And we're going to tell you I just link them out like this. Yes. She says, Good morning, thank you for coming out today for the benefit of anybody watching this video at some future time. >> Today is Friday, September 30th, 2016, and it's raining outside significantly. >> Where has the engineering department is going to hold meeting this morning to discuss building an ark. Ok, that's not prove it, but I do appreciate that everybody resisted the temptation to curl up under the covers, especially secure in the knowledge that they can watch the video later. >> And especially considering that it's Friday and we're talking about stuff but will not appear on Wednesdays, but a little appear odd November's exam. >> So never a bad idea. >> We're gonna talk some more about chiral molecules today. I'll add news report This just in one question we sorta kinda brought up last time, but just really dive into detail about why should the nature handedness, right? For example, amino acids, most of them are the ifconfig, the left-handed form. Why should that be like, could nature decided to go with a well, nobody really knows what the thinking has been for a long time that how chirality got started here on earth is that it came from somewhere else to chiral molecule found in space. >> Maybe a billion years ago, some meteor hit got a few chiral molecules. >> And Peri, argue that anybody wants a copy of this article, the epidemic. Now, before we get back into the subject matter of chapter six, we didn't have a request over one problem from the textbook. >> That request came in by email as problem 3.37 and specifically Part a, and this is fair game for Wednesday. >> The exam problem 3.37 says draw a Newman projections for the most stable conformation of each of the following compounds, looking down the indicated pot. And the question was specifically about Part a. >> And I think this is instructive. >> Do they think people struggle with how to draw a Newman projections? So let's just take a moment and do this. >> What I've done here is to draw this just about exactly like it looks from the textbook. >> And the point is, when it says looking down the indicated by, I'm going to draw this arrow, but you need to do this. Kind of imagine that you're a little stick guy standing right here looking down that arrow and specifically looking down the bond between these two carbon atoms right here. Now we know that the draw, a Newman Projection, you'd have to draw a circle. And what the circle represents is the front carbon atom from the perspective of the observer. So for my little stick guy here looking down this bond, this carbon is in front, closest to stick guy, point of reference. The carbon atom at the other end of that bond that stick guys looking down, I'll refer to as the back carbon app. >> Now to draw a Newman projection, we start by drawing three lines looking something like that. >> And the lines that go all the way to the middle of the circle represent the bonds attached to the flank permanently in this molecule. What three things that are attached to the front carbon atom? Now, for the moment, never mind the back carbon atom. >> Nevermind anything attached to the back carbon atom. >> Let's look at the carbon atom and tell me what three things are attached to it. Now, obviously this is going to bond line. >> Notice that you're not comfortable with bottom line notation. >> It might be difficult to ascertain what three things that hopefully you've gotten comfortable with. Bible I notations. But if not, you can always pencil in more detail in the drawing. So they might tell me what three things are attached to the carbon. >> And hydrogen is one and how many other carbons? >> To me, it is a good idea to pencil in the carbon atoms at the end of this line. There's a methyl group. At the end of this line, there's a methyl group attached to this carbon atom. >> There is a hydrogen atom. >> Remember, carbon must have four box. >> So whenever necessary, pencil in hydrogen atoms to make four bonds. >> So with those three things being attached to the front carbon atom, we can begin our drawing by penciling in two methyl groups and the hydrogen pretty much wherever we want. >> Now I have to do is worry about the back carbon out. >> The problem asks us to draw the most stable confirmation of this molecule. >> Is that going to be staggered or eclipsed. >> Staggered, which means we put in the bonds that end at the perimeter of the circle, which represent the bonds on the back carbon atom. We're going to put them in like this. >> So they are as far apart as they can possibly be from the bond from the carbon atom. >> That's what a staggered conformation looks. >> Okay, forget about the front carbon atom. >> Now focus on the backlog that what three things are attached to the back? >> Permanent yet has a carbon to hydrogen >> This carbon right here, which has three other methyl groups attached to it, and these two hydrogen atoms right here. And if you want, I'll pencil and the rest of this to show that what that other thing attach there is, is a tert-butyl group. So to draw the most stable conformation of the molecule, which of these three things would you like to put the tertiary good group on? Well, let's put it this way. Whereas the one place we shouldn't put the tertiary hooked it up here between the two methyl groups, right? We have no choice but to put it somewhere near one of the methyl groups arbitrarily, I will put it down here. >> But if you had chosen to put it over here, that would be okay. >> Then pencil and the hydrogens, because the logic makes sense. >> More importantly, do both pictures make sense? >> Okay, that's what a Newman Projection is trying to show. >> So if the original picture is unclear, draw everything out and then focus on which is the carbon atom, which is the back carbon atom, and grow your pictures appropriately. >> So please feel free to try the rest of that problem and we'll see how it goes. Alright, so back to conquer six. >> We were talking last time about enantiomers, molecules like these to you that are mirror reflections of each other, but not the same thing. >> Therefore, stereo isomers of each other, but specifically stereo isomers that are mirror images of each other, which is what the word means. And we were talking a little bit last time about the thalidomide situation. A little bit is sort of the classic example of why it's important when drug manufacturers markets some new pharmaceutical to the public. >> But they have not only the right compound, but the right enantiomer of the right compound in the bottom. >> So it's important to be able to separate enantiomers and identify which one is the good stuff. But that's the problem is, as we were suggesting last time, separating enantiomers. And not necessarily the easiest thing to do because he manteia verse for the most part, have identical physical and chemical properties. Say boiling points, say melting points, same solubility in water, same solubility of most other solvents, but there are a couple of differences. One is how they interact with other chiral species. The example we mentioned last time is that a left shoe, which is chiral, fits better on the left split than it does on the right foot. >> Your feet are basically isomers of each other. >> But obviously the interaction or the chiral shoot, but those two anti-American p is going to be different. And the other thing that we're going to talk about for starters today has to do with a phenomenon called the direction of optical rotation, Section 6.8. In your textbook, the leading question is, how has chirality detected in the laboratory? And the answer is, using an instrument that looks something like this called a polarimetry But we'll describe what's going on inside a perimeter. >> Let me call your attention to page 50 in the lecture notes. >> You've probably seen this kind of sinewave picture used to represent the beam of light before. This sort of thing makes it easier to talk about concepts like wavelength and frequency. >> And presumably you've heard about this stuff in your general chemistry course, if not before, light is sometimes called electromagnetic radiation because it's composed of electric and magnetic fields whose intensities oscillate as a function of time. >> The oscillations are represented by sine wave pattern here. >> However, if you look down a beam of light and actually see the electric and magnetic fields oscillating, you would find that they oscillate in all direction. >> So what I've tried to represent here with this circle, which is supposedly a beam of light hitting you in the eyes. And the arrows going every which way represent the electric and magnetic fields oscillating in all directions. So this is an example of normal. But if you now pass that light through a polarizing filter, and this doesn't have to be anything particularly exotic, the lens of an ordinary pair of sunglasses. >> And it'll do the job very nicely. >> What comes out the other side is what's referred to as plain polarized light, in which we have filtered out the light that oscillates in all directions except one. >> And arbitrarily I've got the one that comes through going up and down. >> So the point is now all the oscillations are in one plane. And the interesting part is what happens if you pass a beam of plane polarized light through a test tube that contains a single of a chiral molecule will worry about how we got that single enantiomer into the test tube later. For right now, it's about possibility. >> This is our inquirer eaten arbitrarily chosen. >> And the point is that we have a test tube full of R2 chloro butane, and we pass a beam, a plane polarized light through it. What comes out the other side is still plain polarized light. >> But you'll notice that the plane of polarization has been rotated. >> Specifically, it's been rotated clockwise by drawing probably is not exactly to scale. But if you do this with actual R2 chloro butane, and it gets rotated 23 degrees clockwise or 23 degrees to the right. The point is the fact that chiral molecules rotate the plane of polarized light at all is that chiral molecules are referred to as optically active. When I mentioned the optical rotation up here, that's what I'm referring to. The phenomenon in which you pass a plane polarized light through a test tube that contains a chiral molecule. >> Or the plane of polarization rotates. >> Fancy terminology, since in this case, we rotate the plane of polarization at 23 degrees to the right. The fancy word for any compound that rotates the plane for the right is dextro reliquary and Dexter rotatory compounds get a positive sign on the degree of rotation. >> So positive 23 degrees year BY that rotates the plane of polarized light 23 degrees to the right. >> And if therefore, extra concepts make sense so far. >> Ok, now suppose we did exactly the same experiment with one exception. >> Suppose instead of R two chloro butane, we use f control butane instead. What result would you expect? >> Yep, rotates at 23 degrees to the left. >> And that is exactly correct. >> Enantiomers differ indeed direction, but not the magnitude of optical rotation. >> So wanting and Hebrew are rotates the plane of polarized light 23 degrees to the right. >> The other enantiomer rotates the plane of polarization 23 degrees to the left. >> And the fancy word to describe that is to say that that molecule is Evo, rotatory. So one way we can tell we came anti-matter is which is put it into a polar rubber tube and see which way it rotates the plane for boy, let me call your attention to the equation that appears on page 176 in your textbook. I want you to understand but ignore it is this equation right here, which refers to the specific rotation on the symbol that's used for specific location is the Greek letter alpha in blankets. And what you can see from the rest of this equation is that when you conduct an experiment like this, you observe a particular rotation in degrees. But the rotation that you observe is complicated by a couple of factors that play with just noticing as I look at this and this addition. But somebody that not a spell a word like somebody, I could call him up on that. But anyway, the point is they've divided by a couple of terms here, length and concentration. And here's what that comes down to. >> Picture my books here and move it up. So you can actually see that the length is referred to as the path length through which the light has to crap. >> In other words, if you're doing an experiment like this and you pass the beam unpolarized light through the test tube and it rotates at a certain number of degrees. >> That shouldn't be too surprising to find that if you use a skinny little test tube like this one, you're not going to get as much rotation as you would if you use a bigger test. >> Do you like this one? >> Because we pass the being through more of the material, it's going to have more of a chance to interact with it and rotate that report degrees. >> The other is if you have a relatively dilute solution of this stuff in the testing, It's not going to rotate very much. Have a more concentrated solution of the deaths due that'll know, teeth, etcetera. But the point is, these are a couple of factors you can compensate for to calculate something called the specific optical rotation for any particular competent that we promised, you know, math in this course. He admitted he should understand that it's possible to determine the specific optical rotation where any compound once you compensate for concentration and pathlength. But having understood that, we are not going to ask you to calculate anything like that. >> You'll ever see a problem at the end of the chapter in your textbook when you're tempted to get out a calculator, just ignore that problem. >> Couple of other examples of how this works. In the previous examples, we were using it to chloro butane as our combat. >> What happens if we try with two chloro propane instead? >> The answer is Nothing, because is this a stereo center? O? It is not definition of a stereo set. >> It is an sp3 hybridized carbon atom with four different substituents attached to if any two of them are the same, like the two methyl groups are here. But that's not a stereo set. And in general, compounds that do not have stereo centers are not chiral. So this is an eight chiral molecules that were just talking about the fact that optical activity is a phenomenon that we normally associate with chiral molecules. A, chiral molecules don't do this. If you take a beam, a plane polarized light, and pass it through a sample that contains a chiral molecule, you will not observe any optical rotation, and such a compound is said to be optically. Last example along these lines, suppose we take our two core of butane and F2 portal from the first few experiments and mix them together, 50-50, and then put it in a test tube and run a beam, a plane polarized light through it. Well, maybe not too surprisingly, such a mixture is also optically, even though every molecule in the mixture is chiral, The problem is that half of them rotate the plane of polarized light to the right, and the other half rotate the plane of polarized light to the left. So by the time all of the great rotating molecules and all the electrical getting molecules do their thing. It all cancels out and comes back to 0, a 50-50 mixture of two enantiomers that is optically inactive is referred to as a receive, a mixture of two enantiomers or sometimes just a soulmate. And we will come back to this concept, would find that on the course. But suffice it to say that such a mixture is also optically. Are there any questions about any of the examples that you've seen so far? Alright, so the point is, once you separate a mixture of enantiomers, and we'll talk probably next class period, not Monday, which is a review session for you lecture, which is next Friday about how one goes about actually separating and expert in afterwards. >> But the point is, once you get separated from each other, you could use techniques like this to figure out which is the good stuff and which was not too based on. >> One more point to be made here. And I touch on it briefly here on page 50, but I don't explain the word, so I probably should. In this example, r turns out to be plus Dexter rotatory, and Es turns out to be minus medieval literature. That is true for some compounds. >> Then again, there are other compounds where it's the other way around. >> In fact, there's an example that appears on page 176 in your textbook. >> Here is an S alcohol that happens to be Dexter rotatory, and therefore it's R enantiomer is over. And the point is that to be able to determine that the optical rotation of a compound, you have to do the polarimetry experiment. You can look at the structure of a molecule and figure out whether it's R or S. But that by itself tells you nothing about which way the molecule rotate between R and S is just based on rules that people made up. The molecules don't bow, whether they're ROS, but they do know how they interact with plane polarized light. So just by looking at this structure, that you would not know necessarily that this was a extra. What you would know is that if this was a Dexter rotatory molecule, which is that it's a manteia. We're must be legal rotatory. And the two of them rotate the plane of polarized light the same number of degrees, just anoxic. So R and S and plus or minus related concepts, but one does not tell you what the other one. >> All right, Let's consider this got bound to three guy Bravo pentane stereo centers are there in this molecule and not that end up with the answer to the definition of a stereo center, an sp3 hybridized atom with four different things attached to it, which means carbon number two. >> In the stereo setup, the four different things are Bromine, methyl hydrogen, and the rest of the molecule. >> Likewise, number three is stereo center bromine ethyl hydrogen, Romo Ethnologue. If the others are not, because they all have at least the lighter side. >> So the point is, since any stereo center, it could be either R or S. >> And we have two such various centers and discount them. >> How many different possible stereo isomers exist for this molecule? >> Yes, for the right answer. >> An analogy is sort of like flipping two coins. >> Suppose you're flipping a quarter and Nicole, you know, they can both land either heads or tails. >> How many possible outcomes other? >> Well, they could both land heads, they give up land tails and you could get one head and one tail. But there's two different ways to do that. Order can land heads, tails, quarter gland data with nickel benign. >> And same thing here. >> Here are the four possibilities. >> One possibility, both stereo centers could be r. >> And for that particular stereo isomer, the full name of the compound is two r, three r 23 dB robo painting >> But the possibility they could about the S two S 3S do through that grandma. >> And the other possibilities are 2R 3S, n2, S3 r two r three r as a mirror image. >> The mirror image is to S3? >> Yes. >> These two are enantiomers of equipment. >> That's to demonstrate for a moment that within a workbook that right, one stereocilia. >> Okay. >> The mirror image of what? Right and good. >> These various entries. >> Mirror image on the right hand and a right foot, left foot. >> In other words, to arrive at the enantiomer of a compound that contains more than one stereo center that's upside down and backwards, who've dominated the other picture directly caused them up anyway. >> Okay? Anyway, if the compound has t stereo centers and you'll want to know what its enantiomer looks like. You have to reverse both of them, which is why 2R 3S has as its mirror image to SQ3R. >> So at 2R, 3R to S3, S3 enantiomers of each other and to R3 S into SCR enantiomers of each other. What's the relationship between two R three R and two R three as well. >> They are certainly not enantiomers because the mirror image of two or three hours to F3, F2, Well, they are stereo isomers of Egypt because all of these are stereotypes. >> So the word we're going to introduce at this point to describe this concept is to say that the relationship between 2R, 3R, 2R, 3S di, hysteria words of each other. >> The word enantiomers refers to stereo isomers that are mirror images of each other. >> Where dietary reverse refers to stereo isomers that are not merely or even more simply, stereo isomers are not even, which means that two S three are NTS three F's also constitutes a pair of dice dairy and for that matter, 2R, 3S, n2 S 3S constitutes a fair die. Is there any pair of stereo isomers of this list that is not a pair of enantiomers must, by definition, if error bias, okay, with what the word means, high. We've actually seen this concept before. >> We just didn't use the word dietary embers of time. >> But this would also apply to any molecules that are cis trans isomers. >> For example, s2, the team versus trains to be not the same. >> Obviously stereo isomers, I hope, and not mirror images of each other, so they're not making us. Therefore, by this definition, those do a dice Same deal for CIS one-to-two di methyl cyclohexane versus Franz wanted to play cycle. >> Not the same thing, not mirror images of each other are stereo isolated, so it must be nice. >> So this concept applies to assist cranes isomers like these, but it also applies to it like this. >> What you're comparing two things that are not still making sense. >> Some people are starting to Jake Knapp that they wish they had taken this morning. >> But 20 more minutes hang in there. >> Here is two portal waterfall model six I80 hoc take ONE stereo centers either at this moment at the right end. >> And it gave me a hint by putting asterisks on them. By the way, there's a number of problems at the end of chapter six in your textbook. But I'll ask you to look at a molecule and put asterisks on the stereo centers. >> Okay? Same kind of thing. >> I do want to point out one thing, because this is one place where people get lost. >> It should be obvious that carbon number two is a stereo center. >> The four different things attached to it, or chlorine, methyl hydrogen, whole rest of the molecules. Let's let people get a little bit stuck on line number four is a stereo set. And I think the reason they get stuck is that you have two CH duties there. Well, yes, you do. But the problem is that you have to look at the entire rest of the molecule, the four different things attached at number four or bromine hydrogen, three carbon piece of the chlorine in it for carve a piece of that iodine in it. That's for different things. >> So that is indeed a stereos. >> And likewise, number six has iodine, ethyl hydrogen, this guy. So if each stereo center could be either R or S, how many stereo isomers are possible for this? >> Yeah, yeah. >> Now you're flipping three coins. Quarter, nickel, dime. One possibility. They all land heads, RRR, they've not talk like a pirate. >> Pages it today. >> Second possibility, two heads and one tail. >> But you could do that three different ways are r, S, r, SR. >> Alright. >> Third possibility two fails and one head SFR, SRS, RSS, final possibility all fails. >> Fss total of eight possible stereo isomers. >> You should be able to pick out four pairs of a nanotube learners from this list. The enantiomer of our RS is SSR. >> Let me see if I can do this right side up. >> It's been outward mirror image on the right hand, the right foot, left foot right here, that are reversible. So there's one pair of enantiomers. I'll let you see if you can find the other three. But the main point is once again, any pair of stereo isomers from this list that is not a pair of enantiomers must constitute a pair of dice stereotypies such as RRR at r, R S RS And r SR. >> And there are many others that there's a pattern that's starting to emerge here. >> And we only had one stereo center each molecule. And it's easy, there's only two possible stereo isomers. What's r one? >> This? >> Then we looked at the case where we had two stereos and urge which gave rise to or stereo isomers. And now we add this molecule with three stereo centers. Ate cereal isolates, and a typical molecule that has four stereo centers. How many stereo isomers would you expect? >> They then 165 stereo centers, 32. >> Every time you add another scenario center, you get twice as many possibilities. >> But if you're putting one more point, and in general there's a rule called the two to the nth power rule, which says that if a molecule contains n stereo centers than the maximum possible number of stereo isomers the molecule can have is given by two to the nth power. >> This becomes pretty impressive after awhile. >> Molecule with ten stereo centers, 11024, possible scary laziness. >> And some molecules have more than that. But I do want to point out one other thing. There's a reason for the phrasing maximum number here because it turns out that in some cases, molecules have fewer stereo isomers than what is predicted by the two to the nth power. And to that end, what I want you to do for a moment, consider this molecule. >> That's a little bit better view of what's going on. >> This is q3, dy, bromo butane, four carbons. There's the butane. The red balls represent the bromine atom is the white balls represent the hydrogen atoms. How many stereo centers? This my YouTube one here, the four different things attached or methyl hydrogen bromine, Western molecule, yellow ones here, bromine, hydrogen, methyl molecule. So according to the two to the nth power rule, how many stereo centers, how you stereo isomers this molecule have fought. >> Turns out, though it actually only has three. >> Here's watch, here's a molecule. >> In principle, the same four possibilities as we saw before for 23 diaper. Although painting and many of the same relationship still apply, 2R, 3R and 2S 3S are still enantiomers of each other. >> 2r, 3R and 2R 3S are still by hysteria because each other and so on. The difference is the relationship between 2R, 3S and 2S, 3R. >> And to show you where this is coming from, I want to thank my original to eat anterior molecules here. >> This is going to take more work than I thought. So bear with me for a minute while I remove the green balls. >> And then I'm gonna have to reconstitute this because of what fits with what? >> One moment. >> Alright. >> Alright, there we go. >> So I guess piece them together to make this map. This is what I just bill you tell that by the blue methyl groups there. Compare this with the one I showed you a few moments ago with the black methyl groups. What I want you to do is look at both of this other stuff out of the way. >> It's like that because even if both of these, and other than the fact that one has blue methyl groups and other has black methyl groups, are these two mirror images of each other? You put a mirror right there. >> The mirror image should look like that. But hey, don't take my word for it. Go home. >> All the models get out a mirror with an arrow next to it. >> But now comes the interesting question. Are these two the same thing? >> Absolutely. Now the point is enantiomers or mirror images that are not C and not the same thing because they're not superimposable on each other such that everything lines up. But in this case, the two mirror images actually can be superimposed on each other so that everything lined up. >> So these two are in fact the same. >> So as opposed to the previous example where 2R, 3S into S3 are mirror images of each other. >> In this case, they're the same thing. >> And there's a reason that that's true. >> What does T3 dive robo butane have? That T3 diaper? >> All my painting does not say again, there was a plane of symmetry right here. >> Or if you prefer to see it on the model, I leave my hand to represent a plane of symmetry. >> The point is, what's laptop reflects exactly into what's on bottom. Definition of a plane of symmetry with a plane that divides the molecule into two paths, such that 1.5 is the mirror image of the other. And that's exactly what we have. >> So the model I was just showing you, there's an example of something called a mezzo vaso. >> Compounds are compounds that have stereo centers but also have a plane of symmetry. >> The point is, whenever you have the possibility of having a vaso compound that reduces the number of possible stereo isomers that's predicted by the two to the nth power. >> This situation is typical our three possible stereo isomers. >> But this model, if you are to R3 art, to S3 S, which are enantiomers of each other. >> And the mezzo compound, which you can think of as either 2R, 3S or to S3 are, it doesn't matter because there's the same thing, mirror reflections of each other that happened to be superposing. >> But the point is you would not count this twice because it's really just two iterations of the same thing. So three possible stereo isomers here, R, R, R SS, and made them vaso compounds by definition are a chiral because the definition of a chiral molecule or a chiral, anything else, it's something that is not able to be superimposed on its mirror image. We just show that amazing compound can be super imposed. So basically compounds or a title. And therefore, what is the optical rotation of any mezzo count back again, it's a good practice to go, oh, it's optically inactive. >> Yes, so the numerical value of the optical rotation will be 0, right? Vaso compounds are optically inactive because they are a chiral. >> And one of the other ways that you could see was there a chiral is they can take a plane of symmetry. >> Well, I was showing you these four compounds a few minutes ago that I gave away the answer to what I'm going to make the last question of the morning. >> Which one is the Mesos job back? >> Well, it's pretty obvious that it c, But let's be clear about why nato compounds at both stereo centers and a plane of symmetry. >> Compound a is not a mezzo popping, not because it doesn't have a plane of symmetry, it does because it has no stereos. >> Compound B does not have any stereo centers. Peter. >> Compound D has stereo centers. >> Hydrogen away from you here, hydrogen towards you here. But it does not have a plane of symmetry because we've got plane right here. This methyl group is outdoor. Do this methyl group is away from you. They don't reflect into each other. >> Let's modify and extend it. >> Only compound C as stereo centers. >> Stereo center here, stereo center here. >> And a plane of symmetry where blacks metal into that bowl, H and H, this carbon and this carbon, etcetera. >> Therefore, that's the only one that meets both criteria for being amazing. >> Countdown makes sense. >> Okay, again, when you get around to doing problems at the end of chapter six in your textbook, which will not be for at least a week. I'm sure there will be problems there that say which of the following is a mezzo compound? >> Being able to address that question. >> What two things? Amazing combat as to hack areas, enters and appointments. All right, we're almost done with chapter six. >> We will not finish chapter six, Adele, next Friday because Monday is going to be nothing but b, answering whatever questions you might have about Wednesdays, exam questions over the weekend you prefer, or fee on Monday anyway, later paper would have had a great idea. One piece of information after that I would have printed out yesterday if that doesn't work that way. Yeah. Yeah. Yeah. Good watch. Yeah.
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From Dana Chatellier March 03, 2020
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