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NMr and Assienment MAME:seactrum Gko Chemical shlft #of H"s from Idomi IntegrationIndex ol Hdellclenst Multiplicity of Fof Hs on nelghboring AiDms Splitting (s...

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NMr and Assienment MAME:seactrum Gko Chemical shlft #of H"s from Idomi IntegrationIndex ol Hdellclenst Multiplicity of Fof Hs on nelghboring AiDms Splitting (s, or miWavanumocrIntensityFunctional Group AssienmentCrany the sintura td tgsln peaks t5 esch set 60 puntous:20+2 18+2 # 1

NMr and Assienment MAME: seactrum Gko Chemical shlft #of H"s from Idomi Integration Index ol Hdellclenst Multiplicity of Fof Hs on nelghboring AiDms Splitting (s, or mi Wavanumocr Intensity Functional Group Assienment Crany the sintura td tgsln peaks t5 esch set 60 puntous: 20+2 18+2 # 1



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How many peaks would you expect in the 'H NMR spectrum of 1,4-dimethylbenzene (paraxylene, or $p$ -xylene)? What ratio of peak areas would you expect on integration of the spectrum? Refer to Table 13.3 for approximate chemical shifts, and sketch what the spectrum would look like. (Remember from Section 2.4 that aromatic rings have two resonance forms.) (FIGURE CANNOT COPY)

Hi, everybody. They say I click it filter E. That's the structure. And yeah, oxygen. So peace. Yes. And Belda 2.72 and see Queen Date. They do do a division at, um, then disappear. Yeah, and they'll down. Poor wind 73 and appear says replayed people make a year. Chemical shipped is due to attachment off CST group to the auction item. So Okay, this what the A Similarly, it shows a basic pick. I am by Sage 58. Then we can say we so we can observe. We can hope so. Two signals the little Yeah. I mean, you go related. Integral. Um indeed, Rinse When we used to do in the what? 13 seed that kids, one ends and then my respect. Uh huh. Speak. Thanks a lot.

Well, everyone, today we're doing Chapter fourteen forty to this question asks us that how could he use chemical shift in interrogation data for these Proton nmr spectrum's to distinguish between each pair of the falling compounds and is telling us that the Asian Amar spectrum of each compound only contains single? It's so how can he use the chemical shift and interrogation integration data to distinguish between these two compounds? Well, let's determine the number ofthe chemically unique protons between the first parent. These two right here. So you see, we have three protons and these terminal carbon groups each have three protons for total nine protons here. Then we have another three protons, another terminal carbon here. So since we have two unique, chemically thing protons, we can determine the racial between them. So here we have nine protons here at three SOS of nine that the regeneration which reduces down two, three to one ratio. And specifically it's three blue protons for every one green proton. And what about this one over here? And so this This one has three protons here and three protons here. So there's a three to three ratio of these chemical unique environment of these chemically distinct protons. So you'LL get a one to one ratio. But what about chemical shift values? So we see that the's protons and green are just that adjacent to a carbonnel group while these protons and blue are in Jason too. Miss Esther, option right here over these three proton groups here which all identical haven't oxygen separating thiss Esther option here. So we're going to conclude. Is that because the protons ond right are closer to your election? Negative, Adams, these all these protons would be greater downfield shifted then this protein the protons in this mock you because of that spacer carbon group right here. Now what about for the next pairs a pair? Be so impair. BBS identified that both these molecules having internal in a cemetery so I would too determined the number of unique protons. I know that for once I reports on on the left I seem to report on the right. So if I take my ratio off glue to green protons, I was six to four ratio which produces down two, three to two. However, for this proton for this molecule over here on the right. If I draw only chemically unique protons. I remember then, trying planets Symmetry goes down through the carbon. So we know that there's only one proton either side. So here we have a ratio of to two six which reduces down so three, two, one. So the ratios of the area underneath the peaks of Anna marble differ so the integration will differ. We'Ll have different racial integration, so it will be a distinguished between these two marshals here. So what about this? Last night you weren't here, So if you identify that all these monkeys actually having terrible symmetry as well so we can cut this molecule like this I mean, cut the second molecule like that so I would determine the number of unique Proton groups Or what's true on one side of the mirror will be also true. Another side, they'LL be the same type. So it turns out I also have two types of protons. It's going to exist in a four to six ratio which reduces down to three, two, two. But now what about this monk you here? So if he's seen this carbon is going through the trunk line symmetry So these all the same types remember what's true on one size to another side. So all nine of these protons are actually existing in the same chemical environment, and then these three protons are still exist, and it's unique. Chemical environments are gonna have a ratio of three to nine, which reduces down two, three, two, one. So once again, the ratios the area underneath the NMR peaks will defer for these two molecules, so you'LL be have distinguish them by their integration volumes.


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