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QueSTion 50Which technique; IR or IH NMR will better distinguish the two isomers shown below?IHNMR These molecules are not isomers Neither will distinguish these is...

Question

QueSTion 50Which technique; IR or IH NMR will better distinguish the two isomers shown below?IHNMR These molecules are not isomers Neither will distinguish these isomers

QueSTion 50 Which technique; IR or IH NMR will better distinguish the two isomers shown below? IHNMR These molecules are not isomers Neither will distinguish these isomers



Answers

How would you use NMR (either $^{13} \mathrm{C}$ or $^{1} \mathrm{H}$ ) to distinguish between the following pairs of isomers? (EQUATION CANNOT COPY)

Okay. So for this problem we want to look at what kind of differentiates in molecule in terms of a proton tomorrow Carbonara. And so um what I have drawn already is the three molecules and questions. So one boutin says to Beauty and two method appropriate. Okay. And so obviously all of them have the same molecular formula. Right? So we want to know, you know, which is going to be better in terms of differentiating these three compounds. And so the first thing that we can do probably the easiest thing we do is just count the number of unique signals. Right? So we're going to represent the hydrogen in red. Okay. And so this one Beauty has two hydrogen out here that are going to be unique because one is cysts to the Fo group and one is trans to the F. A group. Okay, there's a hydrogen here. Yeah, hydrogen is here and hydrogen is there. Okay, so we're gonna have five unique hydrogen in our second molecule. Yeah, we have a plan of symmetry. Therefore that A is the same as this and this be is the same as this being. Okay, so only two signals. And finally excuse me. And finally in our 3rd and final molecule again, we have a plane of symmetry, which means this muscle groups hydrogen and this method groups hydrogen is are the same. And when we look at these hydrogen because they see the same thing, no matter where they look based on that plane of symmetry. Again, these are going to represent to specific hydrogen. So it's also going to be too Okay, so the 2nd and 3rd compound are going to be rather difficult to assess by standard means by proton NMR, you could look more heavily at the coupling and differentiate them. But if we really want to go for the ease, we're going to look for just number of signals first. Okay, So I'm gonna take hydrogen is off and let's do the same thing with carbons. Okay, we'll do those in green. So we have one. Yeah, two, three four unique carbons. Thank you. Okay, for our second compound. That plane of symmetry. Still very useful here. It tells us we're still going to have just two unique carbons. Right? And finally, when we look at the last molecule again, we're going to use that Line of symmetry. However, we have one big difference now between the 2nd and 3rd molecule and that is the quaternary carbon at the center here. Although it will be a weak signal, it should still give us a signal and that would give us a very easy way using carbons to differentiate between these three compounds simply by looking at the number of signals. Yeah, interpretation. Mhm.

This question asked. A rich technique brought on anymore or stated in El Ahmar can be effectively used to distinguish between three I summers off booking. So let's go through our each of them in Try to figure out how many signals each kind of in a marble produce. So for the first, the molecules are the first bison. Huh? The proton NMR spectrum will have one too to be for signals and, uh, seated in an immoral spectrum. They also have one do 34 signals. So I am northern down the different types of protons and the different types of carbons for each case. But the second I saw mo, they were proton NMR spectrum give rise to fun Joe Cigna and, uh, C 13 NMR spectrum give rise Do again 12 signals 100 divers God molecules, the broken and the mass spectrum. And you guys do one two signals find, uh, c 13 NMR spectrum of you guys too. Well to three signals. No Brodell in the more spectrum will produce. So signals for two off this I saw you do over this and over this by the settled in animal spectrum aji three unique numbers off signals for each of the I sumers. So in this case, because we want there to be more differences between the three Spectra when we take one of their techniques um Cedo deal in the March, which will give three unique numbers numbers of signals for the three and sumas is the most suitable technique to distinguish these ice almost from each other.

This is the answer to Chapter 19. Problem number 57 Fromthe Smith Organic chemistry textbook. Ah, and this problem says ah used the proton anymore. And I are spectra given below to identify the structures of two. I simmers A and B having molecular formula C four h 802 Um And so, as always, with a problem like this were given, ah, formula and spectral data and asked to come up with a structure, I think that the best way to start is to calculate the hydrogen deficiency index for the formula. Um and so in this case, it's going to be HD. I, uh, is equal to two times the number of carbons, which is four plus two minus the number of Hodgins, which is eight. All of that over too. So we have to over two or one. And so we have one degree oven saturation in this molecule. Um uh, and, uh, given that we have to Oxygen's present, um, I'm going to say that our one degree of on saturation is ah, Corben, you'll and then we probably have a carb oxalic acid or perhaps an Esther, Um or maybe not. But that would account for those two oxygen's s. So then when we go on, look at the I R. Um, for a we have ah, broad peak at 30. 230. 600 wave number. That's always indicative of an O. H. Bond. Um, we have another peek at 2800 to 3000 which is, um, csp three h. That doesn't really tell us much. Every molecule is gonna have them. Um, and then we have this peak. It's 1700 so that is always associated with a carbon oxygen double box. So we do indeed have a carbon. You'll hear. That's where our single degree oven saturation is going to come from on. Then we need the anymore data to tell us what type of carbon you'll we have on to try to help us put the rest of this molecule together. So we have a 2.2 p p. M. Single it that integrates to three. Um, and so this is probably a metal group. Ch three. Um, we have a 2.55 ppm single. It that only integrates to one. Um Mmm. Ingrates. One. We have this. Ohh! Peak in our i r data. And so I'm going to say that this is Ah, this is probably at the O. H. Oven alcohol. And so we have a carbon. You in an alcohol in this molecule, but we don't have a car box. So Gassid or or a nester? So instead, we have ah kee tone or an alto hide. Ah, and an alcohol. I'm so then we have a 2.7 p p. M. Triplet, which integrates to two. And so this is a C H to ah, and we have another triplet that integrates to two at 3.9. So this is probably also a ch two. Um, and it split to a triplet and the single above it is split to a triplet. Ah, and so I'm going to say that this forthe signal because it's the furthest down field is going to be a ch two with an alcohol on it. So one end of the molecule right now looks like this. Think of ch two. Oh h um And then that's going to be attached to a ch Tuas. Well, okay. And then I'm going to say that that's C H to eyes attached to our carbon. You okay? And then our carbon you'll I's gonna be attached to are the mental group that we have. I think that makes sense, right? So the metal group, we know tha doesn't have any neighbors because it's only a single. It the two ch twos that we have would both be split to triplets of the way that I've drawn this. And, um and the O H group is there, So yes, this is Ah, this is a good structure, Given all of the spectral data on this is actually the right answer here. So again, it looks like this one two in the alcohol. So there is our structure for a with the key tone and alcohol in it. It's now we're gonna move to be, um, so we don't need to recalculate the degrees when saturation. It's the same formula. But we can just make a note. Just Oops. Forgot that when when I changed pages, I have to re select my color here. So one degree oven saturation. So again, um, uh, I'm thinking Ah, carbon Neil. And then maybe maybe in Esther or car back selling acid. Or we could we could have another carbon Ellen and alcohol, like we had last time. Ah, And so, looking at the i R data again, we see it at 1700 wave number peak. That's going to confirm the presence of a carbon deal here. Carbon oxygen, double bond on then the peak from 25 to 30. 500. That's kind of broad, but like weirdly misshapen. Ah, that is very characteristic of Ah, car box Cilic acid. Ohh! So we'll call this car box Cilic. Ohh! And so taken together are these two pieces of I R data are definitely making me think that we have car box Selig acid and looking at the anymore. Dad, I'm actually going to skip to the third signal. So this, uh, this chemical shift this far downfield chemical shift 10.7 p p. M. Anything that air? Pardon me. Any time that you really see a chemical shift beyond 10 you should be thinking carb oxalic acid broad single it also characteristic of a carb oxalic acid. Oh, h um, and and the fact that it integrates toe one. So this, uh, this is telling us that we do, in fact, have a car back selling acid in this molecule. So there's that piece that takes care of our our two oxygen's and are one degree oven saturation. Um and so we're left with this peak at 1.6 p. P. M. That is doublet. It integrates to six. And so I'm thinking to methyl groups here, So two equivalent methyl groups, we're gonna give rise to that on. Then we have a peak that is at 2.3 p. P. M. And a split to a septet on integrates toe one. And so the key here is to look at the splitting patterns between these these 1st 2 signals. So these 1st 2 signals are splitting each other, so we have a signal that integrates to six that is split to a double it. And then we have a signal that integrates toe one that is split to a septet. So bye bye n plus one, right, and plus one is two or one plus one is two, which is why we get a double it for the first signal. And six plus one is seven, which is why we get a septet for the second signal. So this is, um Corbin with three substitue INTs and one hydrogen. Um, and two of those three substitutes are going to be methyl groups, so we can put this together like this. So we have one metal group, our second equivalent metal group. That's where our first signal comes from. We have one hydrogen on this carbon. That's the second signal. Ah, and then we have one Maur carbon to assign in this molecule. Because, remember, it was, uh ah. See, a herd C four each, 802 on. And so we can put our carb oxalic acid right here. And so this is a good structure for this molecule. So again, it would look like this, uh, and this is the correct answer here. Okay. And so again, I think the way to approach these start by calculating in HD I look at the i r. Information and try to determine what it's telling you about functional groups in each molecule on. Then look at the NMR information on, and if you interpret it right, it tells you all of the pieces that you have, and then you just need to fit them together in a way that makes sense. And that's the answer to Chapter 19. Problem number 57

And this problem we're looking at utilizing protein anymore to distinguish between ice and more specifically, were asked to look at this too. I summers and see how they're proton signals will be different. And right away we can look right here and look at the symmetry. Obviously, we have one symmetric compound and one that it is. And so here we're only expecting to have one signal versus here where we expected to have one because there's a proton attach here, two entering, so at least three signals we can see from there. So Okay, here, we only have a ch three that looks into nothing. So we're gonna get one single it here we have. Okay, a pattern that we have discussed in previous questions. Where we have if it's a ch two attached to a ch three, we have a triplet quartet, so we're going to have at least one triplet, one quartet and this Proton sees nothing because there is no process next to it. So we're gonna have a single it this triple it, it's going to be around one. This quartet is going to be around two and then this single it Sorry this trip is gonna be around one. This this quartet is gonna be around too. And this one's will be around 3.5. Say what?


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