İzomerler Hakkında Detaylı Notlar

Hello everybody, my name is Iman. Welcome back to my YouTube channel. Today we're continuing our lecture on isomers. Where we left off is on configurational isomers. Now to talk about configurational isomers, we want to start with chirality. So objects that are not superimposable on their mirror images are called chiral objects. And the opposite of this is something called a chirality. Now an example of chirality is your hands. If you place both face down on the table and then you try to take your right hand and place it over your left, you notice that they are not. superimposable. All right. Now, how do we take that and think about it in regards to molecules and compounds? Well, in terms of identification for organic chemistry here for chirality, a chiral center is a carbon with four different groups attached to it. Your first step should always be to identify the number of chiral centers in a molecule. All right. And again, the way you identify that is by finding a carbon that has four different groups attached to it. All right. So once again, an object is considered chiral if its mirror image cannot be superimposed on this original object. This implies that the molecule lacks this internal plane of symmetry. So chiral center is a carbon with four different groups on it. This carbon will be asymmetrical, is going to be an asymmetrical core of optical activity, and it's going to be known as a chiral center. Now, two molecules that have non-superimposable mirror images of each other are called enantiomers. Molecules may also be related as diastereomers. These molecules are... chiral and share the same connectivity but are not mirror images of each other. And this is because they differ at some but not all of their multiple chiral centers. All right, and we're going to dive into enantiomers and diastereomers as well. But before we do that, let's take a look at some of the other enantiomers that we have. Before we do that, we now understand what a chiral carbon is, four different groups attached to a carbon. Let's learn how to identify whether that chiral center gets a R or an S assignment. that requires for us to talk about R and S forms, which is part of objective three. So we're taking a pause from objective two and configurational isomers to go over R and S forms because they will play a really important role in us looking at molecules and trying to decide, are they enantiomers? Are they diastereomers? Are they mesocompounds? We're going to learn what all that means. And being able to look at two molecules and compare and determine whether they're enantiomers or not. their innate tumors, diastereomers, or mesocompounds, requires us to be able to, one, identify chiral centers, and then assign chiral centers R or S configuration, and then comparing and contrasting based off of that. All right, so let's cover R and S configuration. Now that we can identify chiral centers, carbons with four different groups attached to them let's learn how to assign the configuration of the chiral centers now at every chiral center you you will either assign r or s but what does that mean this is an attempt to relay information in the spatial arrangement of atoms at a chiral carbon center. The Kahn-Inglod prelog rules will help us do just that in five steps. The first step is to identify the four atoms directly attached to the chiral center, all right, to that chiral carbon. The second rule is to assign a priority to each atom based on its atomic number. The highest atomic number receives priority and the lowest atomic number receives a priority of four. This lowest atomic number priority is often a hydrogen atom. Now, step three says if you have a tie, if there's a tie, if two atoms have the same atomic number, then what you want to do is move away from the chiral center and look for the first point of difference. All right, one constructing list to compare. Remember that a double bond is treated as two separate single bonds. when you are thinking and trying to determine priority between two groups that tie. So if there's a tie, move on to the next atoms attached until you can assign priority for all groups. Step four says, rotate the molecule so that the fourth priority is on it. Dash. Remember, dashes imply that that atom is going behind the plane of the page. Wedge means it's coming out of the plane of the page. And then step five, determine whether the sequence one, two, three follows a clockwise order. If it does, it's R. Or if it follows a counterclockwise order, then it is assigned S. All right. Now, the question here, all right, the question is here is, well, how do we begin to visualize what this means? All right. How do we begin to visualize assigning priority, for example? And how do we begin to? you know, assign priority? And then how do we begin to make sure that our fourth priority is always on a dash? How do we rotate molecules correctly to do that and then determine the one, two, three sequence? So for example, let's say we have a molecule like this, all right? We have a bromine group, we have a hydrogen. We have a, let's see, a nitrogen here, and it has two methyl groups attached to it, and then that's it. All right. If we're looking at a molecule like that, where is the chiral? Is there a chiral center? That's the first question. The answer is yes. Here's a chiral center. This carbon is chiral because it has four different groups attached to it. It has a bromine. It has a hydrogen. It's attached to a nitrogen, and then it's attached to this carbon as part of a methyl group. All right, so we've found and identified a chiral carbon, a chiral center. Now what we want to do is begin to assign priority to each atom based on its atomic number, on their atomic numbers. So bromine here. is going to have the highest atomic number. It gets priority one. It's followed by nitrogen for two, carbon for three, hydrogen for four. Cool. Now, there's no ties here, thankfully. If there was carbons on both sides, for example, then you would move to the next. um, you would, you would look to what the carbons attached to, and then move out until you can identify a priority. So let's stick to this example and then we'll look at what if it was a carbon and we had a tie in priority. So this is what it is now. Bromine one, nitrogen two, carbon three priority, hydrogen four priority. Now we step four says to make sure that the fourth priority is on a dash. Good for us here. It is on a dash. So what do we do? We ignore it. And then we draw from 1 to 2 to 3. All right, and notice how that is clockwise. All right, that is moving clockwise. So it gets assigned, that chiral center gets assigned R. All right, fantastic. Now let's pretend, let's draw a different molecule. All right, let's draw something a little different. let's say this is a, let's do this. Let's say that this is a chlorine atom. This is hydrogen. This is a methyl group right here. And this is a carbon that's attached to two other carbons. Cool. Let's identify a chiral center. Here's a chiral center right here. This carbon's attached to four different groups, a chlorine, a hydrogen, a methyl, and then this group right here. All right. So cool. That's a chiral center. That, that little dot that's highlighted, that little carbon that's highlighted in green, that's a chiral center. Now the second step is to assign priority. Well, that's a chlorine, that's a hydrogen. This is a carbon right here. It's attached to three hydrogens. This is a carbon as well. It's attached to a carbon over here, a carbon over here, and then there's also a hydrogen. Now let's assign priority. Chlorine obviously gets one. Now we have a tie for two carbons. Hydrogen obviously gets four anyways. The tie here is between these two carbons which one gets to priority which one gets three so they're both carbons So what we want to do is make a list of what these carbons are attached to this carbon is attached to three Hydrogens this carbons attached to two carbons and and one hydrogen, all right? Carbon beats hydrogen, all right? And so this carbon right here gets priority two, and this one gets priority three. Now, thankfully, again, the hydrogen is on the dash, so we ignore it, and then we draw arrow from one to two to three. This is counterclockwise, so this gets, this chiral center gets an S assignment. Now, what if it's not so easy that the hydrogen is, the fourth group is not on a dash? How do we approach that? How do we begin to visualize? how to rotate a molecule appropriately to get a hydrogen on a dash. Well, there's actually a little bit of a trick to this, all right? So there's three different ways that you can imagine how this could play out, and we're going to go over each of them and how to tackle that, all right? So we can identify chiral center, and we can, you know, assign... RS configuration, but for determining, you know, clockwise or counterclockwise, there are a couple hints to keep in mind. All right. First is if your fourth group is on a dash, then you are so lucky. You don't have to, if your fourth priority group is on a dash, you're lucky, you just ignore it. Goodbye, and you draw your arrow from 1 to 2 to 3 and determine if that's clockwise, it gets R, or if it's counterclockwise, it gets S assignment. Okay, that one is the easy one. What if your fourth group, all right, is not on a dash, but it is right next to a dash? All right, look at this fourth group. It's not on a dash, but it's right next to a group that is on a dash. The three priorities on a dash instead of the four, but they're right next to each other. What do we do? What we do is we still figure out our one to two to three sequence. All right, here's our one, here's our two, and here's our three. All right. We determine our 1, 2, 3. We assign it. That's clockwise, so it gets R. And then we give it the opposite configuration. We give it an S because our fourth group was never on a dash. All right, so that is the trick for that. If you're... Fourth group is not on a dash, but it's next to it. You can do this. All right, and here's another example. Same thing. Your fourth group is not on a dash, but it's right next to a group that is. So what you can do is still just draw your 1, 2, 3. That's counterclockwise. That would be given S. No, you give it the opposite because that fourth group was never on a dash, but it was near it. All right, so that's the trick for that. The last thing is, well, if you have a fourth group and it's not on dash, but it's right across from the dash. So here's the fourth group. Here's the dash. What do you do? Well, you just do your one to two to three sequence. All right. And just give it that appropriate configuration. So this is counterclockwise. It gets S and you don't worry about it. And this is because to get the fourth group to the dash when they're across from each other, you're going to do two rotations and they'll ultimately cancel each other out in that way. All right. So those are the three tricks I always use if I don't know how to properly rotate a molecule. So this requires no 3D visualization on your part to appropriately rotate molecules, which could be way faster on a timed exam like the MCAT. All right. So now we can take all this information we learned about chiral centers and RNS configuration to understand and identify stereoisomers. Stereoisomers. but there is one more thing one more thing that i want to cover before we do that before we start talking about what enantiomers are what diastereomers are and what meso compounds are we said we know how to find a chiral center that's a carbon with four different groups and we know how to do rns configuration at a chiral center with those four different groups but what about all right what about Carbon-carbon double bonds. All right, what about cis and trans? That's an important discussion for us to have. All right, so how do we treat double bonds in regards to configuration? Well, this is where we start to talk about E and Z forms. So E and Z nomenclature is used for compounds with polysubstituted double bonds. Recall that simpler double bond-containing compounds, they can use the cis and trans system. All right? But you can, you know, you might also be given or asked to use the E and Z designation. All right. So how do you do this? Well, you identify where your double bond is. So, for example, here's a double bond. All right. You figure out your highest priority groups. All right. So for looking at these. Um, this is going to be, these two are going to be your higher priority groups here, the bromine and chlorine. All right, cool. So you've determined your, your two highest priority groups. You've determined where your double bond is. You're going to draw a line through it and you've determined your two highest priority groups. Now, if your two highest priority groups are on the same side, which they are based off of where you drew a line across through your double bond, if they're on the same side, which these two priority groups are on the same side. All right. This is given Z assignment. All right. But if we look here, here's our carbon double bond. Our first, our first priority group is here in our second priority group here. And what we notice is that they're on. So opposite sides, if they're on opposite sides are two highest priority groups, then this is given an E assignment. All right. E. And then you could proceed with the name. So this is how you do E and Z designation of alkenes. All right. So this was also, again, a little jump into this. third objective, right? We're kind of embedding the third objective as we have a, what I think is a chronological order discussion on configurational isomers. All right. So now we understand R and S, and now we understand what E and Z is as it relates to cis and trans. All right. Fantastic. So now we know how to identify chiral centers. We know if we find a carbon with four different groups, that's a chiral carbon, and then we can designate it an R or S configuration based off of the rules we learned. If we have double bonds, remember we already know cis and trans. But we're going to be focusing on the E and Z designation here. If your two highest priority groups are on the same side of the double bond, then this gets a Z assignment. And if the two highest priority groups are on opposite sides of the double bond, it gets an E assignment. All right, we'll get into more depth in relation to the importance of this. Really, for the rest of the discussion, we're going to be talking more about R and S configuration and how we can... use this idea to define enantiomers, diastereomers, and mesocompounds appropriately. All right. So now we can talk about enantiomers, diastereomers, and mesocompounds. All right. So your first step is always to identify chiral centers and then assign them as R and S. So let's begin talking about enantiomers. Enantiomers are non-superimposable mirror images. They cannot appear identical simply by rotation. Now, how can you tell if mental rotation is difficult for you, right? If you're looking at these two molecules and you're trying to tell if they are enantiomers or not, and you can't mentally rotate it appropriately, especially assigning and changing the wedges and dashes appropriately, how can you tell if these two molecules are enantiomers or not? Well, easy. We're going to rely on R and S configuration assignments to help. All right. So here's how it works for enantiomers if you have only one chiral center. If you have one chiral center and you're looking at two molecules and you're trying to determine if there are enantiomers, first, you find that one chiral center for each of them. And then you're going to work and assign each one R and S. Let's pretend, okay, we're pretending here that this is R and this is S, okay? I'm just pretending here, all right? This one's R and this one's S. They both just have one chiral center and one of them is R and one of them is S. Because they are opposite, all right, one is R and one is S at the chiral center, then they are indeed enantiomers, all right? So that's if there is one chiral center. And you're always going to need at least one chiral center. center to even say or begin to investigate if two molecules are chiral centers. And if there's just one chiral center, then it becomes easy. Assign the chiral center for each of the molecules that you're comparing, each of the two molecules you're comparing. If in one molecule that carbon center is R and the other one is S, then they are enantiomers because they have the opposite R and S assignment. All right. If you're looking at both of these and they're both R, then they're just identical. All right. We're not talking about enantiomers. Now, there could be two chiral centers, right? You can have two molecules that you're comparing and you're trying to determine if they're enantiomers and they both have two chiral centers. All right. In that case, if at all chiral centers between the two molecules, they have opposite R and S assignment. All right. Then they are enantiomers. So let's pretend. all right here's a molecule here's a chiral center bromine chlorine, nitrogen. All right. This one's on a wedge. This one's on a dash. And then you're comparing the same one here. This one's on a wedge. This one's on a dash. I'm not going to assign them here. I just want to draw them. All right. And this is a carbon. If this is a carbon and it has another chiral center. So let's do that. This is attached to a carbon. This is a hydrogen. This is a fluorine. All right. So here, We have two chiral centers. Let me draw them out appropriately. Alright. I'm just drawing stuff so I can get the point across. We're not going to actually investigate them. I'm just trying to get the point across. So let's do this as a wedge. That's bromine-chlorine. This one has a hydrogen and a fluorine. So these two molecules, we're comparing them. We want to know if they're enantiomers or not. They have two chiral centers here and here, here and here. So they have the two same chiral centers. If we were to determine that at this chiral center, you know what? Let me change their colors. Alright, this one's red and this one's red. because they're the same and this one's blue and this one's blue if we decide if we were to figure out what the rns assignment is here and we decided that this is r and we looked at the other molecule at the same chiral center we determined this is s and then we did the same thing with the other chiral center this one is um S and this one is R. All right. Now, because they have because the two chiral centers between the two molecules, because all the chiral centers between the two molecules have opposite R and S assignment at those positions, then they are enantiomers. If they were both identical, like RS for this one and then also RS for this one, these are just going to be identical. Now, when you have two chiral centers, you also have to worry about diastereomers. You also have to consider if they're diastereomers, which has a different rule. The rule that they're opposites here is what gives it away that they're enantiomers. All right. So that's how you use R and S configuration to determine whether, all right, two molecules are enantiomers, whether they have one chiral center or two chiral centers or more. All right. Fantastic. Now, before we get to diastereomers and mesochromosomes, compounds. I want to just scroll over here. All right. And side note on enantiomers, that's going to be important for the MCAT. Now we said enantiomers are superimposable mirror images and thus have opposite stereochemistry at every chiral carbon. All right. They have the same chemical and physical properties, those enantiomers, except for one thing. They differ in regards to how they rotate plane polarized light. All right. They have the same chemical and physical properties except for rotation of plane polarized light and reactions in a chiral environment. Now, optical activity refers to the ability of a molecule to rotate plane polarized light. All right. So you can have a compound that rotates the plane polarized light to the right. All right. It can rotate the plane polarized light to the right. Or you can think of this as clockwise. All right. This is dextrorotary. All right. So that's assigned D. All right. And it's usually labeled as a positive. You can have the light be rotated to the left. This is called counterclockwise. This is levorotatory. All right. And this is identified as L minus usually written or is just labeled as minus. The direction of rotation can't be determined from the structure. of a molecule, you have to actually determine it experimentally. And so that is, it is not related to the absolute configuration of the molecule. Now, the amount of rotation depends on the number of molecules that a light wave encounters. All right. This depends on two factors, the concentration of the optically active compound and the length of the tube through which the light passes. Chemists have set standard conditions of one gram per milliliter for concentration and 1 decimeter for length to compare the optical activities of different compounds. Now, rotations measured at different concentrations and tube lengths can be converted to a standardized specific rotation using this equation right here. This is the specific rotation in degrees. All right, this is the observed rotation in degrees, C is the concentration, and L is the path length. All right. Now, if you were to have a mixture with both positive and minus enantiomers in equal concentrations, they form what's called a racemic mixture. In these solutions, the rotations cancel each other out and no optical activity is observed. All right. Fantastic. So that was an important thing that I wanted to mention, all right, as a side note for enantiomers. All right, now let's also go ahead and define diastereomers. So diastereomers are non-superimposable. And not mirror images. So in other words, non-identical stereoisomers. Now, here's the key for thinking about diastereomers. They're going to have the same molecular formula. They're going to have the same connectivity. and they're going to at least have two chiral centers. All right. So unlike enantiomers, all right, going back to enantiomers, the requirements were, of course, that they have the same molecular formula and the same connectivity. Those are the two conditions of stereoisomers. But enantiomers needed at least one. chiral center. And then you assign RNS to each and compare positions. If at all chiral centers between the two molecules they have opposite RNS assignment, then they are in anteriors. If all are identical, then the molecules are identical. Now when we think about diastereomers, here's the key for that. You have the same molecular formula, you have the same connectivity, you need at least two chiral centers. And how you approach this is you're going to assign RNS to each All right, and then compare positions. So let's look at these drawings right here. All right, let's draw, let's look at these drawings right here, and then let's begin to understand enantiomers and diastereomers. The rule for enantiomers was if they have opposite R and S configuration at each carbon center, chiral center that you're comparing, then they are enantiomers. But with diastereomers, If at some chiral centers that you are comparing between two molecules, you have a mix of the same R and S assignment and opposite, then they are diastereomers. So let's try to do this, all right? I am going to hypothetically assign R and S configurations at these, just so that we can get the point, all right? So this is R, and let's say that this is also R, for example. Okay, so these are both R. Let's say that this is R and this is S. Let's say that both of these are S. And then let's say that this is S and this is R. Okay, now look let's look at this point in this molecule and then this chiral center, right? These are the same chiral carbons in both of these molecules. This one is R and this one is S. Cool. They're opposites at that chiral center. Now let's look at. this chiral center for these two molecules. All right. This one is R and this one is S. They're also opposite at that chiral center. So both of these chiral centers we looked at in these two molecules have opposite R and S assignment. This one had R, this one had S at the same chiral carbon. At this same chiral carbon, this one had R, this one had S. Opposites, opposites at both of those chiral centers. That means these molecules are enantiomers. All right, cool. Let's look at these molecules now. I'm going to circle them in blue. We're just looking at them as we choose, all right? Let's look at these two. If we look at this chiral carbon at both of these molecules, one is S, one is R. That's opposite. Cool. If we look at the second chiral center, oh, this one's R, this one's S. At both chiral carbons, they have opposite R and S assignment. This one's S, this one's R. This one's R, this one's S. Opposite at... the same chiral position, the same chiral carbon. These two are also enantiomers, all right? So these are enantiomers, these are enantiomers. Cool, so we understand enantiomers now. If there is opposite R and S at the same chiral center, for all the chiral centers that you're looking at, all of them have opposite R and S configuration, however many your molecule has, whether it's one, two, three, or more, all right, then that defines... finds in the antiumers. Now, what about diastereomers? All right, so let's do some erasing. All right, let's do some erasing. All right, these are RR. All right, let's look at these, these two. All right, let's compare them. At this chiral carbon for both of these molecules, they're assigned R. They're the same. Cool. If we look at the second chiral carbon in these molecules. One is R and one is S. So at one chiral center, they're the same. At another chiral center though, they're the opposite. This mix of same and opposite is what defines diastereomers. So these two molecules molecules, if we were looking at them, we would define these to be diastereomers. All right. So that is the distinction between Indian tumors and diastereomers based off of the R and S assignments. Now, one more note on diastereomers. Cystrans isomers are a subtype of diastereomers in which groups differ in their position about growth. that double bond, about that immovable bond. So going back to our priority rules, an alkene is Z if the two, if the high priority substituents are on the same side of the double bond, and they're E if they're on opposite sides of the double bond. All right, so that's enantiomers, that's diastereomers. Fantastic. But wait, there's one more thing you have to be aware of. of when you're looking at stereoisomers for molecules that have reflectional symmetry, and that is mesocompounds. Mesocompounds are chiral compounds that have multiple chiral centers. All right, they have a least two chiral centers. All right. Meso compounds have, excuse me, they have an internal plane of symmetry. So they will be optically inactive because Because the two sides of the molecules cancel each other out. Now, I'm going to show you how to identify mesocompounds. All right? I'm going to show you how to identify mesocompounds. This is the rule. If you have at least two chiral centers, all right? If you have at least two chiral centers and you have an internal plane of reflection, you're going to have a mesocompound. plane of reflection, all right, if you have an internal plane of reflection, and then you have also, all right, if your molecule has reflectional symmetry and at least two chiral centers, then the pair of quote-unquote enantiomers with R and S assignment on one and then S and R on the other that you had thought was an enantiomer pair is actually a mesocompound. They are simply rotations. of each other. So for example, look at this molecule. It has two chiral centers. It also has an internal plane. of reflection. Now, if you go ahead and assign these, this might be R and this might be S, and then this might be S and this might be R. So you notice that they're opposites here, and then they're opposites here, and you might think enantiomers. But because there is an internal plane of reflection and you're looking at two chiral centers, these molecules are actually mesocompounds. They are the same molecule, they're just simply rotated. All right, so that's the one catch. All right. You want to be careful because there's an internal plane of reflection. So what you thought was an enantiomer in this case is not. All right. So let's put all these rules together so that we have kind of a workflow for this. All right. If there is one chiral center, all right, and they have opposite RNS configuration, there are enantiomers. If they have the same, then they're identical. That's with one chiral center. If you have two chiral centers and they're opposite RNS configuration at every chiral center, then they're enantiomers. If there are some opposite and some same RNS configurations, then they're diastereomers. And if they're all exactly the same, then they're identical. Now, the only thing is if there's reflectional symmetry, double check for meso compounds. All right. Fantastic. Now, there's one more last thing that we want to cover, and that's Fischer projections. All right. Vertical. So this is. right here, a common depiction of molecules you'll see in biochemistry. All right. Sorry, that took, I had a little brain fart there. I was like, where do they see this? biochemistry. All right. Now, if you were trying to take this and convert it to say bond line, all right, take this Fisher projection, convert it to bond line. That's going to be an important skill to have. All right. And it's going to be important to understand the workflow for this. All right. Now, here's how I do this. Here's my recommendation. All right. We're going to identify these as the ends, and we're going to go. ahead and draw this kind of structure right here. Kind of looks like a table top. All right. We're going to put this on one end and put the other group on the other end. All right. Top group on the left, bottom group on the right. Cool. Then we're going to see how many carbons are in between those two groups. Three. So we're going to draw one, two, three points. All right. One, two, three points. Cool. Then what are we going to do? We're going to draw the groups that are attached to each carbon. So this is going to be numbered our first carbon. This is our second carbon. And that's our third. One, two, three. All right. Now anything on the right gets assigned a dash. So what's on the right at carbon one? An OH group. So we draw it on a dash. Anything on the left gets a wedge. So the hydrogen gets a wedge. Look at carbon two. What's on the right? OH group, alcohol group, we draw it on a dash. What's on the left? Hydrogen, that goes on a wedge. Alright, keep that rule. Alright, third carbon, what's on the right? Hydrogen group, that goes on the dash. What's on the left? Alcohol group that goes on the wedge. Cool. Now, your bond line is not going to look like this. It's going to look more like a zigzag of carbons, right? Because that's the format for bond line. So we don't want to keep it like this. We want to keep this part, but then we want to bring it down here. We want to take this group down here and then we can go back up and then it looks like a normal bond line. So what we're doing is we want to bring this group to a downward position. All right. To do this, we're going to switch these two groups in. their assignment of dashes and wedges. So now what we draw, all right, is this alcohol group stays on a dash. We're going to forget the implicit hydrogens here in our bond line. Now we brought, we took this group, we brought it down and we switched the wedges and dashes when we did that. All right. So the alcohol group is now on a wedge instead of a dash. All right. Over here, we didn't flip anything. So we keep it as is the alcohol groups on the wedge. hydrogen on a dash, but we don't draw our hydrogens. And then we add our CH2OH group at the end. And that's how you go from Fischer projection to bond line. And sometimes, you know, if bond line is your preferred way of looking at things and identifying, you know, RNS configuration, then this is the way to convert from Fischer projection to bond line in order to do that. Everybody has their... preference in regards to that. You can still do it here. All right. You just look at it and you can still do it in the Fisher projection. My preference and what I recommend is convert it to bondline. It's super easy. And then go ahead and approach your RNS configuration from there. All right. So that's all I have for you. All right. That was configurational isomers. So continuing objective two, and then we also covered objective three embedded within each section. All right, let's cover the important points again. We said configurational isomers can only be interchanged by breaking and reforming bonds. All right, we learned how to do RNS configuration based off of five rules. We also learned some quick tricks on how to do that if your fourth group is on a dash and when it isn't on a dash. Then we talked about E and Z forms as well. So if an alkane is Z, an alkane is Z if the highest priority substituents are on the same side of the double bond, and then they're E if they're on opposite sides. All right. And then we talked about enantiomers, diastereomers, and mesocompounds. And we learned how to distinguish whether two molecules we're comparing are enantiomers or diastereomers or even mesocompounds based off of a couple rules that we have written right here all right and then we learned how to convert fischer projection into bond line so that we can go ahead and just do our normal rns configuration as we have learned it All right. With that, we've covered our lecture on isomers. Now, if you want more information and practice, well, we'll do a practice problem video in the next lecture in this playlist. But also, I have an OCHEM1 playlist. Chapter 4 and Chapter 5 are everything that we covered in this lecture. And I do a lot of practice problems there. So if you really need more of a refresher on this content, all right, be honest with yourself. If you need it, go invest in that time. Watching those longer lectures, those more in-depth lectures with even more problems as well, so that you can definitely make sure you understand these concepts for the MCAT. All right, I'll see you in the next video where we do practice problems. Leave any comments, concerns, questions down below. Other than that, good luck, happy studying, and have a beautiful, beautiful day, future doctors.

Hello everybody, my name is Iman. Welcome back to my YouTube channel. Today we're continuing our lecture on isomers.

Where we left off is on configurational isomers. Now to talk about configurational isomers, we want to start with chirality. So objects that are not superimposable on their mirror images are called chiral objects. And the opposite of this is something called a chirality. Now an example of chirality is your hands.

If you place both face down on the table and then you try to take your right hand and place it over your left, you notice that they are not. superimposable. All right. Now, how do we take that and think about it in regards to molecules and compounds?

Well, in terms of identification for organic chemistry here for chirality, a chiral center is a carbon with four different groups attached to it. Your first step should always be to identify the number of chiral centers in a molecule. All right. And again, the way you identify that is by finding a carbon that has four different groups attached to it.

All right. So once again, an object is considered chiral if its mirror image cannot be superimposed on this original object. This implies that the molecule lacks this internal plane of symmetry. So chiral center is a carbon with four different groups on it.

This carbon will be asymmetrical, is going to be an asymmetrical core of optical activity, and it's going to be known as a chiral center. Now, two molecules that have non-superimposable mirror images of each other are called enantiomers. Molecules may also be related as diastereomers.

These molecules are... chiral and share the same connectivity but are not mirror images of each other. And this is because they differ at some but not all of their multiple chiral centers.

All right, and we're going to dive into enantiomers and diastereomers as well. But before we do that, let's take a look at some of the other enantiomers that we have. Before we do that, we now understand what a chiral carbon is, four different groups attached to a carbon.

Let's learn how to identify whether that chiral center gets a R or an S assignment. that requires for us to talk about R and S forms, which is part of objective three. So we're taking a pause from objective two and configurational isomers to go over R and S forms because they will play a really important role in us looking at molecules and trying to decide, are they enantiomers?

Are they diastereomers? Are they mesocompounds? We're going to learn what all that means. And being able to look at two molecules and compare and determine whether they're enantiomers or not.

their innate tumors, diastereomers, or mesocompounds, requires us to be able to, one, identify chiral centers, and then assign chiral centers R or S configuration, and then comparing and contrasting based off of that. All right, so let's cover R and S configuration. Now that we can identify chiral centers, carbons with four different groups attached to them let's learn how to assign the configuration of the chiral centers now at every chiral center you you will either assign r or s but what does that mean this is an attempt to relay information in the spatial arrangement of atoms at a chiral carbon center.

The Kahn-Inglod prelog rules will help us do just that in five steps. The first step is to identify the four atoms directly attached to the chiral center, all right, to that chiral carbon. The second rule is to assign a priority to each atom based on its atomic number.

The highest atomic number receives priority and the lowest atomic number receives a priority of four. This lowest atomic number priority is often a hydrogen atom. Now, step three says if you have a tie, if there's a tie, if two atoms have the same atomic number, then what you want to do is move away from the chiral center and look for the first point of difference. All right, one constructing list to compare. Remember that a double bond is treated as two separate single bonds.

when you are thinking and trying to determine priority between two groups that tie. So if there's a tie, move on to the next atoms attached until you can assign priority for all groups. Step four says, rotate the molecule so that the fourth priority is on it. Dash. Remember, dashes imply that that atom is going behind the plane of the page.

Wedge means it's coming out of the plane of the page. And then step five, determine whether the sequence one, two, three follows a clockwise order. If it does, it's R.

Or if it follows a counterclockwise order, then it is assigned S. All right. Now, the question here, all right, the question is here is, well, how do we begin to visualize what this means?

All right. How do we begin to visualize assigning priority, for example? And how do we begin to?

you know, assign priority? And then how do we begin to make sure that our fourth priority is always on a dash? How do we rotate molecules correctly to do that and then determine the one, two, three sequence? So for example, let's say we have a molecule like this, all right?

We have a bromine group, we have a hydrogen. We have a, let's see, a nitrogen here, and it has two methyl groups attached to it, and then that's it. All right. If we're looking at a molecule like that, where is the chiral? Is there a chiral center?

That's the first question. The answer is yes. Here's a chiral center. This carbon is chiral because it has four different groups attached to it. It has a bromine.

It has a hydrogen. It's attached to a nitrogen, and then it's attached to this carbon as part of a methyl group. All right, so we've found and identified a chiral carbon, a chiral center. Now what we want to do is begin to assign priority to each atom based on its atomic number, on their atomic numbers. So bromine here.

is going to have the highest atomic number. It gets priority one. It's followed by nitrogen for two, carbon for three, hydrogen for four. Cool.

Now, there's no ties here, thankfully. If there was carbons on both sides, for example, then you would move to the next. um, you would, you would look to what the carbons attached to, and then move out until you can identify a priority. So let's stick to this example and then we'll look at what if it was a carbon and we had a tie in priority. So this is what it is now.

Bromine one, nitrogen two, carbon three priority, hydrogen four priority. Now we step four says to make sure that the fourth priority is on a dash. Good for us here.

It is on a dash. So what do we do? We ignore it.

And then we draw from 1 to 2 to 3. All right, and notice how that is clockwise. All right, that is moving clockwise. So it gets assigned, that chiral center gets assigned R. All right, fantastic.

Now let's pretend, let's draw a different molecule. All right, let's draw something a little different. let's say this is a, let's do this. Let's say that this is a chlorine atom. This is hydrogen.

This is a methyl group right here. And this is a carbon that's attached to two other carbons. Cool.

Let's identify a chiral center. Here's a chiral center right here. This carbon's attached to four different groups, a chlorine, a hydrogen, a methyl, and then this group right here. All right.

So cool. That's a chiral center. That, that little dot that's highlighted, that little carbon that's highlighted in green, that's a chiral center. Now the second step is to assign priority.

Well, that's a chlorine, that's a hydrogen. This is a carbon right here. It's attached to three hydrogens.

This is a carbon as well. It's attached to a carbon over here, a carbon over here, and then there's also a hydrogen. Now let's assign priority. Chlorine obviously gets one.

Now we have a tie for two carbons. Hydrogen obviously gets four anyways. The tie here is between these two carbons which one gets to priority which one gets three so they're both carbons So what we want to do is make a list of what these carbons are attached to this carbon is attached to three Hydrogens this carbons attached to two carbons and and one hydrogen, all right?

Carbon beats hydrogen, all right? And so this carbon right here gets priority two, and this one gets priority three. Now, thankfully, again, the hydrogen is on the dash, so we ignore it, and then we draw arrow from one to two to three.

This is counterclockwise, so this gets, this chiral center gets an S assignment. Now, what if it's not so easy that the hydrogen is, the fourth group is not on a dash? How do we approach that? How do we begin to visualize? how to rotate a molecule appropriately to get a hydrogen on a dash.

Well, there's actually a little bit of a trick to this, all right? So there's three different ways that you can imagine how this could play out, and we're going to go over each of them and how to tackle that, all right? So we can identify chiral center, and we can, you know, assign... RS configuration, but for determining, you know, clockwise or counterclockwise, there are a couple hints to keep in mind.

All right. First is if your fourth group is on a dash, then you are so lucky. You don't have to, if your fourth priority group is on a dash, you're lucky, you just ignore it.

Goodbye, and you draw your arrow from 1 to 2 to 3 and determine if that's clockwise, it gets R, or if it's counterclockwise, it gets S assignment. Okay, that one is the easy one. What if your fourth group, all right, is not on a dash, but it is right next to a dash?

All right, look at this fourth group. It's not on a dash, but it's right next to a group that is on a dash. The three priorities on a dash instead of the four, but they're right next to each other.

What do we do? What we do is we still figure out our one to two to three sequence. All right, here's our one, here's our two, and here's our three.

All right. We determine our 1, 2, 3. We assign it. That's clockwise, so it gets R. And then we give it the opposite configuration.

We give it an S because our fourth group was never on a dash. All right, so that is the trick for that. If you're... Fourth group is not on a dash, but it's next to it. You can do this.

All right, and here's another example. Same thing. Your fourth group is not on a dash, but it's right next to a group that is.

So what you can do is still just draw your 1, 2, 3. That's counterclockwise. That would be given S. No, you give it the opposite because that fourth group was never on a dash, but it was near it. All right, so that's the trick for that. The last thing is, well, if you have a fourth group and it's not on dash, but it's right across from the dash.

So here's the fourth group. Here's the dash. What do you do?

Well, you just do your one to two to three sequence. All right. And just give it that appropriate configuration. So this is counterclockwise. It gets S and you don't worry about it.

And this is because to get the fourth group to the dash when they're across from each other, you're going to do two rotations and they'll ultimately cancel each other out in that way. All right. So those are the three tricks I always use if I don't know how to properly rotate a molecule. So this requires no 3D visualization on your part to appropriately rotate molecules, which could be way faster on a timed exam like the MCAT.

All right. So now we can take all this information we learned about chiral centers and RNS configuration to understand and identify stereoisomers. Stereoisomers. but there is one more thing one more thing that i want to cover before we do that before we start talking about what enantiomers are what diastereomers are and what meso compounds are we said we know how to find a chiral center that's a carbon with four different groups and we know how to do rns configuration at a chiral center with those four different groups but what about all right what about Carbon-carbon double bonds.

All right, what about cis and trans? That's an important discussion for us to have. All right, so how do we treat double bonds in regards to configuration?

Well, this is where we start to talk about E and Z forms. So E and Z nomenclature is used for compounds with polysubstituted double bonds. Recall that simpler double bond-containing compounds, they can use the cis and trans system. All right? But you can, you know, you might also be given or asked to use the E and Z designation.

All right. So how do you do this? Well, you identify where your double bond is.

So, for example, here's a double bond. All right. You figure out your highest priority groups. All right.

So for looking at these. Um, this is going to be, these two are going to be your higher priority groups here, the bromine and chlorine. All right, cool.

So you've determined your, your two highest priority groups. You've determined where your double bond is. You're going to draw a line through it and you've determined your two highest priority groups.

Now, if your two highest priority groups are on the same side, which they are based off of where you drew a line across through your double bond, if they're on the same side, which these two priority groups are on the same side. All right. This is given Z assignment. All right. But if we look here, here's our carbon double bond.

Our first, our first priority group is here in our second priority group here. And what we notice is that they're on. So opposite sides, if they're on opposite sides are two highest priority groups, then this is given an E assignment. All right.

E. And then you could proceed with the name. So this is how you do E and Z designation of alkenes. All right.

So this was also, again, a little jump into this. third objective, right? We're kind of embedding the third objective as we have a, what I think is a chronological order discussion on configurational isomers.

All right. So now we understand R and S, and now we understand what E and Z is as it relates to cis and trans. All right. Fantastic. So now we know how to identify chiral centers.

We know if we find a carbon with four different groups, that's a chiral carbon, and then we can designate it an R or S configuration based off of the rules we learned. If we have double bonds, remember we already know cis and trans. But we're going to be focusing on the E and Z designation here. If your two highest priority groups are on the same side of the double bond, then this gets a Z assignment. And if the two highest priority groups are on opposite sides of the double bond, it gets an E assignment.

All right, we'll get into more depth in relation to the importance of this. Really, for the rest of the discussion, we're going to be talking more about R and S configuration and how we can... use this idea to define enantiomers, diastereomers, and mesocompounds appropriately.

All right. So now we can talk about enantiomers, diastereomers, and mesocompounds. All right.

So your first step is always to identify chiral centers and then assign them as R and S. So let's begin talking about enantiomers. Enantiomers are non-superimposable mirror images.

They cannot appear identical simply by rotation. Now, how can you tell if mental rotation is difficult for you, right? If you're looking at these two molecules and you're trying to tell if they are enantiomers or not, and you can't mentally rotate it appropriately, especially assigning and changing the wedges and dashes appropriately, how can you tell if these two molecules are enantiomers or not?

Well, easy. We're going to rely on R and S configuration assignments to help. All right. So here's how it works for enantiomers if you have only one chiral center.

If you have one chiral center and you're looking at two molecules and you're trying to determine if there are enantiomers, first, you find that one chiral center for each of them. And then you're going to work and assign each one R and S. Let's pretend, okay, we're pretending here that this is R and this is S, okay?

I'm just pretending here, all right? This one's R and this one's S. They both just have one chiral center and one of them is R and one of them is S. Because they are opposite, all right, one is R and one is S at the chiral center, then they are indeed enantiomers, all right?

So that's if there is one chiral center. And you're always going to need at least one chiral center. center to even say or begin to investigate if two molecules are chiral centers.

And if there's just one chiral center, then it becomes easy. Assign the chiral center for each of the molecules that you're comparing, each of the two molecules you're comparing. If in one molecule that carbon center is R and the other one is S, then they are enantiomers because they have the opposite R and S assignment. All right.

If you're looking at both of these and they're both R, then they're just identical. All right. We're not talking about enantiomers. Now, there could be two chiral centers, right? You can have two molecules that you're comparing and you're trying to determine if they're enantiomers and they both have two chiral centers.

All right. In that case, if at all chiral centers between the two molecules, they have opposite R and S assignment. All right. Then they are enantiomers.

So let's pretend. all right here's a molecule here's a chiral center bromine chlorine, nitrogen. All right.

This one's on a wedge. This one's on a dash. And then you're comparing the same one here. This one's on a wedge.

This one's on a dash. I'm not going to assign them here. I just want to draw them.

All right. And this is a carbon. If this is a carbon and it has another chiral center.

So let's do that. This is attached to a carbon. This is a hydrogen. This is a fluorine.

All right. So here, We have two chiral centers. Let me draw them out appropriately. Alright.

I'm just drawing stuff so I can get the point across. We're not going to actually investigate them. I'm just trying to get the point across.

So let's do this as a wedge. That's bromine-chlorine. This one has a hydrogen and a fluorine.

So these two molecules, we're comparing them. We want to know if they're enantiomers or not. They have two chiral centers here and here, here and here.

So they have the two same chiral centers. If we were to determine that at this chiral center, you know what? Let me change their colors.

Alright, this one's red and this one's red. because they're the same and this one's blue and this one's blue if we decide if we were to figure out what the rns assignment is here and we decided that this is r and we looked at the other molecule at the same chiral center we determined this is s and then we did the same thing with the other chiral center this one is um S and this one is R. All right.

Now, because they have because the two chiral centers between the two molecules, because all the chiral centers between the two molecules have opposite R and S assignment at those positions, then they are enantiomers. If they were both identical, like RS for this one and then also RS for this one, these are just going to be identical. Now, when you have two chiral centers, you also have to worry about diastereomers.

You also have to consider if they're diastereomers, which has a different rule. The rule that they're opposites here is what gives it away that they're enantiomers. All right.

So that's how you use R and S configuration to determine whether, all right, two molecules are enantiomers, whether they have one chiral center or two chiral centers or more. All right. Fantastic.

Now, before we get to diastereomers and mesochromosomes, compounds. I want to just scroll over here. All right. And side note on enantiomers, that's going to be important for the MCAT.

Now we said enantiomers are superimposable mirror images and thus have opposite stereochemistry at every chiral carbon. All right. They have the same chemical and physical properties, those enantiomers, except for one thing.

They differ in regards to how they rotate plane polarized light. All right. They have the same chemical and physical properties except for rotation of plane polarized light and reactions in a chiral environment.

Now, optical activity refers to the ability of a molecule to rotate plane polarized light. All right. So you can have a compound that rotates the plane polarized light to the right.

All right. It can rotate the plane polarized light to the right. Or you can think of this as clockwise.

All right. This is dextrorotary. All right.

So that's assigned D. All right. And it's usually labeled as a positive.

You can have the light be rotated to the left. This is called counterclockwise. This is levorotatory.

All right. And this is identified as L minus usually written or is just labeled as minus. The direction of rotation can't be determined from the structure. of a molecule, you have to actually determine it experimentally. And so that is, it is not related to the absolute configuration of the molecule.

Now, the amount of rotation depends on the number of molecules that a light wave encounters. All right. This depends on two factors, the concentration of the optically active compound and the length of the tube through which the light passes.

Chemists have set standard conditions of one gram per milliliter for concentration and 1 decimeter for length to compare the optical activities of different compounds. Now, rotations measured at different concentrations and tube lengths can be converted to a standardized specific rotation using this equation right here. This is the specific rotation in degrees. All right, this is the observed rotation in degrees, C is the concentration, and L is the path length. All right.

Now, if you were to have a mixture with both positive and minus enantiomers in equal concentrations, they form what's called a racemic mixture. In these solutions, the rotations cancel each other out and no optical activity is observed. All right.

Fantastic. So that was an important thing that I wanted to mention, all right, as a side note for enantiomers. All right, now let's also go ahead and define diastereomers.

So diastereomers are non-superimposable. And not mirror images. So in other words, non-identical stereoisomers.

Now, here's the key for thinking about diastereomers. They're going to have the same molecular formula. They're going to have the same connectivity. and they're going to at least have two chiral centers.

All right. So unlike enantiomers, all right, going back to enantiomers, the requirements were, of course, that they have the same molecular formula and the same connectivity. Those are the two conditions of stereoisomers. But enantiomers needed at least one.

chiral center. And then you assign RNS to each and compare positions. If at all chiral centers between the two molecules they have opposite RNS assignment, then they are in anteriors.

If all are identical, then the molecules are identical. Now when we think about diastereomers, here's the key for that. You have the same molecular formula, you have the same connectivity, you need at least two chiral centers.

And how you approach this is you're going to assign RNS to each All right, and then compare positions. So let's look at these drawings right here. All right, let's draw, let's look at these drawings right here, and then let's begin to understand enantiomers and diastereomers. The rule for enantiomers was if they have opposite R and S configuration at each carbon center, chiral center that you're comparing, then they are enantiomers. But with diastereomers, If at some chiral centers that you are comparing between two molecules, you have a mix of the same R and S assignment and opposite, then they are diastereomers.

So let's try to do this, all right? I am going to hypothetically assign R and S configurations at these, just so that we can get the point, all right? So this is R, and let's say that this is also R, for example.

Okay, so these are both R. Let's say that this is R and this is S. Let's say that both of these are S.

And then let's say that this is S and this is R. Okay, now look let's look at this point in this molecule and then this chiral center, right? These are the same chiral carbons in both of these molecules.

This one is R and this one is S. Cool. They're opposites at that chiral center. Now let's look at. this chiral center for these two molecules.

All right. This one is R and this one is S. They're also opposite at that chiral center.

So both of these chiral centers we looked at in these two molecules have opposite R and S assignment. This one had R, this one had S at the same chiral carbon. At this same chiral carbon, this one had R, this one had S. Opposites, opposites at both of those chiral centers. That means these molecules are enantiomers.

All right, cool. Let's look at these molecules now. I'm going to circle them in blue.

We're just looking at them as we choose, all right? Let's look at these two. If we look at this chiral carbon at both of these molecules, one is S, one is R. That's opposite.

Cool. If we look at the second chiral center, oh, this one's R, this one's S. At both chiral carbons, they have opposite R and S assignment.

This one's S, this one's R. This one's R, this one's S. Opposite at...

the same chiral position, the same chiral carbon. These two are also enantiomers, all right? So these are enantiomers, these are enantiomers.

Cool, so we understand enantiomers now. If there is opposite R and S at the same chiral center, for all the chiral centers that you're looking at, all of them have opposite R and S configuration, however many your molecule has, whether it's one, two, three, or more, all right, then that defines... finds in the antiumers.

Now, what about diastereomers? All right, so let's do some erasing. All right, let's do some erasing.

All right, these are RR. All right, let's look at these, these two. All right, let's compare them.

At this chiral carbon for both of these molecules, they're assigned R. They're the same. Cool.

If we look at the second chiral carbon in these molecules. One is R and one is S. So at one chiral center, they're the same. At another chiral center though, they're the opposite.

This mix of same and opposite is what defines diastereomers. So these two molecules molecules, if we were looking at them, we would define these to be diastereomers. All right. So that is the distinction between Indian tumors and diastereomers based off of the R and S assignments.

Now, one more note on diastereomers. Cystrans isomers are a subtype of diastereomers in which groups differ in their position about growth. that double bond, about that immovable bond. So going back to our priority rules, an alkene is Z if the two, if the high priority substituents are on the same side of the double bond, and they're E if they're on opposite sides of the double bond. All right, so that's enantiomers, that's diastereomers.

Fantastic. But wait, there's one more thing you have to be aware of. of when you're looking at stereoisomers for molecules that have reflectional symmetry, and that is mesocompounds.

Mesocompounds are chiral compounds that have multiple chiral centers. All right, they have a least two chiral centers. All right.

Meso compounds have, excuse me, they have an internal plane of symmetry. So they will be optically inactive because Because the two sides of the molecules cancel each other out. Now, I'm going to show you how to identify mesocompounds.

All right? I'm going to show you how to identify mesocompounds. This is the rule.

If you have at least two chiral centers, all right? If you have at least two chiral centers and you have an internal plane of reflection, you're going to have a mesocompound. plane of reflection, all right, if you have an internal plane of reflection, and then you have also, all right, if your molecule has reflectional symmetry and at least two chiral centers, then the pair of quote-unquote enantiomers with R and S assignment on one and then S and R on the other that you had thought was an enantiomer pair is actually a mesocompound. They are simply rotations.

of each other. So for example, look at this molecule. It has two chiral centers. It also has an internal plane.

of reflection. Now, if you go ahead and assign these, this might be R and this might be S, and then this might be S and this might be R. So you notice that they're opposites here, and then they're opposites here, and you might think enantiomers.

But because there is an internal plane of reflection and you're looking at two chiral centers, these molecules are actually mesocompounds. They are the same molecule, they're just simply rotated. All right, so that's the one catch. All right. You want to be careful because there's an internal plane of reflection.

So what you thought was an enantiomer in this case is not. All right. So let's put all these rules together so that we have kind of a workflow for this. All right. If there is one chiral center, all right, and they have opposite RNS configuration, there are enantiomers.

If they have the same, then they're identical. That's with one chiral center. If you have two chiral centers and they're opposite RNS configuration at every chiral center, then they're enantiomers.

If there are some opposite and some same RNS configurations, then they're diastereomers. And if they're all exactly the same, then they're identical. Now, the only thing is if there's reflectional symmetry, double check for meso compounds.

All right. Fantastic. Now, there's one more last thing that we want to cover, and that's Fischer projections. All right.

Vertical. So this is. right here, a common depiction of molecules you'll see in biochemistry. All right.

Sorry, that took, I had a little brain fart there. I was like, where do they see this? biochemistry. All right.

Now, if you were trying to take this and convert it to say bond line, all right, take this Fisher projection, convert it to bond line. That's going to be an important skill to have. All right.

And it's going to be important to understand the workflow for this. All right. Now, here's how I do this.

Here's my recommendation. All right. We're going to identify these as the ends, and we're going to go.

ahead and draw this kind of structure right here. Kind of looks like a table top. All right.

We're going to put this on one end and put the other group on the other end. All right. Top group on the left, bottom group on the right.

Cool. Then we're going to see how many carbons are in between those two groups. Three.

So we're going to draw one, two, three points. All right. One, two, three points.

Cool. Then what are we going to do? We're going to draw the groups that are attached to each carbon.

So this is going to be numbered our first carbon. This is our second carbon. And that's our third. One, two, three. All right.

Now anything on the right gets assigned a dash. So what's on the right at carbon one? An OH group. So we draw it on a dash. Anything on the left gets a wedge.

So the hydrogen gets a wedge. Look at carbon two. What's on the right?

OH group, alcohol group, we draw it on a dash. What's on the left? Hydrogen, that goes on a wedge. Alright, keep that rule. Alright, third carbon, what's on the right?

Hydrogen group, that goes on the dash. What's on the left? Alcohol group that goes on the wedge.

Cool. Now, your bond line is not going to look like this. It's going to look more like a zigzag of carbons, right? Because that's the format for bond line.

So we don't want to keep it like this. We want to keep this part, but then we want to bring it down here. We want to take this group down here and then we can go back up and then it looks like a normal bond line. So what we're doing is we want to bring this group to a downward position.

All right. To do this, we're going to switch these two groups in. their assignment of dashes and wedges. So now what we draw, all right, is this alcohol group stays on a dash.

We're going to forget the implicit hydrogens here in our bond line. Now we brought, we took this group, we brought it down and we switched the wedges and dashes when we did that. All right. So the alcohol group is now on a wedge instead of a dash. All right.

Over here, we didn't flip anything. So we keep it as is the alcohol groups on the wedge. hydrogen on a dash, but we don't draw our hydrogens. And then we add our CH2OH group at the end. And that's how you go from Fischer projection to bond line.

And sometimes, you know, if bond line is your preferred way of looking at things and identifying, you know, RNS configuration, then this is the way to convert from Fischer projection to bond line in order to do that. Everybody has their... preference in regards to that. You can still do it here. All right.

You just look at it and you can still do it in the Fisher projection. My preference and what I recommend is convert it to bondline. It's super easy. And then go ahead and approach your RNS configuration from there. All right.

So that's all I have for you. All right. That was configurational isomers.

So continuing objective two, and then we also covered objective three embedded within each section. All right, let's cover the important points again. We said configurational isomers can only be interchanged by breaking and reforming bonds.

All right, we learned how to do RNS configuration based off of five rules. We also learned some quick tricks on how to do that if your fourth group is on a dash and when it isn't on a dash. Then we talked about E and Z forms as well.

So if an alkane is Z, an alkane is Z if the highest priority substituents are on the same side of the double bond, and then they're E if they're on opposite sides. All right. And then we talked about enantiomers, diastereomers, and mesocompounds. And we learned how to distinguish whether two molecules we're comparing are enantiomers or diastereomers or even mesocompounds based off of a couple rules that we have written right here all right and then we learned how to convert fischer projection into bond line so that we can go ahead and just do our normal rns configuration as we have learned it All right.

With that, we've covered our lecture on isomers. Now, if you want more information and practice, well, we'll do a practice problem video in the next lecture in this playlist. But also, I have an OCHEM1 playlist. Chapter 4 and Chapter 5 are everything that we covered in this lecture. And I do a lot of practice problems there.

So if you really need more of a refresher on this content, all right, be honest with yourself. If you need it, go invest in that time. Watching those longer lectures, those more in-depth lectures with even more problems as well, so that you can definitely make sure you understand these concepts for the MCAT.

All right, I'll see you in the next video where we do practice problems. Leave any comments, concerns, questions down below. Other than that, good luck, happy studying, and have a beautiful, beautiful day, future doctors.

Transcript for:İzomerler Hakkında Detaylı Notlar

Transcript for:
İzomerler Hakkında Detaylı Notlar