4

2. Draw the structure of the compound with the #H NMR and IR spectra shown below and the formula CsH120_ 4 % WacknfthMlniclnirien6 (ppm) frequency...

Question

2. Draw the structure of the compound with the #H NMR and IR spectra shown below and the formula CsH120_ 4 % WacknfthMlniclnirien6 (ppm) frequency

2. Draw the structure of the compound with the #H NMR and IR spectra shown below and the formula CsH120_ 4 % Wacknfth Mlniclnirien 6 (ppm) frequency



Answers

Determine the structure of each of the following compounds based on its mass, IR, and 'H NMR spectra.

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

This question asks us to identify the structure off, but come bound that has a molecular formula C eight inch pan. Oh, we're given the IR spectrum for this molecule and the proton NMR spectra for this molecule which we can use to interpret the structure. So if you look at the ir spectra, we see that that is no carbon is stretch Neil 1700. They have numbers. So from that we can tell that this compound doesn't have a Carbonell group. If you go back to the proton NMR spectra, we can see that there Oh, to double its near seven bpm. So pigs knee and seven ppm, usually called response to protons coming from the Benz ending. And, uh, usually we can account for each proton made a Benz injuring if there are no substituted groups. But we see that from the integrations. After two picks near 70 p and we can only account for two plus 24 benzene peaks, the stairs are stacked out of the six protons in a Benz ending, probably to off them have bean substituted. And so there are only 40 minutes and you see how all right how symmetric the do that Blood near seven ppm look. So that factors as that the bends ending in substituted to give it some symmetry. So there are two, uh, there are likely to substituted groups, and they are substituted in a way that will give some symmetry to the molecule. And then if we go to the, uh, upfield any off the NMR spectrum, we see that there are two peaks which are good, single it's and they each correspond to up, integrated to integration off three. So these two can. These two peaks can be coming from two metal groups and the fact that they are single it they're says that they don't have any adjacent protons, uh, on their carbon next to them. And after that, you see that line after single? It's out of the two single. It's in the up billion year off the proton NMR. Spectra up is is downfield. So there's a big neo people in eight ppm so that it's much more down field. Come back to the big Nia to find three bpm. So we have a Electra negative act, um, in this molecule of formula, which is oxygen and no oh, that that could be, uh, oxygen near one of these metal groups, which is making it give a really downfield chemical shift. So far, we have put together some information. There has to be a Benz injuring that has so substituted groups in a symmetrical So that is most likely a I mind for substituted men's in group. This is Bond and the stool three. Well, so that's most likely are 14 substituted benzene and then we have do omit their groups. Outcast Mitt I'm is actually be possibly really close to an oxygen and their boot, not just in two carbons that have hydrogen attached to them. If we give these, uh, these things in mind again with them together and construct destructive So this is the molecule that death, uh, ir spectrum at Leicester Proton NMR, spectra correspond to making call it one Madox e four methyl benzene

This is the answer on to chapter 14. Problem number 67 Fromthe Smith Organic chemistry textbook. Ah, and we're told that treatment of this alcohol that I've drawn here with, um Tosic acid affords two products Eman end with molecular formula, see six each 12 were given an m R. Spectra for these two molecules were asked to draw a structure for them and then to propose a mechanism that explains their formation. Um and so right away, looking at this, we see that there's an alcohol in the starting material and no alcohol product. So this, it occurs to me is probably a dehydration reaction, in which case we are going to have the formation of a double bond. And so these two molecules Eman end probably differ only in the position of the double bond. Um, and we're giving a formula. See six h 12. So I think a good place to begin is to calculate a quick HD I So it's gonna look like this minus 12 there, Um, all of that over too on that equals one. So I think I think I'm right. I think there's one double bond in each of these molecules. on, and we're pretty much talking about, uh, just a difference of where that double bond is going to be. So then we can look at the end of more data and start to try to assign meaning to some of these signals. So we have for m here. These 1st 3 signals are gonna be methyl groups. Um, then we have a multiply it a 2.0, with an integration of two. Ah, and a triplet at 5.1 with an integration of one. Um, OK. And so, um, I'm going to say that this multiplied at two point. Oh, is going to be a methylene group. Um, and then because it's so far downfield at 5.1 ppm, um, I think that this final signal is going to be from an AL keen pray time. Um, so again, 5.1 ppm pretty far downfield on that is consistent with where we would expect Al cane protons to show up. Um, okay. And so putting all of that information together, um, along with the fact that this is just a dehydration reaction so we can start by drawing the backbone of the molecule, which we know from our starting material. So it's gonna look like this. And then somewhere in here is a double bond. Um, Okay, so let's look at splitting. So we have we have our first signal. Ah, 1.0 p p. M. Triplett are the ingrates to three. So we have a method group our that's being split, um, by a methylene group. So what I'm saying is this, uh, this first signal corresponds to this method group. Ah, and this signal for our ch two, uh, is going to be splitting it. And so that has to be there. And it has to be in tax service. We could ask the CH to Gert And so that, to me, means that in this molecule, our double bond is here. Ah, and I know that because this last signal, um, for the Al Keen Proton has to be coming from a single out keen proton right there on that carbon on dso That's the only way to get all of this. Any more information to sort of make sense. Um, okay. And so then looking at end, um, again, I expect the only difference between m an end really to be placement of a double bond. So I'm gonna go ahead and drawback in the backbone of the molecule here, the framework, Um, and then again, start thinking about each of these pieces of an M R data. So this first signal, it's gonna be metal group. Let's skip the second signal. The third signal is gonna be a metal group, okay? And so what I'm seeing here is that there are only two metal groups, so we have two metal groups intact this time instead of three. So that should tell us that this time the double bond has to be a terminal double box because it's sort of getting rid of one of our method groups. So I'm gonna go ahead and put the double bond here this time. Um and so in this case, are we have this 1.8 p p. M. Single it, um and it's a single it. So I'm gonna go ahead and say that it's right there. Ah, and then our other method group, which is split into a triplet, is going to be right there. Um, yeah, and I think that that makes sense. Ah, And then when we look at these others. What we see, um, are to methylene groups here and here. Um and then this time we have a 4.8 p p. M. Doublet that integrates to to h Ah, and this is going to be our al Keen protons this time. This time we have to al keen protons. Right? So this, uh, this signal is going to be coming from the two protons that are on this terminal Al Keen. Um, and then the other two, uh, methylene groups. Um, unfortunately, unfortunately, I only have four colors, so I'll do these two in green. So we have two methylene groups here in here. Um, yeah, and so the one that's definitely split into a triplet. So actually, let me Ah, let me go ahead. And again, I do apologize that I sort of have to Ah, reuse color here. But so this methylene group that split into a triple it is going to be corresponding to these two protons because they only have two neighbors, which is what splits them to a triplet. Um, and then I did that in greed. Okay. And so the other ah, the signal at 1.4 p p. M. That is split into multiple. It is gonna be from these two protons, and it's split into a multiple it because it's being split by the other methylene group to its left and by the metal group to its right. Okay. And so that's That's Eman end. But of course, there is a second port to this problem. So we're also asked, unfortunately to draw a mechanism that accounts for the formation of these two molecules. So we are starting from here with the alcohol right here, Okay. And we are treating this with Tosic acid, so each O. T s. Let's do it with that. So the first step here is going to be the lone pair on this oxygen. One of lone pairs are grabbing this hydrogen and become impregnated. So we now have, uh, pro donated alcohol there, So Oh, each too. Um, and of course, there is a positive charge on this. Oxygen now has three bonds, and so one of these bonds is going to break. The electrons will revert to the oxygen, and it will leave as water. Now, we have a situation where we have a carbo cat eye on So carbo cat eye on aj's gonna be right here. And, um, let me go ahead and draw this hydrogen in here, and so we're actually going to get a rearrangement. So this is a one to hydride shift. Um, so it's gonna look like that. You may also know it as a carbo cat. Eye on rearrangement. Um, and so remember, a tertiary carbo cat eye on is going to be more stable than secondary carbo cat eye on. And so this molecule will rearrange in order to make the more stable carbo cat eye on, which is the tertiary carbo cattle. So there it is now with the positive charge house on a tertiary carbon. Okay. And so basically one of two things can happen from here. Um and really, they're they're the same thing. So we have Ah, hydrogen here. Um, or I guess I'll do this like this. We have the exact same molecule, but we can think about a hydrogen here. And so if, uh, the congregate base of the toxic acid that we started with, um plucks this hydrogen, uh, or new double bond will be terminal. And we will get, um, the structure that was referred to as n on the previous page and and in this problem. So this would be a molecule end, Um, or if the con ticket base of the tosic acid takes this proton instead. Oops. I should put that possible charge. They're the new double bond will form not terminally, but in the interior of the molecule there. And so in that case, we would get the molecule that was referred to as molecule M in this problem on the previous page. So, uh, there we go. There's the mechanism that accounts for the fact that we get both of these molecules in combination. Um, when performing this reaction s So there's the stepwise mechanism that were asked for. And then here again on this page are the structures of M and N and how we derived them. We're doing an HD eye and then looking at an M or data. Ah, and that's the answer to Chapter 14. Problem number 67


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