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Draw (ne structures Compound 16 a. and 16b. (12 pts), clearly Indicate your assignments of all proton resonances pts) : Calculate Ihe Unsaluralion Index of each Com...

Question

Draw (ne structures Compound 16 a. and 16b. (12 pts), clearly Indicate your assignments of all proton resonances pts) : Calculate Ihe Unsaluralion Index of each Compound (2 pts) (20 points total)c =1.35 ppm singlet; 6HCompound 16a CH;NUnsat; Index - (2C+2-H-X+NZ 103 b= 3.00 ppm singlet; 2Ha =7,35 Ppm} multiplet 5H6 ppmDIBAL2) HzO,H;o"d=1.10 ppm; singlet; 6HUnsat Index = (2C+2-H-X+NV2Compound 16h CH,O 9.58 ppm; singlet, 1H b=7.30 ppm multiplet; 5Hc=275 ppm; singlet 2Hppm

Draw (ne structures Compound 16 a. and 16b. (12 pts), clearly Indicate your assignments of all proton resonances pts) : Calculate Ihe Unsaluralion Index of each Compound (2 pts) (20 points total) c =1.35 ppm singlet; 6H Compound 16a CH;N Unsat; Index - (2C+2-H-X+NZ 103 b= 3.00 ppm singlet; 2H a =7,35 Ppm} multiplet 5H 6 ppm DIBAL 2) HzO,H;o" d=1.10 ppm; singlet; 6H Unsat Index = (2C+2-H-X+NV2 Compound 16h CH,O 9.58 ppm; singlet, 1H b=7.30 ppm multiplet; 5H c=275 ppm; singlet 2H ppm



Answers

Match the spectroscopic data in the table to the compounds whose structures are shown below. (FIGURES CAN'T COPY) $$\begin{array}{ll} \text { Compound } & ^{13} \mathrm{C} \text { NMR } \delta / \mathrm{ppm} \\ \hline \mathbf{A} & 25.1,27.1,41.9,211.3 \\ \mathbf{B} & 136.4,187.1 \\ \mathbf{C} & 20.6,178.1 \\ \mathbf{D} & 13.8,22.3,24.9,30.7,47.2,173.8 \\ \mathbf{E} & 162.3 \\ \hline \end{array}$$

This is the answer to Chapter 14. Problem number 65 Fromthe Smith Organic chemistry textbook. Ah, and this problem asks us to propose to structures. Ah, that are consistent with two sets of data that we've been given eso for each structure that we're supposed to propose. We've been given a molecular ion. Um, and I are spectrum in an enemy war spectrum. Okay? And so, starting with J, we're told that Jay has, um, its molecular ion at 72. Um and so from there, we can start to think about some formulas. S 06 carbons equal 72 s 05 carbons on 12 hydrogen ZX could get us to 72. Um, but I'm going to jump around a little bit. Ah, And in light of the fact that J has an i r. Peek at 17. 10 wave number, which is consistent with a carbon oxygen double bond, I'm going to say that there is anoxia gin in Jae. And so when I see the mtz at 72 I think this is going to be C four each aids. Oh, um and I think that because again, this I r tells me that there has to be a carbon oxygen double bond in here. Um, and so if the formula is see for H 80 the degrees of on saturation for that formula are going to be, uh, two times four plus two minus eight. All of that over too. And so that's going thio be one degree oven saturation on dhe. So I think this is, ah, the correct formula. And I think that we have a carbon oxygen double bond in this molecule s. So then we need to look at our anymore. Signals s 1.0 ppm. Triplet. That integrates to three. That's going to be a method group. Um, and I'm gonna skip the next signal on Look at this 2.4 ppm quartet that integrates to two. And that's gonna be a methylene group. Ah, And so these two groups are splitting each other. Uh, right, because we have a quartet on the methylene signal on, So that means it has three neighboring protons on the signal for the metal group split into a triple it, which means that it has to neighboring protons on. So these two are next to each other. They're splitting each other. Um, And then the last signal of the second signal here, the 2.1 ppm single it that integrates to three. Uh, that's going to be a method group with no neighboring Proton because it's a single it. Ah. And so if we put all of this information together, um, what we get looks like this. So here's our Carbonell. And then there's another method group on just to sort of spell this out S o. This signal is going to be this metal group. This signal is going to be this method group, and this signal is going to be this math linger. Okay, so there's, uh there's j and now we just need to do the exact same thing for Kay. So looking uk, we have ah molecular ions at 88. Um, and since looking at the i r for j proof so helpful, let's start with the i r. Here. So this time on the i r. Is a peak between 3200 and 3600 wave numbers. That's a broad peak. Ah, and so this is consistent with an O. H. So we have an alcohol in this molecule Um, And so again, we need to account for the presence of an oxygen. Ah, and so C five, um, case that 60. And then oxygen is gonna be 16. So that's 76. So let's go see five each. 12. Oh, okay. See? Five each 12 0 And so if we do the hydrogen deficiency index for that formula, what we get is two times five plus two minus 12 all of that over to equal zero. So there are no degrees of on saturation in this molecule us and no double bonds, no rings. Just a, uh, chain of carbons with an alcohol in it. Okay. Um, and so looking at our anymore the fourth signal, the 1.6 p p. M. Single it jumps out to me. This is probably, uh, the proton of our hijack see group. Um, Okay, So the 0.9 p p. M. Triplett are that integrates to three. Uh, that's going to be a method group. The next signal, the 1.2 p p. M. Single it that integrates to six is going to be to methyl groups. So there two metal groups in the same environment and they're not split by anything. Um, and so I'm gonna go ahead and guess that they are on our carbon with the alcohol on it. And then, lastly, we have 1.5 ppm. Quartet are that ingrates to two. So this is going to be a methylene group. And so again, notice we have We have a triplet are that integrates to three and a quartet that integrates to two. And so these two signals are splitting one another. Um, okay. And so when we put all of that together, we can put it together like this. Um, so let's do it like this, Okay? And I think actually, drawing it out with letters in this case maybe makes it a little easier to see. So obviously, uh, single for the O h. Eyes is this? Oh, age. Um, we have these two methyl groups that are equivalent. So that's gonna be these two method groups. Uh, and then we have, um, as I said, the methylene and the metal group that are splitting one another. Well, let's be consistent, I suppose. Oh, dear. There we go. Okay. So sorry. All right, so here we go. So this black arrow Obviously this is the methylene group, and then the blue arrow eyes referring to this method, Bert. And so this is a good structure for this molecule that fulfills all of the requirements that we were given on. And so that's the the answer for K. And then the answer again for J looked like this. Um and that's the answer to Chapter 14 problem number 65.

Well, everyone, today we're doing Chapter fourteen on thirty two in this poem assets the job, the structure A compound with the molecular formula of C four h eight. Oh, and that has a signal in this carbon thirteen spectra that is, has a chemical shift value that is greater than one sixty ppm and then also draws I simmer or an ice over with same formula, but the only difference between them and has a signal with lower than one sixty ppm. So we need identify what car type of carbon exists when it's greater than one hundred sixty ppm. And if you look at our chart table correcting a textbook, we know that Carbonell carbons. So this carbon here, the carbon all carbon, produces a chemical shift value that is greater than one hundred sixty. It's over two hundred sometimes, so we know that the initial structure must have a carbon or carbon. And we know that the ice armor that doesn't have a signal above one sixty cannot have a carbon carbon. So we know that this option must be either some sort of ether or some sort of, um, alcohol group or something like that. So let's start with the first example here. So we need to determine how many double bonds or how many degrees on saturation exists in this formless. So using in my equation that helped me out so much as a student the number of double bonds or number the degree of saturation is number of carbons for four minus the number of hey, lights and hydrogen is over two eight hundred zero. He lives over too, minus the number of Nigerians over to which is your over to which is zero plus one is that for months for zero plus one equals one. So that means that we know that we have to have one double bond in our molecule. And for the cardinal case, this is already a total bond. So I'm going to do now is just fill in the best of the structure out with a propos structure like this. So that matches the chemical formula. So we know that we have one carbon one option that was already used. It's now we have C three h ate that we need to satisfy. So this is one, two, three carbon. So now the C three satisfied, and this turmoil. Cardinals three protons destroy macarons. Three shots six. And this one has to says eight. You know that this is also satisfying. Alternatively, I could have drawn something like this. This would also satisfied. But now this an alga. Hyson Aldo hide is different than a carbon all group. So we need this cardinal to be some sort of key tone to produce a signal that is above one sixty ppm and not Aldo Hide. Ah, carbon All group here because I would produce a lower signal. So we know that indeed we do need this Ah, and our group or a group on the other side of the cardinal to produce this key tone so we would have that one six ppm. So that's why we know that this structure is correct and not this Well, what about if the A nice murder has signal lower than one sixty ppm so lower than one six people? And we know that the carbon must be bound to some sort of election Native Adam And what we have here is oxygen. So we know that we can draw on alcohol directly bound to a carpet here so this will produce this carbon signal that's still quite high but will be under one sixty ppm. But a degree on saturation is still one. So if I draw the other three carbons one, two, three. So we have one, two, three, four. I need to add one double bond because I know that this is not going to be a Cardinal carbon here. This will not be a carbon carbon here because they now produce over one sixty. So I know that it just needs to be a double bond somewhere. So if I was a drug dealer bond here, for example, I would see that we have two protons here. One here says three, four, five, six, seven, eight. So this is satisfied satisfies our Muckler equation here. But what about if I do so, bond? Here we have one, two, three, four, five, six, seven, eight. So there's also another possibility that we could have for thiss isom er here for this question

We're just considering some more NMR scenarios here, and we're just gonna be drawing structures based on formulas that we're given, as well as the peaks that may be seen on the spectrum. So fastly I have C two h 60 here where we have one single it present. So we have one peaks, meaning that all of my protons are in the same environment. So given I've six protons, I might want to consider an element of symmetry that might be present. That will cause all of these protons to be chemically equivalent because given that we've got one single it, that means that all six protons are in the same environment. So I would start off with my central oxygen, and then we can just attach our to me 1000 groups either side. And that gives us an element of symmetry that allows us to group both of these groups of protons into the same environment. That gives us one large single it that will have an integration of six there. It has a multiplicity off, a singular as it's stated in the question, and that is because is split by no other protons that are in different environments. So given that Russian out, you can continue with the following examples that we do have and I've drawn up the structures. But in the order that we have chemical connections up on the screen and just for this last example, we do have two possibilities there. Just be mindful of that that both are correct, given the information in the spectrum.

This is the answer to Chapter 19 Problem number 49 from the Smith Organic Chemistry Textbook. And this problem says, draw all residents structures of the continent basis formed by removal of labeled protons A B and C in 13 Psychlo Hex it. Ah, heck seen Die owned and said it s set an allied for each compound rank those protons in order of increasing acidity and explained the order you chose. All right, so starting with 13 cyclo taxi and I own, um, we will just go in order s o the loss of Proton h A. I'm going to give us this molecule. I wish a lone pair and a negative charge there. Um And then there's one resident structure that we can draw. Um, that's going to be a result of the lone pair going there in these electrons kicking up to the oxygen. And so, uh, we can potentially have our double bond there. Single bond there, um, three lone pairs and the negative charge on this oxygen now. So just like that, Hurry. Um, far as h b. So this was a being lost. A sz far as HB being lost. Um there's really only one structure we can draw. So if we I worked, you have hb be snatched off, we would yet this structure where the lone pair and the negative charge are on the carbon that HB was on. And there's not really anywhere for those electrons to go. So that's it. That's the only structure there no resident structures. Which brings us to the loss of H. C. So All right. So if h c were to be the proton they got plucked off, we would have Are the lone pair their native charge there? Um, and we can, uh, d localized that, Um, so, uh, the electrons can either go here and this double bond can can break in. Those electrons can revert to that oxygen, or they could go the other way, and those electrons can revert to those oxygen. So that oxygen. So we have two possible residents structures here. They look like this. So possible. Oh, sorry. That's not right. So potentially, um, this one with the double bond there, Uh, Or if we were to follow the red arrows, we would get this, uh, where the double bond, uh, is here and this is a single bond, and this double bond is still intact. So let me put my my lone pairs of charges in. So if we follow the blue arrows, we get to that. Ah, And if we follow the red arrows, then the extra loan pair in the negative charge is they're okay. And so those are our three residents structures for H. C. So, in order of increasing acidity, remember that the more stable congregate base is going to have the more acidic acid. And so number of residents structures correlates directly with stability. So H B only has one residents structure. So H B is going to be the least stable base and therefore the weakest acid. Um, h ey has two residents structures. So therefore, it's going to be intermediate in terms of content based stability and therefore a city and H C, with three residents structures, eyes going to be the most acidic. Um, so then we just need to do the exact same thing for a set. Tan Allied, I said an allied, uh, here, in terms of A, B and C, Unfortunately for be, there are seven potential residents structures, so this is gonna take me a couple minutes to draw when we get to that point. But let's start with a So if we lose a, um, that's going to result in this molecule. So three. So that's gonna look like this with loan pair and the negative charge on this carbon. Um, and there's really nowhere for that's go. Um, yeah, You know, there's there's no potential resident structures there, So this is each a here, and this only has one residents structure eso we should know from the previous problem. That's probably going to be our our weakest, Uh, we just contact base and our least stable acid. Um, so I'm gonna skip beef right now, and you see, because C is much easier than be eso if we were to lose, um, Proton be this is each be that we're talking about now. So if we were to lose h b, we would Oops. Oh, you know, I just said, I'm going to skip B and do you see? It's said. Then I started to draw B. Oh, Um, okay, so let's talk about HC first. Um and I mean it this time. So if we were to lose HC uh, we would get our loan pair and our negative charge there on this carbon. Um, And there is one potential resident structure that we can draw here so these electrons can add in here to make a carbon carbon double bond. These electrons can revert to this oxygen. Ah, and so we have the residents structure that looks like this double bond there. Single bond to this oxygen, and the oxygen is going to be the seat of our negative charge. Um, And so, uh, now, let's do HB. So I'm gonna put this on its own page because there are going to be, As I said, seven residents structures here, so it's quite a bit, um, so and so here is a deep protein ated, uh, h b. So the nitrogen now has two lone pairs, and it has a negative charge. Um, and so the first resident structure that we can draw eyes going to look like this. So one of these lone pairs can add in here. This double bond one of the bonds can break in. Those electrons can revert to the oxygen. Ah, and that will give us that will give us this with one lone pair on the nitrogen now, and three lone pairs and a native charge on this oxygen. Um, we can conceive of another resident structure here. Very similar where the electrons of these double bonds just move around This aromatic ring. It's nothing really changes, but it is technically a resident structure here. Okay, so there we d'oh! Um, okay. Ah, And so, um, we could also, uh, and I'll use red to show what I mean. So, um instead of that sort of sequence, uh, if instead, um, the lone pair of electrons in the nitrogen added in the other way, we could do that and have that double bond break. And so if we follow the red arrows here, what we get is this. And so now, um, we have a lone pair and a negative charge on this carbon. Um, And from here, we're basically just gonna move these electrons around this ring. Um, so, uh, they'll add in here. These will go here. These will go here. It's going to give us this. Yeah, It's our loan pair and native charger here now. Um Oh, wait. Oh, So that's what I did. I drew Ah, through too many arrows at once. So rather, rather than do that. Ah. Okay, so these electrons should be going to this carbon, and that other double bond is going to be unaffected. So this is what we get for this resident structure. So now, from here now, these electrons could add in here, and these electrons could revert to that carbon. Okay, there we go. All right, So there's our loan pair and our negative charge or other lone pairs in. Okay. And so, uh, from here, Now, um, these electrons can go here. These electrons can go here. Ah, and now we've sort of gotten back to where we started, but the bonds in the aromatic ring are in different places. So if you compare where they were beginning to where they are now, uh, it's different. One to the nitrogen has as two lone pairs. The positive, not a positive charge. Oh, boy s o. The nitrogen has two lone pairs on the negative charge. Once again. Um, but as I said that the bonds are different around the aromatic ring. And so this is, uh, all of this is the potential resident structures. If we remove h b. And so I'm going back to this. Part of the question s so h ey has one resident structure, so it's going to be the least stable contacted base and therefore the least acidic acid. HC has two resident structures s so it's going to be intermediate Lee stable on dso the intermediate acid Ah, and HB when it's removed. Um, there are seven potential resident structures for its consequent base, and so it's going to be the most stable contacted base and therefore the strongest asset. Um, let me just make sure, um yeah, and so that's everything that we were asked in this question. Um, yeah, so a lot of lot of resident structures to draw here. But when you do, draw them all out these answers to make sense. Ah, and it is good practice always to draw a resident structures. A CZ you saw and I messed it up a little bit at one part. Um, yeah. So it's always good to practice on. And that's the answer to Chapter 19. Problem number 49


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