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7.45 7NMR online # of Signals pdfFor Homewone Nwmhc 0l Sunan Ic each rCsonance Inteeratlon as descrnadlntina vldeoNctAIl" 2 (Ilc -enoxypronlx MIlunc SunciheIhc...

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7.45 7NMR online # of Signals pdfFor Homewone Nwmhc 0l Sunan Ic each rCsonance Inteeratlon as descrnadlntina vldeoNctAIl" 2 (Ilc -enoxypronlx MIlunc SunciheIhctieacl -enoty-L-nicthulpronine14 dlethyleycksietane (KL-S-ciht/--IncIhoxtcpIall->-Ollc 2Ixptopyl- $-mcth} Ibutanolc ucld

7.45 7 NMR online # of Signals pdf For Homewone Nwmhc 0l Sunan Ic each rCsonance Inteeratlon as descrnadlntina vldeo NctAIl" 2 (Ilc -enoxypronlx MIlunc SunciheIhctieacl -enoty-L-nicthulpronine 14 dlethyleycksietane (KL-S-ciht/--IncIhoxtcpIall->-Ollc 2Ixptopyl- $-mcth} Ibutanolc ucld



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Describe the $^{1} \mathrm{H}$ NMR spectrum you would expect for each of the following compounds, indicating the relative positions of the signals: a. $\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$ b. $\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}$

Everyone today we're doing Chapter fourteen problem twenty And this problem asked us how many signals are present in the Proton and Amar Spectra for each of the falling molecules, What's the splitting? Observed each signal. So let's start with a this one over here. So we knew right away that these terminal carbons of three protons each Then we can all child and we know that for my doctoral, discarding has four bonds was no protons attached to it was carbon should have two bones here to hydrants here. So if I'm interested in identifying the spotting patterns on the number of peak for each pose on the first day I need do is identify the number of unique, chemically distinct protons I have. So we know that this is Sigma Bond here, So these three protons can rotate and you kind of get a blurred effect of all three d. So these all act as if they're in the same chemical environment. Next, we have thes two protons here which is bound to the same carbon. And then we have finally are alcohol proton here. So if we start from the left, if you see that we're looking at one of these examples, we see that we're interested in these three protons and I want to know the spotting patterns. So neighbor plus one. So I'm looking at the neighbor carbon. Here we have one neighbor carbon. And how many protons attached shows? One zero protons wasn't zero plus one. So we're gonna have a single it here. So these nine protons going to be represented by a single peak NMR Spectra. Well, what about these two protons here? Well, these protons two protons are bound to a central carbon as bouncing auction and bound to a carbon. The oxygen So in thing to remember is oxygen's and nitrogen protons do not split other signal and they're not swayed by adjacent protons. So even though this options bound to croton, you can on Lee do splitting from carbons and not oxygen's or nitrogen. So that's something a key thing to keep in mind. So even though the yes, there is a proton attaches auction is not involved with sporting thes h h is off the carbon. So don't other hope is to look at this carbon here, which is also neighboring the carbon that we're interested in. But there's no protons off this carbon. So neighboring equals zero plus one equals a single. It's so these two protons what he calls a single it. And once again, protons of a mites or a means nitrogen and oxygen is do not split too right away. We know that this also be a single it. So we have three distinct protons, so we're gonna have three signals, each of them being a single it. Now, if you go on to second problem and we draw out, all are pros arms. So now we see that we have three unique different types of protons because each of these protons exists in its own chemically distinct environments. So we know that we're gonna have three protons from carbons. But then we also going to consider that we have this alcohol close. So in total, we have four different types of protons. So now that we're gonna have four unique and, um, our signals. But what about the peaks? Course went each one will if you look at this terminal protons here and off this general carbon. We see that this carbon is neighborhood one other carbon here attach this. Cartman, we have two protons. So any clothes too? So we do. And plus one, two plus one equals three. So we know that these three protons are gonna be represented by a triplet? No. What about Thies too? Full talks Well, these two protons are bound to central carbon is bound to one type of protons on the left and another different opals on on the right was about to different types of protons. And if you guys remember from previous examples, there's an equation that we gotta multiply the facts by each so using the M plus one rule. So the equation for a number of total possible number peaks will get for these two protons is and plus one times M plus one. So I am is just a different types of protons and him and are different. I support Aunt Emma's the red proton and is the proton. So if I follow the rule, we have hundred protons neighboring on the left of three plus one is less. Four times were, but, um so we have another name in carbon on the right and his naming carbons bound to two more proton. So we have two plus four nickels three. So this one will have a maximum possible off twelve peak. So this will be a multiplex here. And what about thes two protons down here? Will these two protons are neighbored by auction member oxygen? Doesn't matter if it has a hygiene or not. These do not at least are not involved in splitting. So we can tell that out And on the left side we have a carbon that has two protons bound to its neighboring equals two plus one equals three. So we'LL have a triplet here for these two protons right here on Obviously this hydrant is not involved The auction's hygiene non involved in explaining So we have a single it here. So until we have a triplet triplets England at a multiplex correspondent Ford chemically distinct protons. And I just look at this last example here So once again draw your different protons So we see that we have three types of protons blue, green and black So start with blue. So we see that people to these protons are neighborhood by one carbon And this carbon has one puts on neighbor So we have n plus one. So one plus one equals two. So these six protons will be represented by two peaks being a doublet. Now, what about the central carbon? Well, this central carbon is neighboring. Two carbons and a nitrogen. Nitrogen is not involved in splitting, so we can just ignore that. But these two carbons are each about three protons. So because we have two different neighbors, but they're not, actually and chemically distinct environments these air identical neighbors, as you can see it by the both being blue colored these identical neighbors. So we do not use this equation. We just use the end plus one equation and we consider all of these protons equipment so they all are and the end type of protons there. We don't have any m types of protons in this example because these two neighbors off the central carbon are identical. So because of that, we see that we have six neighbors. Plus, once we get seven, so we get seven peak signal for this middle carbon here and obviously these two protons once again, nitrogen and action prisons do not split, so we just get a single it for this one. So in total, we have three different types of peaks, one being doublet, one being having seven peaks of one, being having one peak here for nation protons

18 Annual in is a cyclic conjugated hydrocarbon with 18 carbons and 18 hydrogen. And the double bonds are arranged in such a way that Six of the hydrogen is point into the interior of the ring and 12 of the hydrogen is point to the exterior in aromatic compounds. The magnetic ring current is such that the exterior hydrogen R. D. Shielded. That's a mark of Arum Ettus city and we think of benzene hydrogen how their D shielded compared to regular alkaline hydrogen. However, the magnetic ah field lines indicate then that the interior hydrogen should be highly shielded. So 18 annually has to NMR signals One that's d shielded to about 9.25 parts per million. That would represent the 12 hydrogen is on the exterior of the rain. And then a signal that um in the negative part per million values highly shielded, which would represent the six hydrogen pointing to the interior of the

This is the answer to Chapter 14. Problem number three from the Smith Organic Chemistry textbook. Ah, and this problem asks us how many one h and M or signals Does each compound show on DSO? Remember that each different type of hydrogen, each different proton in a molecule, eyes going to give its own signal. Um and so the number of different types of hydrogen is going to determine the number of signals on DSO. For a there's only one type of hydrogen. So all six of these protons are equivalent and so there's only gonna be one signal for all six of them. So then for B ah, we have two types. So we have these ch three types, if you will. So they're going to give a single signal on. And then the two protons here on the central carbon are also going to give a single signals. That is two signals s 02 different types of hydrogen Sze. So then, for si ah, we also see two different types of hydrogen Sze Sofer, See again, we have the metal protons on the ends on. Then we have these equivalent methylene protons inside, so we will also get to signals here for C. Okay, so then for D? Um, again, we're only gonna get two kinds of signals. So we have, um, six of these metal group protons on each side. Um, so they're going to give a single signal 12 of those protons, and then we have these two meth eyeing protons in there, so just ch And so they're also going to give a signal. So once again, for Di, two different kinds of protons to signals. Okay, so then, um e uh, each of these is a different type of proton. So, um, looking at B, c and D, What we saw basically, was that there there was, ah, plane of symmetry through the middle of each molecule so we could divide each of these molecules neatly in half. And that's Ah, how we would, um that's why um, the protons, the either side were equivalent. So that's not the casing e. There's not a plane of symmetry here because it's It's an Esther. Um, and so the right side is gonna be different than left side. I said these protons are unique. These protons are unique. These protons are unique. And these protons are unique. And so the end result of that is going to be four signals. So four different types of protons are they'll each give their own signal. So four anymore signals I'm S o. F is a similar case, so I would just again just highlight, um, each of these are different. Um, and then these six here. So the two metal groups are all going to be a single signal. So again, four signals for F. Okay, Uh, and then so looking at G, what we see in G's that will get a single signal for this method group. So for these three protons and then the presence of this chlorine means that each of these CH two groups is different because the first CH two group will be adjacent to the chlorine, and then the next one will be one away from that, and then one away from that. So on and so forth. So the answer here is actually gonna be eight signals Now, in real life, would you be able to distinguish all eight of those signals? Probably not. It would probably all run together on the spectrum, but for the purpose of this problem. You would, uh, there should be eight to state signals on. And if you had a powerful enough magnet, you probably could get those to resolve. Um okay and then sew for H. Um, h is gonna be a similar case. So even though there are these two ch two groups, one of them is immediately adjacent to this oxygen in the alcohol and the and the other is not. So we're gonna have a signal for these three. A signal for these two. Ah, a different signal for these two. Um and then the alcohol proton is also going to give rise to a signal s. So the answer to H eyes once again is four. And so that's the way to approach this problem. We just need to identify how many different types of protons are in each of these molecules. And then each different type of proton is going to give rise to its own signal. So we counted different types, and that's how many signals we should expect. And that's the answer to Chapter 14. Problem number three

This is the answer to chapter 20 to problem number 87 Fromthe Smith Organic chemistry textbook. Ah, And this problem says the Proton anymore. Spectrum of two claro Seta Meid shows three signals at 4.0 to 7.35 and 7.60 ppm. What? Protons have rise to each signal? Explain why three signals are observed. Okay, s so the 4.2 ppm protons, um or P P m signal, uh, comes from these two protons. Um, and then the 7.35 ah, and 7.60 ppm. Signals come from thes two protons. Um, and the reason that three signals are observed instead of just two signals, um is because of the residents structure. For this, a meid that looks like this, the electrons move, um, in the fashion that I just drew. And so we get a resident structure that looks like this. Okay, um, and so, uh, this structure contributes significantly to the hybrid. So significant contribution to resonance hybrid. Uh, and since it's such a significant contribution, we do need to take into account. Um, you know, the features of this resident structure. Ah, and so because in this structure there's a carbon nitrogen double bond. Um, there is limited rotation, so carbon nitrogen, double bond, uh, limits rotation around that bond. Ah. And so, uh, the two protons on the nitrogen are sort of locked into their environments. And so you have one that could be considered sis. Uh, t the oxygen. Ah, and then one that could be considered sis two. Ah, the the c l c h to great. Ah, and so they are actually in different environments. And since they're in different environments on their stuck in those environments because there is very limited rotation around that carbon nitrogen bond because it has significant double bond character, the result of that is that they're not equivalent. And so they each give their own signal. Ah. And so that is the answer to chapter 20 to problem number 87


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