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32) The 1H-NMR spectrum shown here belongs to which of the following molecules? (multiplet at ~2.7 Ppm is . septet)PpM...

Question

32) The 1H-NMR spectrum shown here belongs to which of the following molecules? (multiplet at ~2.7 Ppm is . septet)PpM

32) The 1H-NMR spectrum shown here belongs to which of the following molecules? (multiplet at ~2.7 Ppm is . septet) PpM



Answers

The $^{1} \mathrm{H}$ NMR spectra of two carboxylic acids with molecular formula $\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2} \mathrm{Cl}$ are shown below. Identify the carboxylic acids. (The "offset" notation means that the farthest-left signal has been moved to the right by the indicated amount to fit on the spectrum; thus, the signal at 9.8 ppm offset by 2.4 ppm has an actual chemical shift of $9.8+2.4=12.2 \text { ppm. })$

Hillary won tonight were doing Chapter fourteen, Home fifty two, and this problem gives us the carbon thirteen anymore Kinkle shift values and asked us to identify which carbon atoms gives rise to each of these signals in the carbon thirteen spectrum. So way having a lot of thinking about him here is our alcohol group, and we have straight uncle chain carbon. So we know that the look on with drunk on Electra native characteristics of the auction and will remove the doctrine density away from the carbon nucleus, making it that the's carbons will resonate at a higher ppm value or larger, wonderful ship ways. We know that this carbon has directly bounded to the oxygen is gonna have the highest chemical shift value, and that's going to decrease the further away we get from the oxygen's. We know that the one closest will be the highest value Shame sixty two, then the one at the very anything. The lows value that's the furthest toys was the most shielded carbon. That's only nineteen. So actually the fourteen one the one next was going to be nineteen. What? And this one's gonna be thirty five, So using the same logic. We can determine the chemical shit values for this ad molecule here. So we have an alga hide carbon. So's Carbonell. Carbon is gonna have the highest coming ships right away. And you can see the cardinal Carbon is that tool five chemical shift value. And then the lost chemical chef value is going to be the ones first away from this and other these two metal protons here and these were both equivalent. So it's gonna be the lows, Val, Just sixteen. That leaves us the central Carmen with forty one ppm. What? No, we need to determine the chemical chef values for this molecule. So this was kind of different because we haven't out King Bond here. So what We know from our King Bonds that they have relatively high chemical shift values, So one could even say that is going to be higher than the carbon that's bound to an alcoholic is we're looking from this example above this carbon is bound to alcohol my sixty two personal line by line. All of a sudden, something that has a double bond. Too much from the bottom is two hundred five. So you say that don't bond that's not about to let you make it about. And some are going to be between sixty five sixty two and two are five. That's what we have. We have something that's one third team and one for three. So we know that these two values will be the values for the carbons attached to our Al Kane double bond here. But which one is going to be for which we are one thirty one forty three? So because of that electron would pulling affect our electro negative on? We can say that the carbon that's closer to his lobster negative, Adam or sister we're just closer to it is gonna have the higher ppm about. So we know right off the bat, this car in here. This is goingto have a higher ppm value off the Al King bond because it's closer to this saturated alcohol group here. All this carbon on the other side of Valentin, you don't have a lower ppm value and using the same ration off top one, we see that we have a sixty nine ppm. Values weaken right away to terminate. That's going to be the carbon that's bound to our alcohol and in the carbon that's furthest away. Its signal bond is going to be the lowest complicit, which is twenty three.

This question were given three proton NMR spectra corresponding to three. I service with the molecule a formula see, for it's nine Biard and they're supposed to take a look at each spectrum and figure out which I cinema it corresponds to. So if we look at the spectrum A we see that the most don't feel signal, which is we have four ppm has integrated issue off one so off for that to be true. Mm. The most likely structure on the part of the structure should be something like this. A single proton attached to a carbon that is attached to the rooming in the molecule because the roominess only, um, component in the molecule that can give rise to a, um down through signal. So that's what we learned from the most downfield signal. And then if we take a look at the rest of the spectrum, we see that there are about three more weeks that we need to account for. That does us. There are three main types of hydrogen is in this molecule, and then we also have three more carbons to a control. So out of the three touches our hydrogen, we see that, um, to off them closer to each other in chemical shifts while one of them has it more field, chemical ship. So for that to be true, the structure has to regiments look like this. So the two near signals were becoming from these three protons and thes two protons, which are what pretty close to the bro mean by the most Upton signal becoming from these three metal Brokaw's equivalent to each other. So this is the soxer corresponding to the spectrum? A. And then if the I wanted the second spectrum, we see that the most downfield signal has an integral ratio off to. So that should be coming from something like this. Two protons that that are attached or carbon that is directly ties to the rooming. And then we have dreamer signals to control. And again, three more carbon still count for. And, um, we also see that the three signals are kind of spread out, and, uh, the, uh, decreasing in the India chemical shift. So that usually tells us that does that linear structure here. So, me, you're a structure like this. They can't account for the integrations as well. This method, a group that is further away from the broom means you grass to the most up the signature. This will have Ah, um, more down for your signal. And this will have a more downfield signals. Compacted that. And these two protons of the address to the most down spacing. And so we can come for the second spectrum using the structure. Then if being warned of the third spectrum, you see that the most down filled signal is again coming from Don't throw darts. So again, we started this and then our there's only two more signals. And one of them has our integration ratio off six, which tells us that it's probably from two metal groups and the singer integration is from her me time group. So we can put all that together like this. Eunjo, please. Metal hydrogen czar equivalent to each other. Mr Vida, Signal that is giving nice during integrated issue off mine be be the signal that has integrated as your mind and, uh, the most down physic never come from this Matalin Protons. So this is the structure corresponding to the spectrum seat

This is the answer to Chapter 14 problem number 31 from the Smith Organic Chemistry Textbook. And in this problem, we're told that a compound of molecular formula C four h 802 shows No, I are peaks at 363,200 or 1700 wave numbers on. We're also told that there's one Proton anymore Signal and born carbon atom our signal for this molecule. And so, given just that information were asked to determine the structure here. Ah. And so, as I've said before, whenever we're given a man like you would formula on and asked to determine a structure, a good place to start is to calculate the hydrogen deficiency index on DSO. Remember, that's gonna be two times the number of carbons plus to minus the number of hydrogen Sze on. Then all of that is over, too. So in this case, it's eight plus two is 10 minus eight is two over two gives us on HD eye of one. And so we know that there's one ring or one double bond in this molecule, doctor. Then let's look at it. The other information that we're given and see exactly what that tells us. So lack of a NY our peak between 3200 and 3600 wave numbers tells us that there's no alcohol in this molecule because that's where we would expect the peak for an alcohol to show by I. R. Um, And the lack of the peak at about 1700 wave numbers tells us that there are no carbon kneels in this molecule, because again, that's where we would expect. Ah, carbon, you'll peak too show, um And so with with no alcohols and no carb Neil's. But to oxygen's in this molecule as per the molecular formula, we should be pretty sure that what we have are ethers. Um, and so then thinking about our other spectral information, there's only one, uh, proton Peak. So all the protons here are gonna be the same. They're gonna be equivalent. Uh, and so likewise, there's only one carbon anymore signal. So all the carbons are also gonna be equivalent. So we have all equivalent protons, all equivalent carbons, Uh, and we have to oxygen's to account for with no alcohols and no carb kneels and one degree of on saturation in this molecule and So all of that information taken together means that there's really only one structure that we can come up with that satisfies this criteria. And so this is 14 dioxane, uh, just drawn it there. Um, all of the protons and this molecule are equivalent off. The carbons in this molecule are equivalent. This fits our molecular formula. It fits our hydrogen deficiency index one that we calculated on. Obviously, there are no alcohol and no carbon deals here. Um and so this is the right answer on again. The way that I arrived at this was to start by calculating an HD I on then, considering the eye, our information that we were given that really tells us what we don't have. Ah, and then considering the fact that the single and a more peak for each of the proton and the carbon enim ours mean that all of the hydrogen zehr equivalent and all the carbons are equivalent. And really, this is the only structure that we could come up with. And so that's the answer to Chapter 14. Problem number 31

This is the answer to Chapter 19. Problem number 61 from the Smith Organic Chemistry textbook. In this problem gives us the three molecules a, B and C that I've drawn here. Andi gives us three sets of carbon 13 and a more spectra numbered 12 and three now would have drawn, uh, written out in red to the right. And we were asked to match the spectra to the molecules. Um And so in order to do this, we need to remember that Carbon 13 anymore tells us about types of carbons in a molecule. So for each type of equivalent carbons, we get one signal s. So when we look at a what we see here is that there are three types of carbons, so we have type one, which is gonna be the three methyl groups on. I'm just arbitrarily assigning numbers. Here we have type two, which is the Tetra Hydro carbon in the middle there. And then we have type three, which is the carbon of our car box set like acid. So, looking at the spectra, a spectrum number two has three signals in it. And so for a theano, sir, is going to be? Spectrum number 23 signals three types of carbons on. There we go. So then, looking at B, every single one of these carpets is unique. So 12 34 and five for the carb oxalic acid carbon s 05 unique carbons here. So we need a signal or ah, spectrum with five signals in it. Ah, and so that's spectrum number one s o. B. Is gonna be one. And then looking at sea, we have one type of carbon in these two method groups. Their equivalent. So they're only gonna give one signal. Ah, second type of carbon would be the carbon to which those metal groups are attached. We have a methylene carbon next. Ah, and then we have 1/4 type of carbon in our car box Selic acid. Ah! And so this is going to correspond to spectrum number three, which has four signals on. We have four types of carbons in molecule. See a suspect from three for sea. So again, that's ah, a spectrum to be a spectrum one and C is spectrum through again. The way that we arrived at that answer was to count the number of unique carbons in each of these molecules and then match that up with the spectrum that had that many signals. And that's the answer to Chapter 19 problem number 60.


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