well in this case that's going to mean that r turns to s and s turns to r and so what i'm going to do is have the hydrogen trade places with another group and technically i could have a trade place with either the wedge or the dash and and i put it in a configuration where i could assign it like one of these two so but usually i just trade it with the dash and go from there so in this case what we're going to do is we're going to leave the ethyl group alone and we're going to leave the wedged chlorine alone so however i'm gonna where the hydrogen used to be that's where i'm gonna put the methyl group and where the methyl group used to be that's where i'm gonna put the hydrogen now we have to keep in mind here that these are not the same molecule if you have any two groups exactly two groups trade place on a chiral center it automatically inverts it and it's the opposite form so these are not the same they're opposites and again this one's still hard to assign but this one's not that number four priorities in the back that's easy to assign and so chlorine is number one carbon's number two this methyl group is number three and one two three hydrogen's four so but that's a right handed turn and this is r and if having substituted you know having two group swap places gives me the inverted stereochemistry where it's r then the original just like we already had said must indeed be s so that's kind of the way i look at it so again i didn't rotate a bond here if i if all you do is rotate a bond you still have the same molecule i didn't rotate a bond i had two groups and i detached them and then reattached them in the opposite places and again if you have two groups swap places on a chiral center you get the opposite version but once again it was really easy to assign this one right-handed turn lowest priority in the back it was r which means that the original must have been s cool if you've already got this down in one of the other two ways fantastic and again looking at it from a different perspective so you can get the number four priority going away from you really is the fastest way but again this last one i find is just the one that students tend to screw up the least when taking organic chemistry as sophomores all right so now we're going to look at a couple of little more difficult examples and professors are notorious for going over rather simplistic examples in class and then asking much harder ones on the exam so i'm gonna try and ask some questions here that might be a little more representative of what you might encounter on an exam here in your organic chemistry course so in this case this first one's definitely got a chiral center right here it's the only chiral center in the molecule so and that chiral carbon right there is bonded to four different things now three to those four things are all carbon atoms so but not drawn in here there is a fourth bond in this case the fourth group is a hydrogen and then comparing carbon to hydrogen the hydrogen definitely is going to have the lowest priority so that's going to be priority number four now you'll find that a lot of organic molecules your lowest priority is a hydrogen and a lot of students get in the habit of being like well just make sure the you know find your hydrogen well it doesn't have to be hydrogen just whatever your lowest priority is whether it's height or not so that's kind of again how they define things and when it's in the back that's when a right-handed turn means are and a left-handed turn means s so life is good here getting your having your your number four priority already be on the dashed position that's where we want it but assigning priority for the others we got a three-way tie with carbon-carbon and carbon in this case similar to what we've done before we'll then look at what three additional atoms are these carbons bonded to well this carbon right here is bonded to this carbon he's also bonded to these two hydrogens and so we'll list this as carbon hydrogen hydrogen listed in descending order of atomic number so for this carbon down here he's bonded to the sulfur and then he also is bonded to a couple of hydrogens not drawn in and so we'll list this as sulfur hydrogen hydrogen and then finally this carbon right here he's got a double bond to an oxygen here a single bond to an oxygen here but he's not bonded to three additional atoms he's only bonded to two additional atoms and so to take into this into account what you do when you've got a double bond is you actually count each bond separately so this counts as one bond oxygen this counts as a second bond auction and then we have yet another bond to an oxygen and so we'll list this out as oxygen oxygen oxygen those first two oxygens being accounted for by the fact that it's a double bond to a single oxygen this way we're always comparing three things now we get here and we can see that there are some differences here and a lot of students will ask me hey chad what's better three oxygens or a sulfur and two hydrogens who's going to get the higher priority so and that's not the right question the con angled pre-log system says when you're going to compare this now you just look for the first place where they're different and so what i'm really comparing here is oxygen to sulfur and then on this one to carbon and it's there's no ties anymore and so in this case sulfur's got the highest atomic number out of those three and so he's going to cause the carbon he was attached to to be priority number one so oxygen is the second highest atomic number so it's going to cause the carbon he was attached to to be priority number two and then finally the lowest of the three is the carbon and so the carbon he's attached to is gonna be number three and then we'll make our circle one two three so in this case that is a left-handed turn and a left-handed turn when your number four priority in the back corresponds to s so this is the s isomer here or we'd say that this molecule is in the s configuration or this chiral centers in the s configuration cool so got one trickier example out of the way and this next one's gonna get even a little trickier from there so we take a look at this one here we've got a single chiral center in this molecule as well and that's right there and in this case just like the last one that chiral carbon is bonded to three other carbon atoms and then he's also bonded to a hydrogen and that hydrogen will infer must be to a wedge bond and that hydrogen once again is going to be priority number four so and once again your number four priority doesn't have to be hydrant it often will but it doesn't have to be so but wherever it is we can make some determinations here because it's on a wedged bond that means if you see a right-handed turn from this side that really corresponds to a left-handed turn if you crawled in behind the board and so a right-handed turn from this perspective with your number 4 as a wedge doesn't mean r it means s and a left-handed turn would mean r so not s cool so just want to kind of put that out there before we start here but assigning priorities for these three carbons now is going to be a little bit of a pain in the butt so i'm going to start with this bottom one because he's kind of the easiest and in addition to being monitored this carbon he's bonded to these two hydrogens and so we're just going to list carbon hydrogen hydrogen again a descending order of atomic number now this carbon right here is bonded to one hydrogen and then he's got a double bond to that carbon right there and so with a double bond to that carbon we'd count it as one bond of carbon second bond to carbon bond to hydrogen and so we'd list him out as carbon carbon hydrogen where these two carbons are actually both that carbon right there all right so then we move over to this carbon right here and he's also bonded to one hydrogen but he's bonded to this carbon and he's bonded to this carbon as well and so we'll list him out as carbon carbon hydrogen as well and we see one point of difference so these two there's no point of difference but these two compared to this one we're really comparing the hydrogen here to the carbon right here and so this hydrogen is definitely going to lose if you will and so we knew these were going to be priorities one two and three and we just figured out that therefore he's going to have to be at least the carbon he's attached to is going to be priority number three but we've still got a tie between these two carbons even based on what they're bonded to so then we have to go further in the chain and say well then what are these carbons bonded to like this one is represents these two or these two i guess represent this one so and then these two carbons are these two and say well what are they bonded to as well let's do the one on the right first so this one right here this carbon right here let's draw this in with yet another color is bonded to two hydrogens and that is bonded to this carbon and so we list it out as carbon hydrogen hydrogen and then this carbon right here is bonded to two hydrogens and it's also bonded to that same carbon as well and so its three things are still going to be carbon hydrogen hydrogen okay so taking this a little further then so this carbon right here again is both of these carbons so and things are going to get a little bit funny so he's bonded to two hydrogens but he's not bonded to anything else at all so what do we do because there's got to be three things and so what they do in this case is they count the pi bond back to the previous atom and so we'll actually count a bond back to the previous carbon as part of that pi bond so instead of just saying h it'll actually be along with this carbon yet again and so this is going to be c h h and this one also is going to be chh because they're again representing both that carbon and lo and behold this is not looking good because we still have a tie so now we have to do the green carbons well in this case the green carbon was the previous one that was blue and it's done he's already been taken care of we have nowhere else to go with this we're done assigning designations you can't go back up the chain you'd have to continue further out so there's nothing else to do here but for this one we're now looking at the green carbon and the green carbon here is bonded to a couple of hydrogens and then depending on who you talk to is bonded to one other carbon so when we're coming from this carbon over here out to here he was bonded to this carbon and these two h's so and when we were coming from this carbon out to here he was bonded to this carbon in two h's and so regardless of which one you look at here he's gonna be bonded to a carbon and two h's so we had further designations to go we have nothing to compare that to and they therefore these are definitely going to be nothing to compare it to and so as a result took all this work to finally realize that this was priority number one and this was priority number two and then we'll make our circle one two three and we can see that once again this is a left-handed turn and as long as your four priorities in the back a left-handed term means s but in our case our number four priority is the wedge bond that's coming out of the board and so a left-handed turn is going to mean r instead we're looking at the molecule from the wrong side and if we crawled in behind the board what looks to us like a left-hand turn from this side would look like a right-handed turn from the other side cool so these are a couple of the trickier examples i could give you so and again this might be more representative of some of the harder ones you might see on an exam so now we're going to take a brief look at naming molecules that have chiral centers and first thing you want to do is just kind of name it as if you didn't know it had any chiral centers and things of this sort and we'll find out we just have to add a designation to it at the beginning of the name so in this case longest continuous chain here would be the one the only carbon chain all the carbons essentially and we'd number it from right to left so and we'd see that we've got a chlorine at position two and the parent chain is gonna be called hexane for a six carbon chain and so we would name this as two chloro hexane now the problem is that name is not sufficient because while that is indeed two chlorohexane so is that that's also two chlorohexane but these are not the same molecule they're enantiomers and so somehow we can't give two molecules exactly the same name as they're not the same molecule and so that's why we got to come up with a way of including the r and the s in the name so let's get him out of here so in this case we'll go through and assign priorities once again and for the chiral center that is right here he's the guy who bonded to four different groups chlorine's number one the hydrogen that's in the back will definitely be number four and then between these two carbons this carbon's bonded to three h's this carbon right here is going to be bonded to this carbon and two h's and we see that carbon is going to be hydrogen and so this will be priority number two this will be priority number three and if we go from one to two to three we'll see that that's a right-handed turn and a right-handed turn when your four priorities in the back corresponds to r so we know this is the r configuration and how do we put that in the name well that's always going to go at the beginning of the name and it's going to go at the beginning of the name in parentheses and so here we'll just say r 2-chlorohexane so when you have just a single chiral center of your molecule just put rs in parentheses at the beginning of the name done now if you've got more than one chiral center not only will you have to stick the designation at the beginning but you'll also have to give me the carbon location where it's located assuming in most cases in pretty much every case you're likely to see it's going to be on a carbon so once again if we numbered this it would be one two three four five six and so notice this is my numbering for naming it not any kind of priorities or anything like this and so in this case we would call this two chloro alphabetical order here if you recall so two let's save myself a little room i'll drop down here so two chloro three methyl and then the parent chain is just hexane and once again we're going to save some room for some parentheses at the beginning but this molecules actually got two chiral centers so it's got the one at two here with the chlorine on it it's got four different groups but this one also is going to be and so we'll assign priorities separately here so for this carbon right here chlorine is definitely number one there once again is a hydrogen attached that's definitely going to be number four and between these two carbons this one's bonded to three h's this one right here is bonded to a carbon up here a carbon over here and then one h that's not drawn in so it'd be c c h and he's definitely going to win then so one and then two three and so if we go around our circle i can see that this is a right-handed turn and it's with the number four priority in the back which means it is r okay so so far so good and rather than trying to muck up this diagram even further i'm just going to redraw it one more time so we can assign the other chiral center and so our other chiral center is this one right here he's bonded to four different groups as well and this one's going to be a little more challenging now he is bonded to a hydrogen and that hydrogen is on a wedged bond but he's going to be priority number four because the other three atoms are all carbons carbon carbon carbon it'll be a three-way tie but hydrogen's definitely gonna be priority number four and once again because it's a wedged bond now it'll be opposite of normal so a right-handed turn would actually correspond to s and a left-handed term would correspond to r because we're looking at the molecule from the wrong side all right to compare these three carbons so this carbon over here is bonded to a chlorine a carbon and then also to one hydrogen so a chlorine a carbon and then to that hydrogen right there so we list it as chlorine carbon hydrogen so this carbon over here in addition to being bonded to those two hydrogens is bonded to a carbon as well and so we list it as carbon hydrogen hydrogen and then finally obviously that carbon right there is just bonded to all three of the h's of that methyl group so h h h and we see the point of difference here and that point of difference chlorine is going to be carbon which is going to be hydrogen and so as a result the carbon the chlorine is bonded to is going to be number one this carbon over here is going to be number two and this carbon up here is going to be number three and as we go from one to two to three it's a right-handed turn and once again because our four priority is on the wedge instead of a right-handed term meaning r it's actually going to mean s so now when we go to our designations a lot of students just want to do this they want to come out here and be like well there it is it's rs or sr whatever so well it turns out that's not good enough because then i don't know which one is which and things of a sort and so what you do out here is kind of put them in numerical order and in this case the chiral center with the chlorine is at position two so on our parent chain and so we're going to say and he gave a designation of r so we're going to say 2r and then you put a comma so and then the chiral center at that position 3 ends up being s and so then you'll say 3 s cool and we only have to worry about this when you have more than one chiral center if you only have one chiral center don't include the number just say r or just say s it's when you have multiple designations to give out here that you gotta start including locations where it just gets confusing on which one you're talking about and so this is 2r 3s 2-chloro 3-methyl hexane now if you found this lesson helpful consider giving me a like and a share or if you just feel sorry for like short bald guys consider giving me a like and a share for that as well if you've got questions feel free to leave them in the comment section i'm on there pretty religiously and if you're looking for practice problems or you want the study guides that go with this check out my premium course on chatsprep.com