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(Jpts ) Which one of the PC NMR spectra is for the compound shown t0 the right?...

Question

(Jpts ) Which one of the PC NMR spectra is for the compound shown t0 the right?

(Jpts ) Which one of the PC NMR spectra is for the compound shown t0 the right?



Answers

Which compounds give one singlet in the $^1H$ NMR spectrum?

This question were given four different substituted benzene dealings. Doubly substituted benzene rings and there are also given in the more spectrum, and there are asked to choose which compound off the full. Does this NMR spectrum corresponding? So these are data structures that, given I've drawn them here. So let's try to figure out the answer to this question. Looking at the NMR spectrum, we can tell that there are going he two hydrogen types coming from the compound because there are only two peaks other than the zero beak. So let's try to write down that information. So there's a peek at its seven ppm or near seven ppm. And then there's a big yeah, 2.3 baby him so usually a peak near seven BP and the result from benzene hydrogen ins. So this has to be coming from Ben's in Hyde Regions and ah, the peak. Me and two point is coming from the substituted groups in the Benz engine, so we can, um, count the number off benzoyl hydrogen is and these compounds it's going to be for each because these are the benzene traditions, and so we have to robes attached in each benzene, uh, compound. So there are four. Hydrogen is remaining, which is the case for all four of these compounds. So there are only four hydrogen coming from the Ben's ending, and that should breathing case with this NMR spectrum. So we can say that that are Ben's dealings are benzene, hydrogen, say, and, uh, when we look at the integrations off the to bigs, we don't have values for the integrations itself. But because the peak height is proportional to the area under the peak or the integration off the peak, we Condell that the signal, corresponding to 2.3 ppm, has more than the number of benzene hydrogen. So that is full right? So we have do see that the number of hydrogen is coming from the substituted groups is more than full all right to get that. So if we put all this information together but began, he's in The do is count the number of hydrogen in the substituted groups to figure out which compound this animal corresponds to. So the first compound we only have you'll substitute and hard agents relative to the four hydrogen is coming from the bench building in the second compound. We have six. Hydrogen is coming from the substituted groups relative to the full. Um, hydrogen is coming from the benzene ring. So here we have four hydrogen. It's which is the same number of hydrogen is coming from the benzene ring. And here we have toe hydrogen, which is less than the number of hydrogen is coming from the Arbenz changing. So from the integration issue, we can again toe that because the height of the 200 peak is greater than the height off the seven BP and peak there have to be more than full. Hydrogen is coming from the substituted groups, and only this compound satisfies that. Tried it. It has six. So you can have full all s. You have to have mood and full. And this is the only compound that satisfies that criteria. So this is the compound that and Neymar corresponds to

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

Well, everyone today were doing chump for fourteen Problem forty eight on this one asks us that referring to approach the compounds in problem forty three which one of these compounds give on ly one carbon and amar spectra one signal in a carbon atom or specters. So we know that each signal or each peak and a carbon atom are spectrum is from a chemically distinct, unique carbon. So in order to have only one signal, we need a molecule that only has one type of carbon that exists in a unique environment. So a good way to start by determining that is determining if you're molecule hasn't terrible in the symmetry. So we know that this monkey was in token of symmetry. So does this. So does this. So is this. And actually, if you look on everything except this month, you watch you have some sort of internal plane of symmetry and there may be multiple internal layers of symmetry. For example, this is this thiss so that didn't really help us in this situation, That's always a good place to start. So next time you determine is are all the substitutes on our molecules the same So if you're looking, for example, this molecule this carbon is adjacent to oxygen's While this carbon is adjacent to a carbon and oxygen this carbon ization to a carbon and oxygen And this carbon is Jason four carbons. So we know that right away these are not unique. Carbon signals they're gonna be isn't one, two, three for five unique signals here because they're all chasing to some different type of some city. So because of that, we know that we can cancel this one out. Well, how about this one or this one? This carbons attached to three Hye Jin's and a carbon and this carbons attached creation the carbon So of these two carbons are actually the same, so they exist in the same chemical environment. So we know that this small killer here is one of the compounds that only give us one. And, um, our signal. Well, what about this monkey hell down here? We know that this car was attached to two Hydrogen is one Roman. One carbon. This card is attached to two carbons. One Hye Jin won, bro. Me too. Right away. These two carbons air difference wasn't knocked included. Well, what about this, Carmen will. This car was balanced to Hodgins and it'LL bound to carbon. This covers ballot to to Adela bonded to one carbon and is bounded to one hundred one carbon signal bonds. We know that these two cartons air difference. This is not involved. One of this monk, you hear this carbon is bound to hydrogen ZX two carbons. This cardinal announce two hundred two carbons. This higher carbon is also behind two agents to carbon, and so is all the other one. So we know that all of these carbons are in a unique, the same unique environment. So they all considered one type of carbon. So this mark you here will give us one. So you know a lot about this. Well, this car was bounced. Three protons would carbon. This carbon is triple wantinto one carbon and also sigma bond to another carbon. So these two carbons right away are different and chemical environments and use the same logic and see that this is true for this one for this one, This one and this also, in essence, using that same methodology determining well if the substitutes are the same, if they are seen then you have the same type of carbons. If they're not the same loon, you'LL have unique types of carbons and we see that this problem you have these two molecules that have the same institutions meaning that they won't show one peak in our and them are carbon anymore Spectra.

In this question, they are asked to identify the structure that gives rise to the, um, given in the marsh spectrum. And they also do it at the molecule that corresponds to the given animal spectrum has a molecule formula, See, servant. Which 14? Oh, to be see that there are twice as much as hydrogen is combated the number of carbons. So we have this formula and if we only have twice as much as, um, Bajans instead off, uh, to end last two there are in is the number off carbons. It means that there is a single on saturation that caused to hide regions to go missing in the structure. So let's keep in mind that does. Oh, that's probably a double bond or a cycle in the structure. So if you take a look at the at the NMR spectrum, we see that doesn't speak neo from so 10 bpm. So that, um, se 13 NMR spectrum peak is usually from a carbon in a group a peek me A 210 oh is observed in a C 13 NMR spectrum, some carbon air group. So it's also keep in mind that we have a carbon, her group. So if we had down saturation, um, that, um, saturation could be coming from the carbon, his group. And then, um so it's pretty much the same thing that we could get out in the two first points that I just mentioned. So I'm also receded. There are only three peaks in the spectrum, out of which one is for the carbonate group. So that does us that there are only three types of carbons in this molecule. So the other two bigs besides the Carbonell Big, um, much field in the spectrum. And ah, one of them is a doublet and our divine Issa corkage. So the doublet should most probably be coming from no sietch group. So I got, uh, in plus minus one last one equals two and the clock that should be coming from most probably met their group. So are three. Passman is for corresponding to a card. It So these are the only three types of carbons presenting this molecule of carbon. It see edge, carbon and method carbons. And, uh, because there are certain conference in this structure, which we can tell from the molecule a formula they have dual own account for all of those carbons using under these three structures. And there's only von carbon Yakob because there is only one oxygen in this molecules of farming. So the arrested the peaks should be coming from, um, or merely ch carbons and method carbons. So this molecule consists off single Carbonell Group, and the rest of the molecule has only CH groups and CH three groups. So they are also able to tell that all the CH groups and all of the material groups in this molecule are equivalent to each other because they only give rise to one beak for each kind. So do to fulfill all that cried here, we have to come up with this image extraction that has from equivalent CH and Mr Groups. So Destructor Bill. I agree with that statement office. All those statements service through CH loops, and they are identical to each other so they'll go tab equivalent protons and all the material protons are going to be equivalent to each other because they're all in the same environment and the same follows for the carbons. Though methane garbine is identical, do the endemic dying carbon and ultimately groups have the same type of cover because of doubt. Um, I didn't good environment. So we also have seven carbons in the structure and 14 Hydra Jin's Inman oxygen. So we have accounted for the structure that corresponds to the given Proton NMR spectrum.


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