Lecture 15. Assigning Tetrahedral Stereogenic Centers
Nowick, James S.
Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 Good morning. Good morning. So today I'd like to actually pick up on the question that was asked by the young woman over there in the previous lecture on how we assign and how we name our tetra-egyl-stereogenic center. And there's a nomenclature system called the nomenclature that basically allows us to assign what's called right-handed or left-handed stereochemistry. That's the convention that everyone knows of it. As long as we're all on the same page, that we all have the same way of talking about the molecule in the stereochemistry. Just like if everyone calls a three-carbon alkane butane or butane instead of prokane. So we're going to learn this nomenclature system today. We're also going to get more familiar with molecules with multiple tetra-egyl-pacemetric centers. Alright, the basic nomenclature is really very simple. Nomenclature consists of R and S to designate the stereochemistry. If you think of R, its Latin term is rectus, which means right. And we're going to learn that that also means clockwise. The left-handed notation is from Latin sinister. And we're going to learn that as synonymous with counter-clockwise. Alright, what I'm going to try to do just as in nomenclature naming alveoli is to do is to do the same thing as nomenclature naming alkanes and various other compounds. There's a series of rules to follow. I'm going to try to take us through those rules and I'm also going to make it as simple as possible. Now there's a lot of neat to stereochemistry. For those of you who have been seeing, there's some fun, there's some three-dimensionality. It's really a deep topic. I have a book that's about 1400 pages just on stereochemistry. And so one of the things that I want to do is to digest the topic and break it just at a simple level into some small, light-sized pieces. So come along with me on this journey. Alright. So, first thing we need to do is identify the tetrahedral stereogenic center. That's pretty easy. So usually at carbon atom, we have four different things on it. We start with the word identical. I am. Different. Different. Different. Different. Different. And then what we need to do is to rank the substituents. Alright. Let's take an example and we're going to take the two butanol that we used before as an example. So I'll rank that again. So pretty obviously the carbon with the hydrogen, the hydroxyl group, the methyl group, and the ethyl group is the carbon, the only carbon in two butanol that has four different substituents on it. So this is A, and in this case the only tetrahedral stereogenic center. Alright. Now what we need to do is to figure out the priorities of the substituents. The first way we rank priority, the most important way is by atomic number. Highest atomic number gets you the highest priority. The lowest atomic number gets you the lowest priority. Kind of makes sense. You're looking at the periodic table. Oxygen is at 8, carbon is at 6, hydrogen is at 4. So for priorities, it's pretty easy to see that our OH, the hydrogen on here is going to be 1. I'll just put an arrow here, highest atomic number. And our hydrogen then is going to be 4. I'll put an arrow here, the lowest atomic number. Now we're faced with the situation of the metal and the ethyl group, and they're both carbon. So our first ranking is at the atoms attached, they're the same, they're both carbon. So what we do now is we move outward away from the tetrahedral stereogenic center. And we identify what's attached to those carbon. So both of these are carbon, all right, C and C here. But attached to this carbon is 3H is H, H, H. Attached to this carbon, we don't count this carbon anymore, we're just moving outward. Attached to this carbon, we're just moving outward. Attached to this carbon moving outward is another carbon in two hydrogens. So this one is C, H, and H. And now we just look, okay, they were both the same atomic number. But moving outward now, we have a higher atomic number, carbon 6, attached to this carbon. Then we do to this carbon. So this substituent, the ethyl group gets a ranking of 2 and the methyl group gets a ranking of 3. So the first rule we can say is higher atomic number, is higher atomic number, higher atomic number leads to higher priority. And realize that when you then are comparing atoms of the same atomic number, you just move out. And that rule is going to get you pretty far. Just like in nomenclature, there's a lot of stuff that can be really arcane. I'm going to try to take us through the simplest and most relevant stuff. Now the next operation once you've ranked the substituents is just to look down the bond from the carbon atom, from the tetrahedral stereogenic center to the lowest priority atom. Get going, everybody. Ok. All in all. Thanks for coming here. I'm going to do my own pump. And then what we do is determine the direction 1, 2, 3 going around as it clockwise or counterclockwise. And then we weigh this 1 parallel and weigh this 1 parallel n with x trades. Question? I'm still a little bit confused. How do you tell the difference in priority between the ethyl group and the methyl group? Ah, question. How do we tell the difference in priority between the methyl group and the ethyl group? Both the ethyl group and the methyl group are carbonated. So we know they're both at the same level there of their atomic number. So now we have to break the time. And to break the time, what we do is we move outward along the chain and say what's connected to each carbon? Connected to the methyl group is three hydrogens, all atomic number one. Connected to the ethyl group, we're just moving out of it so I'm not counting the tetra-needral asymmetric center, is a carbon, the terminal methyl group on the other side, and two hydrogens making up the methylene group. So now we say on the tiebreaker, the ethyl group has the carbon on the right, has a carbon attached to it, and that trumps hydrogen attached to the other. So that gives us an assignment of two on the priority of the ethyl group and three on the priority of the methyl group and, of course, group four on the hydrogen. Other questions? These are good. These are really important to be taking about this. Ah, what happens if you have the same groups on either side? So let us pretend for a moment that we had two ethyl groups. Somebody, what would we do in this case? It's not a stereogenic center. If they're the same group, then we don't have a tetra-needral stereogenic center. We need to have four different substituents. So in that case, we have a plane of symmetry in this molecule in three-pentadol, and as a result, we know immediately that we don't have a chiral molecule. Other questions? All right, next thing is visualization. What we want to do is site down, look down the bond to, from the carbon atom to the hydrogen atom in two-butadol. And so I'm going to draw things out, and here's where, if you're not yet proficient with visualizing, here's where models can really help you out. So here's our OH is one, our ethyl is two, our methyl is three, and I have this hydrogen going back. So we're looking down the bond to it, and I go one, two, three. Our direction is clockwise, and that means our tetra-needral stereogenic center is R. So we call this compound R2-butanol. We always, by convention, put the R in the parenthesis. Let me do this with some video projection, because as I said, we need to become proficient at this, and it takes a little bit of practice. Let's see if I have a model of R2-butanol. So I'm going to do this two ways. We'll take it slow. I'm going to show you how I do it, and then for people in the back, we'll do it on the video projection. So here's the model that I built. Please, if you have your models, pop yours together. No better time to get your exercise and visualizing it than right now. All right. So, hold up your models. I want to see more of them. Still a neighbor in auto-papery or a makeup friend. We just had a seminar speaker who was an undergraduate at UCI in 1995 or 1998. She's now a professor at the University of Toronto, and she was trying to decide whether to take the chemistry general, the Chem51 or Chem52 course. She ended up taking Chem52. Not only did she end up getting inspired to go on in chemistry, organic chemistry, because that was what she also met the man who became her husband in that class. So go to your neighbor. Go to your neighbor. Go to your model. And borrow a model. All right. So I'm going to do this here, and then I'm going to do it up on the video projector. So here's my two-butanol, and you notice we're looking down the CH bond. We go one for the oxygen, two for the ethyl, three for the metal that's clockwise. Let me put that up on the video as well, just to make it maybe a little easier to answer. No. Working from the lamp. All right. So, here we are. You notice, and this is the case, because if you don't site down the bond, you'll site in the wrong direction. You'll get it exactly backwards. So our tetrahedral stereogenic center is in front, where my finger is. The hypochism is the lowest priority substituent. Is it back where my finger is? We have the oxygen atom. We're getting a mirror projected here. Yes. Ah, right. All right. Yes. This is the one case where I think video ends up being a 1, 2. That's weird. Okay. We're going to stick to this on the model. So again, one for the oxygen. Everyone's looking down. Here's the oxygen. It's the same as on the blackboard. Here's the ethyl group. Here's the methyl group. And so, from your point of view, one, two, three is clockwise. So this is telling you where our hydrogen is. The question is, is this telling us which direction? Well, if you could take it to waves. Here, I've given us a molecule, and I've said which wave is it. If I said draw S, then you draw it in a particular wave. And the way I always do it when I'm asked to draw S or draw R is I go ahead and I simply draw one, because I don't know off the top of my head. Check it. And if I end up not getting it right, then the trick is you just swap the two substituents. So if I go ahead and simply swap two, not three, not one, but two substituents, then that's what we call inverting the tetrahedral stereogenic center, or inverting the chirality of the molecule. And now I know that this molecule is the opposite. So this one is S to butanol. Now you can do this a couple of waves. One way is you can simply do the rotation in your head. You can go ahead. I'm going to swap them on my plastic models. You can go ahead and do the rotation in your head and spin the molecule around. So in other words, I can go ahead and say, alright, I will just spin this around like so. And then when I've spun it around, I'm citing down the CH bond, one, two, three, it's counterclockwise. But the other way you can do it is you can say, well, let's draw in this way, but I know if I reverse things, if instead of citing down the bond to the lowest priority substituent, I'm citing from the bond with the lowest priority substituent in front instead of in back, then everything's reversed. So in other words, you look at this molecule as I'm holding it now with the hydrogen pointing right out at you and the oxygen and the methyl group and the methyl group. The way I deal with this is simply I say, well, it's all backwards. So when I go one, two, three, oxygen, methyl, methyl, I'm going clockwise, but we're citing in the wrong direction on the bond, which means in reverse it's counterclockwise and the steric chemistry is S. You can do it either way in three different things, all equivalent that we've done. One is we've gone and said we can rotate the molecule and simply make a new drawing. Another way is to say, well, we know we've got this molecule by simply swapping the oxygen and the hydrogen and we knew when it was like this, it was R because we were citing down the bond. So if we swap two substituents, we invert the stereocenter and we go from R to S. Or you can go ahead and say, well, I'm citing it backwards, so what should be clockwise is counterclockwise and what should be counterclockwise is clockwise. Questions or thoughts? And of course, always the fallback is build a plastic molecular model. And when you're learning, when you're starting to see these drawings on the blackboard for the first few times and they're not yet clear, that tangible model, that model you can hold in your hand, grasp the substituents, look down the bond, that is really the thing that's going to get these blackboard drawings into your brain. So let me write that take off. All right, so let me write that take home message here just so you can get it in your notes. Swapping two substituents. Swaps the stereochemistry and then let's try another example. All right, example I want to try right now is two chloro-tetrahydropyrene. And just like we did with two butanol, we're going to determine whether the drawing that I'm about to make is R or is S. We're going to assyllate the absolute stereochemistry. So tetrahydropyram, I've mentioned before, is a five-membered cyclic ether containing one oxygen. This is the tetrahydropyram ring structure. The oxygen gets numbered for one position, so the chlorine is at the two positions. That's our priority, that's just numbering the chain. So here's the structure, let me show you how I think about it. Now one way I think about it is to start by simply drawing in my hydrogen. Where am I drawing my hydrogen? What's the tetrahydro-stereogenic center? The carbon with the chlorine on it. And that hydrogen is pointing out because the chlorine is pointing back. Which substituent has the highest priority? Chlorate. So that gets us to sign number one. Which one is the second priority? Oxygen. Third priority? Carbon. Fourth priority? Hydrogen. Which way, if we look at the molecule as it's drawn, are we looking down the bond to the lowest priority substituent? No, it's reverse. Two things we can do. One is we can flip the molecule over. All we do is rotate it 180 degrees, I'm going to pick it up and flip it over here. Now the hydrogen is going back, we're citing down the bond to the lowest priority substituent. We can go one, two, three and say this is counterclockwise and this molecule is therefore S2-chlorotendrohydropyram. Another thing that we can do is simply make a physical model. Our oxygen is just the tetrahedral construction with carbon, with the red pieces just like carbon. The trick with the oxygen is those two extra sticks on it represent lone pairs. Chlorine by convention, since chlorine is green, chlorine is typically represented by a green ball. Chlorine is a yellow-green gas, of course very poisonous. You smell it with bleach or with swimming pool water. And you can put on your chlorine and make the model. And then you don't have to worry about any sort of drawing on the blackboard. You just say, well, okay, I'll just look down the bond. The bond is pointing back. Here's the chlorine, here's the oxygen, here's the carbon, we're counterclockwise. So you never have to worry about being able to visualize. And if you're proficient at this, what you can do is just keep it all in your head and say, oh, well, the hydrogen is pointing out, which means all the rules are backwards. So since the hydrogen is pointing out and we're going chlorine, oxygen, carbon, clockwise, that's really counterclockwise because everything's backwards. So counterclockwise means s. And the more you practice, the more you're going to be doing this automatically. But at this point, find something that's comfortable for you and practice. Whether it's starting with the models or physically doing the rotation on the blackboard or simply mentally keeping track of everything on the blackboard. All right. Couple of more little rules. So couple of more little rules. One of them is how we treat double bonds, triple bonds. And we're going to pretend something. All right. Here's an example where we have a double bond and they have an OH going out and a hydrogen going back. What we're going to do is pretend that this carbon here, the one I'm pointing to, the one of the double bond, instead of being a double bond is bonded to a second carbon and did own for this one. So what we're going to do is pretend that our structure is like this. All I've done is to go ahead and I'll draw on the hydrogens too. All I've done is to say a double bond counts as being bonded to not one carbon but two carbons. In other words, on this substituent then we say this substituent, instead of counting as having one carbon attached to it, counts as having two attached to it. So this carbon counts as having a carbon and a hydrogen attached to it. Whereas this substituent here counts as having a carbon, a hydrogen and a hydrogen attached to it. Does that make sense? So the idea for dealing with double bonds is simply just a way of keeping track. Say two bonds to another carbon counts as two carbons. I have it just in my mind just draw a second carbon but you can also say, alright, well we're going to count this carbon twice so we're going to say that this carbon, since it's got one bond to a carbon and another bond to a carbon, it counts as being attached to carbon, carbon and hydrogen. This carbon counts as being attached to one carbon and two hydrogen. And so our prioritization at this point is that the OH group is number one, the hydrogen is number four and by this convention the double bonded carbon now counts as number two and the single bonded carbon counts as three. So when we assign stereochemistry we're siding down the CH bond, we're going clockwise and so this molecule counts as being our stereochemistry. In other words, this molecule counts as R3 methyl 1Nt. Nomenclature here, we haven't learned yet. This is a nomenclature for alkanes, it says that the double bond is at the one position. Obviously the 3 methyl is just like our nomenclature for alkanes, it says the methyl group is, whoops, ox. Actually the example I was going to do for that was simply same issue here. So I'm going to skip the nomenclature of this molecule because this molecule becomes, pentene becomes one pentene if you want to know what becomes R1 pentene 3-ball. So the main point, don't worry about that nomenclature, the main point is for our stereochemistry. Alright, one more example with carbon-carbon bond, so we set a double bond, counts as having two carbons bound to a carbon. A carbon-carbon triple bond counts as having three carbons bound to a carbon, yeah. Yes, say thank you. Why do I just prioritize this carbon? Because we have one, so the question is why do I just prioritize this carbon? We have one tetrahedral stereogenic center, I look at the four substituents attached to it and I say we have chlorine first priority, oxygen second priority, carbon third priority, hydrogen fourth priority. So I never have to go out, we never have to have our little runoff election saying there are two carbons attached. If I had a molecule like this, let's say with an isopropyl group and a methyl group attached, then I'd have to start to say ah, we have two carbons, but which carbon is higher priority? So it would be one for the chlorine, two for the oxygen, three for the carbon with two carbons attached to it, and four for the carbon with only hydrogen attached to it, does that make sense? As long as we are only dealing with elements of different atomic number, we never have to go on further down the chain. And of course with one tetrahedral stereogenic center, we're only worrying about what's attached to one carbon. What? Could this carbon be the third substituent, but it's not attached to the tetrahedral stereogenic center? Everything depends on what's attached to the tetrahedral stereogenic center. Let's try two more examples with double and triple bonds. So start with a triple bond, and let's take this compound here. So just as with a double bond, we pretended that we had two carbons attached, one for each of the bonds. With a triple bond, we pretend that we have three carbons attached. So in other words, now we're in a situation where we go and say this is like, I'll again write, pretend, carbon bound to a carbon, but with two carbons attached. Oops, do we net all of here? And so now in prioritizing, obviously hydrogen is the lowest priority substituent. Obviously the hydrogen gets priority of more. The carbon that's the triple bond, we count as having three carbons attached to it. So we count this as having carbon, carbon, and carbon attached to it. We count this, the ethyl group, as having carbon, hydrogen, and hydrogen attached to it. And we count the methyl group as having hydrogen, hydrogen, and hydrogen attached. So by this little convention, by this little way of pretending here, now we go first priority for the acetylene group. The triple bond and carbon. Second priority for the ethyl group. Third priority for the methyl group. And so this stereocentric, clockwise, is R. And of course you don't have to have any hydrogens attached to your tetrahydro-stereogenic center. You could have all carbons attached. You could have a carbon atom with a methyl group, an ethyl group, a propyl group, and a bule group. And you would just keep moving along the chain. The methyl group would be lowest priority. The ethyl group would be next lowest priority as you move down the chain. The propyl group would be next. And the bule group would be highest priority. So you just continue along comparing until you get a difference. Let's try one more example with a double bond. But this time I want to make it a double bond to oxygen. And so I'll draw an aldehyde. And I'll give us an alcohol group. And this aldehyde has the same type of conundrum. We have a tetrahydro-stereogenic center. I put my finger onto it. We have a hydrogen. The hydrogen is lowest priority. Conveniently we're siding down the bond to the hydrogen. Now we have three carbons. A methyl group has H and H attached to it. The carbon with the OH group has OH and H attached to it. The carbon that makes up the aldehyde also has an oxygen and a hydrogen attached to it. But just as we pretended in the carbon double bond that a carbon-carbon double bond meant that there were two oxygens attached, two carbons attached, in the case of a carbon-oxygen double bond, we pretend that we have two oxygens attached. In other words, we say it's OH and H. We treat our aldehyde and we treat this as equivalent to with the carbon having two oxygens attached. And so this particular example, our priorities go 4, 3, 2, and 1. And so our stereochemistry here is counterclockwise. We're conveniently siding down the bond. It's S stereochemistry. All right, when you have two isotopes and almost invariably we're talking hydrogen and deuterium here, you can also have a tetrahedral stereogenic center. And so I'll give us one example where now we have two isotopes of hydrogen attached to one carbon. One of them is deuterium. Deuterium is H2. It's heavy hydrogen. Regular hydrogen is H1. Not surprisingly, higher atomic mass counts as higher priority. And so our deuterium gets a priority of 3. And now we go 1, 2, 3. This isotope or this molecule with isotopes has an or stereochemistry to it. All right. What is deuterium? Deuterium is, I'll explain while I race the board, deuterium is an isotope of hydrogen. It's present in one part, in 7,000, in everything, in water, in your body, all over the place. And deuterium is simply often put in as a label into molecules. So you can make molecules with one deuterium in it. It's called heavy hydrogen. All right. Last time we drew some structures of cyclo pentanol, of methyl cyclo pentanol, and I want to show you a trick. So last time I permuted the stereochemistry and drew out four different molecules. And I called them A, B, C, and D. And I think I drew them like this. I just want to show you how we can very quickly assign the stereochemistry to all four of these molecules of two methyl one cyclo pentanol. The trick is we only have to assign stereochemistry to one of them. And then I just watched what I do with the stereocentrics. And remember, this molecule was two methyl one cyclo pentanol. And I'm just going to start with the molecule on the left. And in my previous lecture where I had all of these and I projected them, I called it molecule A. And I'm going to start with the stereogenic center at the one position. The hydrogen is priority one. The hydrogen is priority four. We now have to look at the two atoms attached to that stereogenic center. The carbon where my finger is has a carbon and two hydrogens attached to it. The carbon where my finger is here has two carbons and one hydrogen attached to it. In other words, the carbon at the two position is priority two and the carbon over here is priority three. We're citing down to the bond, to the hydrogen, we're citing down the bond to the lowest priority substituent. So this substituent is R. That's this stereocenter. I'm going to redraw the molecules, same molecule, but now we're going to focus on the other stereogenic center. On this stereogenic center now, our hydrogen is still lowest priority, lowest atomic number. Now we have three different carbons attached to it. Carbon, carbon and carbon. The methyl group has all hydrogens attached to it, three hydrogens. This group, the one is the one position of the ring, has an oxygen, a carbon and a hydrogen. Oxygen is higher priority than anything else in the molecule. This one here has a carbon and two hydrogens. This substituent becomes priority one, priority two, priority three and so this substituent, this stereochemistry is S. This allows us to assign all of our stereochemistry. This molecule is one R, two S and then two methyl, one cyclopentadol. The next molecule, I've inverted both of the stereocenters. I've swapped the two substituents, the hydroxy and the methyl group. This molecule is one S, two R. The third molecule, we kept the original stereochemistry at the top one but swapped the second. So this molecule is one R and then two R. And the last molecule, we've swapped them around again. Let's see, is that right? Yes. So this is one R, two R and the last one is one S, two S.
 Good morning.  Good morning.  So today I'd like to actually pick up on the question  that was asked by the young woman over there in the previous lecture  on how we assign and how we name  our tetra-egyl-stereogenic center.  And there's a nomenclature system called  the nomenclature that basically allows us to assign  what's called right-handed or left-handed  stereochemistry. That's the convention that everyone  knows of it. As long as we're all on the same page,  that we all have the same way of talking about the molecule in the stereochemistry.  Just like if everyone calls a three-carbon alkane  butane or butane instead of prokane. So we're going  to learn this nomenclature system today. We're also going  to get more familiar with molecules with multiple  tetra-egyl-pacemetric centers. Alright, the  basic nomenclature is really very simple.  Nomenclature consists of  R and S  to designate the stereochemistry.  If you think of R, its Latin term is rectus, which means  right.  And we're going to learn that  that also means clockwise.  The left-handed  notation is from  Latin sinister.  And we're going to learn that  as synonymous with counter-clockwise.  Alright,  what I'm going to try to do  just as in nomenclature naming  alveoli is  to do is to do the same thing  as nomenclature naming alkanes and various other compounds.  There's a series of rules to follow. I'm going to try to take  us through those rules and I'm also going to make it as simple as possible.  Now there's a lot of neat to stereochemistry.  For those of you who have been seeing, there's some fun, there's some three-dimensionality.  It's really a deep topic. I have a book that's about 1400 pages just  on stereochemistry. And so one of the things that I want to do  is to digest the topic and break it just at a simple  level into some small, light-sized pieces. So come along  with me on this journey. Alright.  So, first thing we need to do is  identify the tetrahedral stereogenic center.  That's pretty easy. So usually  at carbon atom, we have four different things on it.  We start with the word identical.  I am.  Different.  Different.  Different.  Different.  Different.  And then what we need to do is to rank  the substituents.  Alright. Let's take an example  and we're going to take the two butanol  that we used before as an example.  So I'll rank that again.  So pretty obviously the carbon  with the hydrogen, the hydroxyl group,  the methyl group, and the ethyl group  is the carbon, the only carbon in two butanol  that has four different substituents on it. So this is  A, and in this case the only  tetrahedral stereogenic center.  Alright. Now what we need to do  is to figure out the priorities of the substituents.  The first way we rank priority,  the most important way is by atomic number.  Highest atomic number gets you the highest priority.  The lowest atomic number gets you the lowest priority.  Kind of makes sense. You're looking at the periodic table.  Oxygen is at 8, carbon is at 6, hydrogen  is at 4. So for priorities, it's  pretty easy to see that our OH, the  hydrogen on here is going to be 1. I'll just put  an arrow here, highest atomic number.  And our hydrogen then is going to be 4.  I'll put an arrow here, the lowest  atomic number.  Now we're faced with the situation of the metal  and the ethyl group, and they're both carbon.  So our first ranking is  at the atoms attached, they're the same, they're both carbon.  So what we do now is we move outward  away from the tetrahedral stereogenic center.  And we identify what's attached to those carbon.  So both of these are carbon, all right, C  and C here. But attached to this carbon  is 3H is H, H, H.  Attached to this carbon, we don't count this carbon anymore,  we're just moving outward. Attached to this carbon,  we're just moving outward.  Attached to this carbon moving outward is another carbon  in two hydrogens. So this one is C, H, and H.  And now we just look, okay, they were both  the same atomic number. But moving  outward now, we have a higher atomic number, carbon  6, attached to this carbon. Then we do  to this carbon. So this  substituent, the ethyl group gets a ranking of 2  and the methyl group  gets a ranking of 3.  So the first rule we can say  is higher atomic number,  is higher atomic number,  higher atomic number  leads to higher  priority.  And realize that when you then  are comparing atoms of the same atomic number,  you just move out. And that rule is going to get you  pretty far. Just like in nomenclature, there's a lot of stuff  that can be really arcane. I'm going to try to take us through the simplest  and most relevant stuff. Now the next operation  once you've ranked the substituents is just to  look down the bond from the carbon atom, from the  tetrahedral stereogenic center to the lowest  priority atom.  Get going, everybody.  Ok.  All in all.  Thanks  for coming here.  I'm going to do my own  pump.  And then what we do is determine the direction 1, 2, 3 going  around as it clockwise or counterclockwise.  And then we weigh this 1 parallel and weigh this 1 parallel n with x trades.  Question?  I'm still a little bit confused.  How do you tell the difference in priority between the ethyl group and the methyl group?  Ah, question.  How do we tell the difference in priority between the methyl group and the ethyl group?  Both the ethyl group and the methyl group are carbonated.  So we know they're both at the same level there of their atomic number.  So now we have to break the time.  And to break the time, what we do is we move outward along the chain and say what's connected to each carbon?  Connected to the methyl group is three hydrogens, all atomic number one.  Connected to the ethyl group, we're just moving out of it so I'm not counting the tetra-needral asymmetric center,  is a carbon, the terminal methyl group on the other side, and two hydrogens making up the methylene group.  So now we say on the tiebreaker, the ethyl group has the carbon on the right, has a carbon attached to it,  and that trumps hydrogen attached to the other.  So that gives us an assignment of two on the priority of the ethyl group  and three on the priority of the methyl group and, of course, group four on the hydrogen.  Other questions? These are good. These are really important to be taking about this.  Ah, what happens if you have the same groups on either side?  So let us pretend for a moment that we had two ethyl groups.  Somebody, what would we do in this case?  It's not a stereogenic center. If they're the same group, then we don't have a tetra-needral stereogenic center.  We need to have four different substituents. So in that case, we have a plane of symmetry in this molecule  in three-pentadol, and as a result, we know immediately that we don't have a chiral molecule.  Other questions?  All right, next thing is visualization. What we want to do is site down, look down the bond  to, from the carbon atom to the hydrogen atom in two-butadol.  And so I'm going to draw things out, and here's where, if you're not yet proficient with visualizing,  here's where models can really help you out. So here's our OH is one, our ethyl is two,  our methyl is three, and I have this hydrogen going back. So we're looking down the bond to it,  and I go one, two, three. Our direction is clockwise, and that means our tetra-needral  stereogenic center is R. So we call this compound R2-butanol. We always, by convention, put the R  in the parenthesis. Let me do this with some video projection, because as I said, we need to become  proficient at this, and it takes a little bit of practice. Let's see if I have a model of R2-butanol.  So I'm going to do this two ways. We'll take it slow. I'm going to show you how I do it,  and then for people in the back, we'll do it on the video projection. So here's the model that I built.  Please, if you have your models, pop yours together. No better time to get your exercise  and visualizing it than right now.  All right. So,  hold up your models. I want to see more of them. Still a neighbor in auto-papery or a makeup friend.  We just had a seminar speaker who was an undergraduate at UCI in 1995 or 1998. She's now a professor  at the University of Toronto, and she was trying to decide whether to take the chemistry general, the Chem51 or Chem52 course.  She ended up taking Chem52. Not only did she end up getting inspired to go on in chemistry, organic chemistry,  because that was what she also met the man who became her husband in that class. So go to your neighbor.  Go to your neighbor. Go to your model. And borrow a model.  All right. So I'm going to do this here, and then I'm going to do it up on the video projector.  So here's my two-butanol, and you notice we're looking down the CH bond. We go one for the oxygen,  two for the ethyl, three for the metal that's clockwise. Let me put that up on the video as well,  just to make it maybe a little easier to answer.  No.  Working from the lamp.  All right. So, here we are. You notice, and this is the case, because if you don't site down the bond,  you'll site in the wrong direction. You'll get it exactly backwards.  So our tetrahedral stereogenic center is in front, where my finger is.  The hypochism is the lowest priority substituent. Is it back where my finger is?  We have the oxygen atom.  We're getting a mirror projected here. Yes. Ah, right. All right. Yes. This is the one case where I think video ends up being a 1, 2.  That's weird. Okay. We're going to stick to this on the model. So again, one for the oxygen.  Everyone's looking down. Here's the oxygen. It's the same as on the blackboard. Here's the ethyl group.  Here's the methyl group. And so, from your point of view, one, two, three is clockwise.  So this is telling you where our hydrogen is.  The question is, is this telling us which direction? Well, if you could take it to waves.  Here, I've given us a molecule, and I've said which wave is it. If I said draw S, then you draw it in a particular wave.  And the way I always do it when I'm asked to draw S or draw R is I go ahead and I simply draw one, because I don't know off the top of my head.  Check it. And if I end up not getting it right, then the trick is you just swap the two substituents.  So if I go ahead and simply swap two, not three, not one, but two substituents, then that's what we call inverting the tetrahedral stereogenic center,  or inverting the chirality of the molecule. And now I know that this molecule is the opposite.  So this one is S to butanol. Now you can do this a couple of waves.  One way is you can simply do the rotation in your head. You can go ahead.  I'm going to swap them on my plastic models. You can go ahead and do the rotation in your head and spin the molecule around.  So in other words, I can go ahead and say, alright, I will just spin this around like so.  And then when I've spun it around, I'm citing down the CH bond, one, two, three, it's counterclockwise.  But the other way you can do it is you can say, well, let's draw in this way, but I know if I reverse things,  if instead of citing down the bond to the lowest priority substituent,  I'm citing from the bond with the lowest priority substituent in front instead of in back, then everything's reversed.  So in other words, you look at this molecule as I'm holding it now with the hydrogen pointing right out at you  and the oxygen and the methyl group and the methyl group.  The way I deal with this is simply I say, well, it's all backwards.  So when I go one, two, three, oxygen, methyl, methyl, I'm going clockwise,  but we're citing in the wrong direction on the bond, which means in reverse it's counterclockwise and the steric chemistry is S.  You can do it either way in three different things, all equivalent that we've done.  One is we've gone and said we can rotate the molecule and simply make a new drawing.  Another way is to say, well, we know we've got this molecule by simply swapping the oxygen and the hydrogen  and we knew when it was like this, it was R because we were citing down the bond.  So if we swap two substituents, we invert the stereocenter and we go from R to S.  Or you can go ahead and say, well, I'm citing it backwards, so what should be clockwise is counterclockwise  and what should be counterclockwise is clockwise.  Questions or thoughts?  And of course, always the fallback is build a plastic molecular model.  And when you're learning, when you're starting to see these drawings on the blackboard for the first few times  and they're not yet clear, that tangible model, that model you can hold in your hand,  grasp the substituents, look down the bond, that is really the thing that's going to get these blackboard drawings into your brain.  So let me write that take off.  All right, so let me write that take home message here just so you can get it in your notes.  Swapping two substituents.  Swaps the stereochemistry and then let's try another example.  All right, example I want to try right now is two chloro-tetrahydropyrene.  And just like we did with two butanol, we're going to determine whether the drawing that I'm about to make is R or is S.  We're going to assyllate the absolute stereochemistry.  So tetrahydropyram, I've mentioned before, is a five-membered cyclic ether containing one oxygen.  This is the tetrahydropyram ring structure.  The oxygen gets numbered for one position, so the chlorine is at the two positions.  That's our priority, that's just numbering the chain.  So here's the structure, let me show you how I think about it.  Now one way I think about it is to start by simply drawing in my hydrogen.  Where am I drawing my hydrogen? What's the tetrahydro-stereogenic center?  The carbon with the chlorine on it.  And that hydrogen is pointing out because the chlorine is pointing back.  Which substituent has the highest priority?  Chlorate.  So that gets us to sign number one. Which one is the second priority?  Oxygen. Third priority? Carbon. Fourth priority? Hydrogen.  Which way, if we look at the molecule as it's drawn, are we looking down the bond to the lowest priority substituent?  No, it's reverse. Two things we can do. One is we can flip the molecule over.  All we do is rotate it 180 degrees, I'm going to pick it up and flip it over here.  Now the hydrogen is going back, we're citing down the bond to the lowest priority substituent.  We can go one, two, three and say this is counterclockwise and this molecule is therefore S2-chlorotendrohydropyram.  Another thing that we can do is simply make a physical model.  Our oxygen is just the tetrahedral construction with carbon, with the red pieces just like carbon.  The trick with the oxygen is those two extra sticks on it represent lone pairs.  Chlorine by convention, since chlorine is green, chlorine is typically represented by a green ball.  Chlorine is a yellow-green gas, of course very poisonous. You smell it with bleach or with swimming pool water.  And you can put on your chlorine and make the model.  And then you don't have to worry about any sort of drawing on the blackboard.  You just say, well, okay, I'll just look down the bond. The bond is pointing back.  Here's the chlorine, here's the oxygen, here's the carbon, we're counterclockwise.  So you never have to worry about being able to visualize.  And if you're proficient at this, what you can do is just keep it all in your head and say, oh, well, the hydrogen is pointing out,  which means all the rules are backwards.  So since the hydrogen is pointing out and we're going chlorine, oxygen, carbon, clockwise,  that's really counterclockwise because everything's backwards.  So counterclockwise means s.  And the more you practice, the more you're going to be doing this automatically.  But at this point, find something that's comfortable for you and practice.  Whether it's starting with the models or physically doing the rotation on the blackboard  or simply mentally keeping track of everything on the blackboard.  All right.  Couple of more little rules.  So couple of more little rules.  One of them is how we treat double bonds, triple bonds.  And we're going to pretend something.  All right.  Here's an example where we have a double bond and they have an OH going out and a hydrogen going back.  What we're going to do is pretend that this carbon here, the one I'm pointing to, the one of the double bond,  instead of being a double bond is bonded to a second carbon and did own for this one.  So what we're going to do is pretend that our structure is like this.  All I've done is to go ahead and I'll draw on the hydrogens too.  All I've done is to say a double bond counts as being bonded to not one carbon but two carbons.  In other words, on this substituent then we say this substituent,  instead of counting as having one carbon attached to it, counts as having two attached to it.  So this carbon counts as having a carbon and a hydrogen attached to it.  Whereas this substituent here counts as having a carbon, a hydrogen and a hydrogen attached to it.  Does that make sense?  So the idea for dealing with double bonds is simply just a way of keeping track.  Say two bonds to another carbon counts as two carbons.  I have it just in my mind just draw a second carbon but you can also say,  alright, well we're going to count this carbon twice so we're going to say that this carbon, since it's got one bond to a carbon  and another bond to a carbon, it counts as being attached to carbon, carbon and hydrogen.  This carbon counts as being attached to one carbon and two hydrogen.  And so our prioritization at this point is that the OH group is number one, the hydrogen is number four  and by this convention the double bonded carbon now counts as number two  and the single bonded carbon counts as three.  So when we assign stereochemistry we're siding down the CH bond, we're going clockwise  and so this molecule counts as being our stereochemistry.  In other words, this molecule counts as R3 methyl 1Nt.  Nomenclature here, we haven't learned yet.  This is a nomenclature for alkanes, it says that the double bond is at the one position.  Obviously the 3 methyl is just like our nomenclature for alkanes, it says the methyl group is, whoops, ox.  Actually the example I was going to do for that was simply same issue here.  So I'm going to skip the nomenclature of this molecule because this molecule becomes,  pentene becomes one pentene if you want to know what becomes R1 pentene 3-ball.  So the main point, don't worry about that nomenclature, the main point is for our stereochemistry.  Alright, one more example with carbon-carbon bond, so we set a double bond, counts as having two carbons bound to a carbon.  A carbon-carbon triple bond counts as having three carbons bound to a carbon, yeah.  Yes, say thank you.  Why do I just prioritize this carbon? Because we have one, so the question is why do I just prioritize this carbon?  We have one tetrahedral stereogenic center, I look at the four substituents attached to it and I say we have chlorine first priority,  oxygen second priority, carbon third priority, hydrogen fourth priority. So I never have to go out,  we never have to have our little runoff election saying there are two carbons attached.  If I had a molecule like this, let's say with an isopropyl group and a methyl group attached,  then I'd have to start to say ah, we have two carbons, but which carbon is higher priority?  So it would be one for the chlorine, two for the oxygen, three for the carbon with two carbons attached to it,  and four for the carbon with only hydrogen attached to it, does that make sense?  As long as we are only dealing with elements of different atomic number, we never have to go on further down the chain.  And of course with one tetrahedral stereogenic center, we're only worrying about what's attached to one carbon.  What?  Could this carbon be the third substituent, but it's not attached to the tetrahedral stereogenic center?  Everything depends on what's attached to the tetrahedral stereogenic center.  Let's try two more examples with double and triple bonds.  So start with a triple bond, and let's take this compound here.  So just as with a double bond, we pretended that we had two carbons attached, one for each of the bonds.  With a triple bond, we pretend that we have three carbons attached.  So in other words, now we're in a situation where we go and say this is like, I'll again write, pretend, carbon bound to a carbon,  but with two carbons attached.  Oops, do we net all of here?  And so now in prioritizing, obviously hydrogen is the lowest priority substituent.  Obviously the hydrogen gets priority of more.  The carbon that's the triple bond, we count as having three carbons attached to it.  So we count this as having carbon, carbon, and carbon attached to it.  We count this, the ethyl group, as having carbon, hydrogen, and hydrogen attached to it.  And we count the methyl group as having hydrogen, hydrogen, and hydrogen attached.  So by this little convention, by this little way of pretending here, now we go first priority for the acetylene group.  The triple bond and carbon.  Second priority for the ethyl group.  Third priority for the methyl group.  And so this stereocentric, clockwise, is R.  And of course you don't have to have any hydrogens attached to your tetrahydro-stereogenic center.  You could have all carbons attached.  You could have a carbon atom with a methyl group, an ethyl group, a propyl group, and a bule group.  And you would just keep moving along the chain.  The methyl group would be lowest priority.  The ethyl group would be next lowest priority as you move down the chain.  The propyl group would be next.  And the bule group would be highest priority.  So you just continue along comparing until you get a difference.  Let's try one more example with a double bond.  But this time I want to make it a double bond to oxygen.  And so I'll draw an aldehyde.  And I'll give us an alcohol group.  And this aldehyde has the same type of conundrum.  We have a tetrahydro-stereogenic center.  I put my finger onto it.  We have a hydrogen. The hydrogen is lowest priority.  Conveniently we're siding down the bond to the hydrogen.  Now we have three carbons.  A methyl group has H and H attached to it.  The carbon with the OH group has OH and H attached to it.  The carbon that makes up the aldehyde also has an oxygen and a hydrogen attached to it.  But just as we pretended in the carbon double bond  that a carbon-carbon double bond meant that there were two oxygens attached,  two carbons attached, in the case of a carbon-oxygen double bond,  we pretend that we have two oxygens attached.  In other words, we say it's OH and H.  We treat our aldehyde and we treat this as equivalent to with the carbon having two oxygens attached.  And so this particular example, our priorities go 4, 3, 2, and 1.  And so our stereochemistry here is counterclockwise.  We're conveniently siding down the bond.  It's S stereochemistry.  All right, when you have two isotopes and almost invariably we're talking hydrogen and deuterium here,  you can also have a tetrahedral stereogenic center.  And so I'll give us one example where now we have two isotopes of hydrogen attached to one carbon.  One of them is deuterium. Deuterium is H2.  It's heavy hydrogen. Regular hydrogen is H1.  Not surprisingly, higher atomic mass counts as higher priority.  And so our deuterium gets a priority of 3.  And now we go 1, 2, 3. This isotope or this molecule with isotopes has an or stereochemistry to it.  All right. What is deuterium?  Deuterium is, I'll explain while I race the board, deuterium is an isotope of hydrogen.  It's present in one part, in 7,000, in everything, in water, in your body, all over the place.  And deuterium is simply often put in as a label into molecules.  So you can make molecules with one deuterium in it.  It's called heavy hydrogen.  All right. Last time we drew some structures of cyclo pentanol, of methyl cyclo pentanol,  and I want to show you a trick.  So last time I permuted the stereochemistry and drew out four different molecules.  And I called them A, B, C, and D. And I think I drew them like this.  I just want to show you how we can very quickly assign the stereochemistry to all four of these molecules of two methyl one cyclo pentanol.  The trick is we only have to assign stereochemistry to one of them.  And then I just watched what I do with the stereocentrics.  And remember, this molecule was two methyl one cyclo pentanol.  And I'm just going to start with the molecule on the left.  And in my previous lecture where I had all of these and I projected them, I called it molecule A.  And I'm going to start with the stereogenic center at the one position.  The hydrogen is priority one. The hydrogen is priority four.  We now have to look at the two atoms attached to that stereogenic center.  The carbon where my finger is has a carbon and two hydrogens attached to it.  The carbon where my finger is here has two carbons and one hydrogen attached to it.  In other words, the carbon at the two position is priority two and the carbon over here is priority three.  We're citing down to the bond, to the hydrogen, we're citing down the bond to the lowest priority substituent.  So this substituent is R. That's this stereocenter.  I'm going to redraw the molecules, same molecule, but now we're going to focus on the other stereogenic center.  On this stereogenic center now, our hydrogen is still lowest priority, lowest atomic number.  Now we have three different carbons attached to it.  Carbon, carbon and carbon.  The methyl group has all hydrogens attached to it, three hydrogens.  This group, the one is the one position of the ring, has an oxygen, a carbon and a hydrogen.  Oxygen is higher priority than anything else in the molecule.  This one here has a carbon and two hydrogens.  This substituent becomes priority one, priority two, priority three and so this substituent, this stereochemistry is S.  This allows us to assign all of our stereochemistry.  This molecule is one R, two S and then two methyl, one cyclopentadol.  The next molecule, I've inverted both of the stereocenters.  I've swapped the two substituents, the hydroxy and the methyl group.  This molecule is one S, two R.  The third molecule, we kept the original stereochemistry at the top one but swapped the second.  So this molecule is one R and then two R.  And the last molecule, we've swapped them around again.  Let's see, is that right? Yes.  So this is one R, two R and the last one is one S, two S.
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Good morning. Good morning. So today I'd like to actually pick up on the question that was asked by the young woman over there in the previous lecture on how we assign and how we name our tetra-egyl-stereogenic center. And there's a nomenclature system called the nomenclature that basically allows us to assign what's called right-handed or left-handed stereochemistry. That's the convention that everyone knows of it. As long as we're all on the same page, that we all have the same way of talking about the molecule in the stereochemistry. Just like if everyone calls a three-carbon alkane butane or butane instead of prokane. So we're going to learn this nomenclature system today. We're also going to get more familiar with molecules with multiple tetra-egyl-pacemetric centers. Alright, the basic nomenclature is really very simple. Nomenclature consists of R and S to designate the stereochemistry. If you think of R, its Latin term is rectus, which means right. And we're going to learn that that also means clockwise. The left-handed notation is from Latin sinister. And we're going to learn that as synonymous with counter-clockwise. Alright, what I'm going to try to do just as in nomenclature naming alveoli is to do is to do the same thing as nomenclature naming alkanes and various other compounds. There's a series of rules to follow. I'm going to try to take us through those rules and I'm also going to make it as simple as possible. Now there's a lot of neat to stereochemistry. For those of you who have been seeing, there's some fun, there's some three-dimensionality. It's really a deep topic. I have a book that's about 1400 pages just on stereochemistry. And so one of the things that I want to do is to digest the topic and break it just at a simple level into some small, light-sized pieces. So come along with me on this journey. Alright. So, first thing we need to do is identify the tetrahedral stereogenic center. That's pretty easy. So usually at carbon atom, we have four different things on it. We start with the word identical. I am. Different. Different. Different. Different. Different. And then what we need to do is to rank the substituents. Alright. Let's take an example and we're going to take the two butanol that we used before as an example. So I'll rank that again. So pretty obviously the carbon with the hydrogen, the hydroxyl group, the methyl group, and the ethyl group is the carbon, the only carbon in two butanol that has four different substituents on it. So this is A, and in this case the only tetrahedral stereogenic center. Alright. Now what we need to do is to figure out the priorities of the substituents. The first way we rank priority, the most important way is by atomic number. Highest atomic number gets you the highest priority. The lowest atomic number gets you the lowest priority. Kind of makes sense. You're looking at the periodic table. Oxygen is at 8, carbon is at 6, hydrogen is at 4. So for priorities, it's pretty easy to see that our OH, the hydrogen on here is going to be 1. I'll just put an arrow here, highest atomic number. And our hydrogen then is going to be 4. I'll put an arrow here, the lowest atomic number. Now we're faced with the situation of the metal and the ethyl group, and they're both carbon. So our first ranking is at the atoms attached, they're the same, they're both carbon. So what we do now is we move outward away from the tetrahedral stereogenic center. And we identify what's attached to those carbon. So both of these are carbon, all right, C and C here. But attached to this carbon is 3H is H, H, H. Attached to this carbon, we don't count this carbon anymore, we're just moving outward. Attached to this carbon, we're just moving outward. Attached to this carbon moving outward is another carbon in two hydrogens. So this one is C, H, and H. And now we just look, okay, they were both the same atomic number. But moving outward now, we have a higher atomic number, carbon 6, attached to this carbon. Then we do to this carbon. So this substituent, the ethyl group gets a ranking of 2 and the methyl group gets a ranking of 3. So the first rule we can say is higher atomic number, is higher atomic number, higher atomic number leads to higher priority. And realize that when you then are comparing atoms of the same atomic number, you just move out. And that rule is going to get you pretty far. Just like in nomenclature, there's a lot of stuff that can be really arcane. I'm going to try to take us through the simplest and most relevant stuff. Now the next operation once you've ranked the substituents is just to look down the bond from the carbon atom, from the tetrahedral stereogenic center to the lowest priority atom. Get going, everybody. Ok. All in all. Thanks for coming here. I'm going to do my own pump. And then what we do is determine the direction 1, 2, 3 going around as it clockwise or counterclockwise. And then we weigh this 1 parallel and weigh this 1 parallel n with x trades. Question? I'm still a little bit confused. How do you tell the difference in priority between the ethyl group and the methyl group? Ah, question. How do we tell the difference in priority between the methyl group and the ethyl group? Both the ethyl group and the methyl group are carbonated. So we know they're both at the same level there of their atomic number. So now we have to break the time. And to break the time, what we do is we move outward along the chain and say what's connected to each carbon? Connected to the methyl group is three hydrogens, all atomic number one. Connected to the ethyl group, we're just moving out of it so I'm not counting the tetra-needral asymmetric center, is a carbon, the terminal methyl group on the other side, and two hydrogens making up the methylene group. So now we say on the tiebreaker, the ethyl group has the carbon on the right, has a carbon attached to it, and that trumps hydrogen attached to the other. So that gives us an assignment of two on the priority of the ethyl group and three on the priority of the methyl group and, of course, group four on the hydrogen. Other questions? These are good. These are really important to be taking about this. Ah, what happens if you have the same groups on either side? So let us pretend for a moment that we had two ethyl groups. Somebody, what would we do in this case? It's not a stereogenic center. If they're the same group, then we don't have a tetra-needral stereogenic center. We need to have four different substituents. So in that case, we have a plane of symmetry in this molecule in three-pentadol, and as a result, we know immediately that we don't have a chiral molecule. Other questions? All right, next thing is visualization. What we want to do is site down, look down the bond to, from the carbon atom to the hydrogen atom in two-butadol. And so I'm going to draw things out, and here's where, if you're not yet proficient with visualizing, here's where models can really help you out. So here's our OH is one, our ethyl is two, our methyl is three, and I have this hydrogen going back. So we're looking down the bond to it, and I go one, two, three. Our direction is clockwise, and that means our tetra-needral stereogenic center is R. So we call this compound R2-butanol. We always, by convention, put the R in the parenthesis. Let me do this with some video projection, because as I said, we need to become proficient at this, and it takes a little bit of practice. Let's see if I have a model of R2-butanol. So I'm going to do this two ways. We'll take it slow. I'm going to show you how I do it, and then for people in the back, we'll do it on the video projection. So here's the model that I built. Please, if you have your models, pop yours together. No better time to get your exercise and visualizing it than right now. All right. So, hold up your models. I want to see more of them. Still a neighbor in auto-papery or a makeup friend. We just had a seminar speaker who was an undergraduate at UCI in 1995 or 1998. She's now a professor at the University of Toronto, and she was trying to decide whether to take the chemistry general, the Chem51 or Chem52 course. She ended up taking Chem52. Not only did she end up getting inspired to go on in chemistry, organic chemistry, because that was what she also met the man who became her husband in that class. So go to your neighbor. Go to your neighbor. Go to your model. And borrow a model. All right. So I'm going to do this here, and then I'm going to do it up on the video projector. So here's my two-butanol, and you notice we're looking down the CH bond. We go one for the oxygen, two for the ethyl, three for the metal that's clockwise. Let me put that up on the video as well, just to make it maybe a little easier to answer. No. Working from the lamp. All right. So, here we are. You notice, and this is the case, because if you don't site down the bond, you'll site in the wrong direction. You'll get it exactly backwards. So our tetrahedral stereogenic center is in front, where my finger is. The hypochism is the lowest priority substituent. Is it back where my finger is? We have the oxygen atom. We're getting a mirror projected here. Yes. Ah, right. All right. Yes. This is the one case where I think video ends up being a 1, 2. That's weird. Okay. We're going to stick to this on the model. So again, one for the oxygen. Everyone's looking down. Here's the oxygen. It's the same as on the blackboard. Here's the ethyl group. Here's the methyl group. And so, from your point of view, one, two, three is clockwise. So this is telling you where our hydrogen is. The question is, is this telling us which direction? Well, if you could take it to waves. Here, I've given us a molecule, and I've said which wave is it. If I said draw S, then you draw it in a particular wave. And the way I always do it when I'm asked to draw S or draw R is I go ahead and I simply draw one, because I don't know off the top of my head. Check it. And if I end up not getting it right, then the trick is you just swap the two substituents. So if I go ahead and simply swap two, not three, not one, but two substituents, then that's what we call inverting the tetrahedral stereogenic center, or inverting the chirality of the molecule. And now I know that this molecule is the opposite. So this one is S to butanol. Now you can do this a couple of waves. One way is you can simply do the rotation in your head. You can go ahead. I'm going to swap them on my plastic models. You can go ahead and do the rotation in your head and spin the molecule around. So in other words, I can go ahead and say, alright, I will just spin this around like so. And then when I've spun it around, I'm citing down the CH bond, one, two, three, it's counterclockwise. But the other way you can do it is you can say, well, let's draw in this way, but I know if I reverse things, if instead of citing down the bond to the lowest priority substituent, I'm citing from the bond with the lowest priority substituent in front instead of in back, then everything's reversed. So in other words, you look at this molecule as I'm holding it now with the hydrogen pointing right out at you and the oxygen and the methyl group and the methyl group. The way I deal with this is simply I say, well, it's all backwards. So when I go one, two, three, oxygen, methyl, methyl, I'm going clockwise, but we're citing in the wrong direction on the bond, which means in reverse it's counterclockwise and the steric chemistry is S. You can do it either way in three different things, all equivalent that we've done. One is we've gone and said we can rotate the molecule and simply make a new drawing. Another way is to say, well, we know we've got this molecule by simply swapping the oxygen and the hydrogen and we knew when it was like this, it was R because we were citing down the bond. So if we swap two substituents, we invert the stereocenter and we go from R to S. Or you can go ahead and say, well, I'm citing it backwards, so what should be clockwise is counterclockwise and what should be counterclockwise is clockwise. Questions or thoughts? And of course, always the fallback is build a plastic molecular model. And when you're learning, when you're starting to see these drawings on the blackboard for the first few times and they're not yet clear, that tangible model, that model you can hold in your hand, grasp the substituents, look down the bond, that is really the thing that's going to get these blackboard drawings into your brain. So let me write that take off. All right, so let me write that take home message here just so you can get it in your notes. Swapping two substituents. Swaps the stereochemistry and then let's try another example. All right, example I want to try right now is two chloro-tetrahydropyrene. And just like we did with two butanol, we're going to determine whether the drawing that I'm about to make is R or is S. We're going to assyllate the absolute stereochemistry. So tetrahydropyram, I've mentioned before, is a five-membered cyclic ether containing one oxygen. This is the tetrahydropyram ring structure. The oxygen gets numbered for one position, so the chlorine is at the two positions. That's our priority, that's just numbering the chain. So here's the structure, let me show you how I think about it. Now one way I think about it is to start by simply drawing in my hydrogen. Where am I drawing my hydrogen? What's the tetrahydro-stereogenic center? The carbon with the chlorine on it. And that hydrogen is pointing out because the chlorine is pointing back. Which substituent has the highest priority? Chlorate. So that gets us to sign number one. Which one is the second priority? Oxygen. Third priority? Carbon. Fourth priority? Hydrogen. Which way, if we look at the molecule as it's drawn, are we looking down the bond to the lowest priority substituent? No, it's reverse. Two things we can do. One is we can flip the molecule over. All we do is rotate it 180 degrees, I'm going to pick it up and flip it over here. Now the hydrogen is going back, we're citing down the bond to the lowest priority substituent. We can go one, two, three and say this is counterclockwise and this molecule is therefore S2-chlorotendrohydropyram. Another thing that we can do is simply make a physical model. Our oxygen is just the tetrahedral construction with carbon, with the red pieces just like carbon. The trick with the oxygen is those two extra sticks on it represent lone pairs. Chlorine by convention, since chlorine is green, chlorine is typically represented by a green ball. Chlorine is a yellow-green gas, of course very poisonous. You smell it with bleach or with swimming pool water. And you can put on your chlorine and make the model. And then you don't have to worry about any sort of drawing on the blackboard. You just say, well, okay, I'll just look down the bond. The bond is pointing back. Here's the chlorine, here's the oxygen, here's the carbon, we're counterclockwise. So you never have to worry about being able to visualize. And if you're proficient at this, what you can do is just keep it all in your head and say, oh, well, the hydrogen is pointing out, which means all the rules are backwards. So since the hydrogen is pointing out and we're going chlorine, oxygen, carbon, clockwise, that's really counterclockwise because everything's backwards. So counterclockwise means s. And the more you practice, the more you're going to be doing this automatically. But at this point, find something that's comfortable for you and practice. Whether it's starting with the models or physically doing the rotation on the blackboard or simply mentally keeping track of everything on the blackboard. All right. Couple of more little rules. So couple of more little rules. One of them is how we treat double bonds, triple bonds. And we're going to pretend something. All right. Here's an example where we have a double bond and they have an OH going out and a hydrogen going back. What we're going to do is pretend that this carbon here, the one I'm pointing to, the one of the double bond, instead of being a double bond is bonded to a second carbon and did own for this one. So what we're going to do is pretend that our structure is like this. All I've done is to go ahead and I'll draw on the hydrogens too. All I've done is to say a double bond counts as being bonded to not one carbon but two carbons. In other words, on this substituent then we say this substituent, instead of counting as having one carbon attached to it, counts as having two attached to it. So this carbon counts as having a carbon and a hydrogen attached to it. Whereas this substituent here counts as having a carbon, a hydrogen and a hydrogen attached to it. Does that make sense? So the idea for dealing with double bonds is simply just a way of keeping track. Say two bonds to another carbon counts as two carbons. I have it just in my mind just draw a second carbon but you can also say, alright, well we're going to count this carbon twice so we're going to say that this carbon, since it's got one bond to a carbon and another bond to a carbon, it counts as being attached to carbon, carbon and hydrogen. This carbon counts as being attached to one carbon and two hydrogen. And so our prioritization at this point is that the OH group is number one, the hydrogen is number four and by this convention the double bonded carbon now counts as number two and the single bonded carbon counts as three. So when we assign stereochemistry we're siding down the CH bond, we're going clockwise and so this molecule counts as being our stereochemistry. In other words, this molecule counts as R3 methyl 1Nt. Nomenclature here, we haven't learned yet. This is a nomenclature for alkanes, it says that the double bond is at the one position. Obviously the 3 methyl is just like our nomenclature for alkanes, it says the methyl group is, whoops, ox. Actually the example I was going to do for that was simply same issue here. So I'm going to skip the nomenclature of this molecule because this molecule becomes, pentene becomes one pentene if you want to know what becomes R1 pentene 3-ball. So the main point, don't worry about that nomenclature, the main point is for our stereochemistry. Alright, one more example with carbon-carbon bond, so we set a double bond, counts as having two carbons bound to a carbon. A carbon-carbon triple bond counts as having three carbons bound to a carbon, yeah. Yes, say thank you. Why do I just prioritize this carbon? Because we have one, so the question is why do I just prioritize this carbon? We have one tetrahedral stereogenic center, I look at the four substituents attached to it and I say we have chlorine first priority, oxygen second priority, carbon third priority, hydrogen fourth priority. So I never have to go out, we never have to have our little runoff election saying there are two carbons attached. If I had a molecule like this, let's say with an isopropyl group and a methyl group attached, then I'd have to start to say ah, we have two carbons, but which carbon is higher priority? So it would be one for the chlorine, two for the oxygen, three for the carbon with two carbons attached to it, and four for the carbon with only hydrogen attached to it, does that make sense? As long as we are only dealing with elements of different atomic number, we never have to go on further down the chain. And of course with one tetrahedral stereogenic center, we're only worrying about what's attached to one carbon. What? Could this carbon be the third substituent, but it's not attached to the tetrahedral stereogenic center? Everything depends on what's attached to the tetrahedral stereogenic center. Let's try two more examples with double and triple bonds. So start with a triple bond, and let's take this compound here. So just as with a double bond, we pretended that we had two carbons attached, one for each of the bonds. With a triple bond, we pretend that we have three carbons attached. So in other words, now we're in a situation where we go and say this is like, I'll again write, pretend, carbon bound to a carbon, but with two carbons attached. Oops, do we net all of here? And so now in prioritizing, obviously hydrogen is the lowest priority substituent. Obviously the hydrogen gets priority of more. The carbon that's the triple bond, we count as having three carbons attached to it. So we count this as having carbon, carbon, and carbon attached to it. We count this, the ethyl group, as having carbon, hydrogen, and hydrogen attached to it. And we count the methyl group as having hydrogen, hydrogen, and hydrogen attached. So by this little convention, by this little way of pretending here, now we go first priority for the acetylene group. The triple bond and carbon. Second priority for the ethyl group. Third priority for the methyl group. And so this stereocentric, clockwise, is R. And of course you don't have to have any hydrogens attached to your tetrahydro-stereogenic center. You could have all carbons attached. You could have a carbon atom with a methyl group, an ethyl group, a propyl group, and a bule group. And you would just keep moving along the chain. The methyl group would be lowest priority. The ethyl group would be next lowest priority as you move down the chain. The propyl group would be next. And the bule group would be highest priority. So you just continue along comparing until you get a difference. Let's try one more example with a double bond. But this time I want to make it a double bond to oxygen. And so I'll draw an aldehyde. And I'll give us an alcohol group. And this aldehyde has the same type of conundrum. We have a tetrahydro-stereogenic center. I put my finger onto it. We have a hydrogen. The hydrogen is lowest priority. Conveniently we're siding down the bond to the hydrogen. Now we have three carbons. A methyl group has H and H attached to it. The carbon with the OH group has OH and H attached to it. The carbon that makes up the aldehyde also has an oxygen and a hydrogen attached to it. But just as we pretended in the carbon double bond that a carbon-carbon double bond meant that there were two oxygens attached, two carbons attached, in the case of a carbon-oxygen double bond, we pretend that we have two oxygens attached. In other words, we say it's OH and H. We treat our aldehyde and we treat this as equivalent to with the carbon having two oxygens attached. And so this particular example, our priorities go 4, 3, 2, and 1. And so our stereochemistry here is counterclockwise. We're conveniently siding down the bond. It's S stereochemistry. All right, when you have two isotopes and almost invariably we're talking hydrogen and deuterium here, you can also have a tetrahedral stereogenic center. And so I'll give us one example where now we have two isotopes of hydrogen attached to one carbon. One of them is deuterium. Deuterium is H2. It's heavy hydrogen. Regular hydrogen is H1. Not surprisingly, higher atomic mass counts as higher priority. And so our deuterium gets a priority of 3. And now we go 1, 2, 3. This isotope or this molecule with isotopes has an or stereochemistry to it. All right. What is deuterium? Deuterium is, I'll explain while I race the board, deuterium is an isotope of hydrogen. It's present in one part, in 7,000, in everything, in water, in your body, all over the place. And deuterium is simply often put in as a label into molecules. So you can make molecules with one deuterium in it. It's called heavy hydrogen. All right. Last time we drew some structures of cyclo pentanol, of methyl cyclo pentanol, and I want to show you a trick. So last time I permuted the stereochemistry and drew out four different molecules. And I called them A, B, C, and D. And I think I drew them like this. I just want to show you how we can very quickly assign the stereochemistry to all four of these molecules of two methyl one cyclo pentanol. The trick is we only have to assign stereochemistry to one of them. And then I just watched what I do with the stereocentrics. And remember, this molecule was two methyl one cyclo pentanol. And I'm just going to start with the molecule on the left. And in my previous lecture where I had all of these and I projected them, I called it molecule A. And I'm going to start with the stereogenic center at the one position. The hydrogen is priority one. The hydrogen is priority four. We now have to look at the two atoms attached to that stereogenic center. The carbon where my finger is has a carbon and two hydrogens attached to it. The carbon where my finger is here has two carbons and one hydrogen attached to it. In other words, the carbon at the two position is priority two and the carbon over here is priority three. We're citing down to the bond, to the hydrogen, we're citing down the bond to the lowest priority substituent. So this substituent is R. That's this stereocenter. I'm going to redraw the molecules, same molecule, but now we're going to focus on the other stereogenic center. On this stereogenic center now, our hydrogen is still lowest priority, lowest atomic number. Now we have three different carbons attached to it. Carbon, carbon and carbon. The methyl group has all hydrogens attached to it, three hydrogens. This group, the one is the one position of the ring, has an oxygen, a carbon and a hydrogen. Oxygen is higher priority than anything else in the molecule. This one here has a carbon and two hydrogens. This substituent becomes priority one, priority two, priority three and so this substituent, this stereochemistry is S. This allows us to assign all of our stereochemistry. This molecule is one R, two S and then two methyl, one cyclopentadol. The next molecule, I've inverted both of the stereocenters. I've swapped the two substituents, the hydroxy and the methyl group. This molecule is one S, two R. The third molecule, we kept the original stereochemistry at the top one but swapped the second. So this molecule is one R and then two R. And the last molecule, we've swapped them around again. Let's see, is that right? Yes. So this is one R, two R and the last one is one S, two S.
1400 pages just on stereochemistry. And so one of with the situation of the metal and the ethyl attached to carbon, carbon and hydrogen. This