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Il Show company L Mars, 1 # ahinoman [680 0f its L MeMs plain{ group of 100. 8 Find 1and 1 points) A 3 Is this ample 2 Tcslt consiucfcd 5 1 1 9 3 1 1 73,44 5 3 [pcobabilitie ' 0l Uccn 1 1points Find Ihe protbility 1 1probabillty of exactly 1Excd (Use Excel Google) Google)1

il Show company L Mars, 1 # ahinoman [680 0f its L MeMs plain { group of 100. 8 Find 1 and 1 points) A 3 Is this ample 2 Tcslt consiucfcd 5 1 1 9 3 1 1 73,44 5 3 [pcobabilitie ' 0l Uccn 1 1 points Find Ihe protbility 1 1 probabillty of exactly 1 Excd (Use Excel Google) Google) 1



Answers

Match the $^{13}$C NMR data to the appropriate structure.
Spectrum [1]: signals at 14, 22, 27, 34,181 ppm
Spectrum [2]: signals at 27, 39, 186 ppm
Spectrum [3]: signals at 22, 26, 43, 180 ppm

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.

Hello on today's reading, Chapter 17 from 56 and this promises to propose a structure consistent with each set of data. So Compound A has a smuggler. Formal attire absorption on this agent mark data. So let's determine 1st 2 degrees on saturation for left platform. A. So we know that the degrees on saturation a number of double bonds equals no double ones equals the number of carbons, which is eight minus number. Hodgins plus hey, lights over too. So it's 10 over to when she was five, minus the number of night vision's over to zero or two plus one to a minus. Five is three plus one equals four. So now we know here we have four double bombs of four degrees on saturation. Well, what does the eye or die tell us? The iron data tells us that we haven't S P too. And SB three hybridized carbon hydrogen bond. What is the h anymore? Data tells us all. We have a triplet which has two neighbors at 1.4, the court said, representing 300 Jin's. Let means three neighbors at 3.95 and we have a multiple. It's representing five versions at 6.827 point three. So we know that these five hundred's right away. We're going to be aromatic so right away. Let's jar and aromatic benzene ring. And we know that these air matter because they fall in the range of 7 to 8 ppm, and only that we know that there's going to be five hundreds on this aromatic ring. So now we know that one of these subsequent is going to actually be the rest of our morning to create this molecule. Not only that, I just satisfied my four degrees on saturation. Three of them are from the old bond. The last the 4th 1 is from the cyclical structures. We have 1234 1st cyclical structure. So now we know that we completed re satisfied with degrees on saturation. So now let's look at this triple this splitting pattern. Will we have five more protons they were interested in? We have a triplet that is neighbored to we have to hundreds and neighbor to three. So it's very obvious that if you have 300 Jin's that our neighboring too then these three hundred's what actors triplet, just like this And these 200 Jin's well, our neighboring three Hydra Jin's. So there will be a quartet, so we know that this will be our structure. But hold on. We just forgot this oxygen. So we know she needed at oxygen in between the bending ring and these groups to satisfy our molecular formula because we know that the benzene ring is not involved in any sort of splitting. So it's okay to at that auction here. So now actually draw and oxygen. And then I combined this. So this becomes my overall structure. So we have oxygen than an ethnic group honors benzene ring. Although I wouldn't be so bill. It's determined to use on saturation again. So number of double bonds and the degree of saturation equals the number of carbons. Just nine minus. Number of hydrogen is plus. Hale is over, too. You have 10 over too, minus number of nitrogen over 20 nations over two plus one. You have nine minus five four plus one because five degrees on saturation. So what desired data tell us, was 16 69 peak? Well, we have a carbon all group in this molecule now. The anymore Dad, we have a single of the 300 single with your visions Doublet to hundreds and Dublin two engines. But these two doubles are in the range of aromatic compound 2 7 to 8 ppm. So we know that we probably have an aromatic ratings. Come on, cure. So let's draw this aromatic friend. And right before we know that cyclical structures one document saturation, we get 23 40 reason saturation. So we know that the carbon all group will be the fifth degree oven saturation. So let's draw that cardinal group right off the benzene. Right? And I'm doing this. Why? Well, because I know that I have three hydrogen. Is that air single it? So I know that by having a cardinal group like this, these 300 Jin's are gonna be neighboring a carbon that has zero on the hydrogen attached to it. So the neighboring will be 00 plus one equals a single that we see here. But now I still have an extra oxygen. I need to add another ch three to this molecule. So what I do is I can add another oxygen, another ch three. Because I know that the speech. There is also a single it and we don't want any splitting to occur with the hydrogen on spending ring like here, here. So we had the auction first, and then I see each tree because oxygen and nitrogen is cannot be involved in any splitting. So these three hundreds will show up as a single it. But either way, I could draw this. A molecule is could be this, but it could also be something like this. We have their mattering with our cardinal group, but now we'll add the auction here and in the triple it or three. Hodgins will be here, and this will still be a single it that I'm Ethel Group will be here off this carbon because now it's neighboring car in this one does not have any hydrants either. So this would still be a single it was given. This data is actually hardest distinction between these two, but it's going to be one out of these two forms. Either one is acceptable

Ganic chemistry textbook. Ah, and this problem gives us two sets of data and asks us to propose a structure consistent with each set of data. Um, and so the first place to start here, Um, so looking, looking at a, um, first thing to do, I think, is to calculate an HD I, uh And so for a that's going to be two times six plus two minus 12. All of that over too. Us. That comes out to two over, too, Which equals one. So there's one degree oven saturation here. So then looking at the I r. Um, data that we have, um, this peak, it's 17. 40. Wave number represents a carbon oxygen double bond s o a. Cardinal. So there is our single degree oven saturation. Um, so then the proton anymore doesn't tell us too much? Um, we have two signals. So there are only two types of protons in this molecule. Um, and both of the signals. Pardon me. Both of the signals are single. It's so nothing is is splitting. Anything else? Um, so none of these protons have neighboring protons. Um, and then the other thing to be aware of eyes the carbon anymore has four signals on. And that tells us there are four types of carbons in this molecule. Ah, and the other thing that we can know this peach at 180 ppm roughly. That's gonna be our carbon Neil Carbon. Um, so it's just further confirmation that we have Carbonell. Ah, And so the only way to really draw this so that it fits all of these criteria Ah, is going to be toe have a turn. Beautiful group here. Um, and then that ter beautiful group is going to be attached to the carbon Yun. Ah, and this is an Esther, and it is a metal Esther. Um and so I are two proton signals, Um, in the proton anymore. One of those signals is from the nine protons of the T beautiful group. And the other signal is gonna be the three protons of the Method group on the other side. Um, none of them split anything or split by anything. Um, and we have our Carbondale. And so that agrees with all the data that we have for a So then, for B will just do exactly the same thing. So to get started on B. We'll start by doing an HD I So for be that's going to look like this s so it's gonna be too time. Six plus two minus 10. All of that over, too. Eso it ends up being four over to, which is too. So we have two degrees oven saturation here on. And so then, looking at the I R. Signals, we have a peek at roughly 3000 wave number, which doesn't tell us much. That's the old ah SP three carbon hydrogen bond. So that's president Many, many molecules. However, the picket 3300 is an SP hybridized carbon bound to a hydrogen. And so that tells us that we haven't sp hybridized carbon in this molecule. Um and then we have a peek at 2150 wave number on that is gonna correspond to a carbon carbon triple bond. So there's our CSP hybridize carbon. Um, so we have ah, carbon carbon triple bond here. That's gonna count for both of our degrees oven saturation that we calculated in the HD I, um And then the other thing T note is the carbon and a more spectrum has four signals So four types of carbons here, Um, And again, just like in part A. We have six carbons, but only four carbon anymore signals. Um and so this should indicate once again that we probably have a t beautiful group. So all three of the metal groups in in a t beautiful group, um, are the same. And so those three carbons air going to give one signal, the central carbon is going to be a second type of signal. Um, and then we need to account for our carbon carbon triple bonds that we know that we have So that here c h okay. And so, uh, so this is the correct structure for B. Um, And again, the way to solve these is just, ah, to start by doing HD eye and then to just take whatever information you're given a piece at a time and try to figure out what it means on then put all of those pieces together. And that's the answer to Chapter 14. Problem number 73

The geometry of the coalition and it can be predicted Officer managing behavior isn't? Which of the following? Is that correct? Yes, it is correct because magnetic behavior can be seen by the general first is the correct And I see him for two minutes involved the S two hi creation. Yes and I plus to have electronic conference and that is really 840 and seeing is that a strong field again that will pairing. So it will be the yesterday for all of us and it is within a group. Do you cannot value follows. It is incorrect because they don't know what they do for us five days then for me then P. D. C. LLC and for whom I am very scared. Ford is also correct. So correct. Option will be 1, 2 and four Palestine, correct.


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