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Consider = molecule containing lhe non-equivalent prolons: Ha & HB & 3c. Provide 1 splitling UCC(on graph PJpET) undthen draw the multiplet for prolon Ie in...

Question

Consider = molecule containing lhe non-equivalent prolons: Ha & HB & 3c. Provide 1 splitling UCC(on graph PJpET) undthen draw the multiplet for prolon Ie in lhe spice below . Label the Lype ol _multiplet for Ha (doublet o[ doublets etc nEM Ip LPPm Coupiing Lonsiant 'Jer 4L LUHE

Consider = molecule containing lhe non-equivalent prolons: Ha & HB & 3c. Provide 1 splitling UCC(on graph PJpET) undthen draw the multiplet for prolon Ie in lhe spice below . Label the Lype ol _multiplet for Ha (doublet o[ doublets etc nEM Ip LPPm Coupiing Lonsiant 'Jer 4L LUHE



Answers

Draw structures corresponding to the following IUPAC names: (a) $(Z)-2$ -Ethyl-2-buten-1-ol (b) $3-$ Cyclohexen-1-ol (c) trans-3-Chlorocycloheptanol (d) $1,4-$ Pentanediol (e) 2,6 -Dimethylphenol (f) o-(2-Hydroxyethyl)phenol

In this problem got boned by separate components. Anyone? It is having a cuter group. There is. It's a triple pane dream, I think. And there are no guns. It's a and let's be so here. Everything we know here hided in it. See No Khomeini And the more signals would be everything for this morning. We know it'll complete the number off signals. We have to begin to play the enrollment in which youthful bones are you taking? I felt a business cycle. Then more your I'm go. But I didn't get all more door this morning. You're religious? No, I said fare home. So we will have a fleet of rupiah or the second goblin. They have the blowing. It could golden. Well, maybe you picked it with you on me. Thanks for painting is one big dog that it's is sexy. It's and it's me. You know, when we look at this morning tour, we have each company's having actually, I could go on and hoping it and all these actually could you ever heard of him are independent and Roman. So each of the problem in this morning really so a separate big the anymore. So one to okay. Or flight six. So when? Eight night, my insignificant. We're before every 90 minutes from this one put together Must That is because of the there's, you know, topic programs you got seeing here. No, this pleading pattern on the bench a were disappearing as a doublet of doublet at one point Hato. So they're speaking, but it's a doublet of doublet. So they're sitting back and realizing you have did the company. What do you mean, A and D. C? I like a drink on the go 8.2 Hutz, it's then they see big words 8.2 hurts. That means this distance is 8.2 hooks and then we have it's de on me. You yet? I know the name. Okay, on the they give me the couple in with them in these two Pro Bowls. It doesn't give me that Bobo on this to work Regional road owns. So the couple in with you in well, on the well, big coupling Constant is given us like pointy hoods. So we make, like 13 hoods. So we get hole Cygnus for three hoods, so pick up news. Uh, like this. Okay. Never picked up doubling

Let's solve this problem from complex irons and coordination compounds where we have to draw the possible structure for these complex species. The first one is pretty sealed for two negative. It has a square planer. Geometry and structure can be shown here, platinum in the center as the central atom surrounded by four, including legions. Overall, there is a too negative charge on this complex. The next one is facial CEO at Tool Whole Tries, and it's three whole tries to positive this exhibits of federal geometry, where cobalt is the central item surrounded by six like Games three and it's three molecule, which are on the same face, and three water molecules, which are on the other phase oral. It carries a to positive charge. The third one also exhibit octahedron geometry, where chromium is the central atom, surrounded by 582 molecules and one cl ligand. Overall, this species carries a to positive charge, so this is the plausible structure of the third one, and these all are the plausible structure. The given complex species

Well, everyone, today we're doing Chapter fourteen on thirty two in this poem assets the job, the structure A compound with the molecular formula of C four h eight. Oh, and that has a signal in this carbon thirteen spectra that is, has a chemical shift value that is greater than one sixty ppm and then also draws I simmer or an ice over with same formula, but the only difference between them and has a signal with lower than one sixty ppm. So we need identify what car type of carbon exists when it's greater than one hundred sixty ppm. And if you look at our chart table correcting a textbook, we know that Carbonell carbons. So this carbon here, the carbon all carbon, produces a chemical shift value that is greater than one hundred sixty. It's over two hundred sometimes, so we know that the initial structure must have a carbon or carbon. And we know that the ice armor that doesn't have a signal above one sixty cannot have a carbon carbon. So we know that this option must be either some sort of ether or some sort of, um, alcohol group or something like that. So let's start with the first example here. So we need to determine how many double bonds or how many degrees on saturation exists in this formless. So using in my equation that helped me out so much as a student the number of double bonds or number the degree of saturation is number of carbons for four minus the number of hey, lights and hydrogen is over two eight hundred zero. He lives over too, minus the number of Nigerians over to which is your over to which is zero plus one is that for months for zero plus one equals one. So that means that we know that we have to have one double bond in our molecule. And for the cardinal case, this is already a total bond. So I'm going to do now is just fill in the best of the structure out with a propos structure like this. So that matches the chemical formula. So we know that we have one carbon one option that was already used. It's now we have C three h ate that we need to satisfy. So this is one, two, three carbon. So now the C three satisfied, and this turmoil. Cardinals three protons destroy macarons. Three shots six. And this one has to says eight. You know that this is also satisfying. Alternatively, I could have drawn something like this. This would also satisfied. But now this an alga. Hyson Aldo hide is different than a carbon all group. So we need this cardinal to be some sort of key tone to produce a signal that is above one sixty ppm and not Aldo Hide. Ah, carbon All group here because I would produce a lower signal. So we know that indeed we do need this Ah, and our group or a group on the other side of the cardinal to produce this key tone so we would have that one six ppm. So that's why we know that this structure is correct and not this Well, what about if the A nice murder has signal lower than one sixty ppm so lower than one six people? And we know that the carbon must be bound to some sort of election Native Adam And what we have here is oxygen. So we know that we can draw on alcohol directly bound to a carpet here so this will produce this carbon signal that's still quite high but will be under one sixty ppm. But a degree on saturation is still one. So if I draw the other three carbons one, two, three. So we have one, two, three, four. I need to add one double bond because I know that this is not going to be a Cardinal carbon here. This will not be a carbon carbon here because they now produce over one sixty. So I know that it just needs to be a double bond somewhere. So if I was a drug dealer bond here, for example, I would see that we have two protons here. One here says three, four, five, six, seven, eight. So this is satisfied satisfies our Muckler equation here. But what about if I do so, bond? Here we have one, two, three, four, five, six, seven, eight. So there's also another possibility that we could have for thiss isom er here for this question

This is the answer to Chapter 17. Problem number 29 fromthe Smith Organic chemistry Textbook. This problem asks us to draw a structure corresponding to each name. Okay, Uh, and so for a were asked to draw paradise chloral benzene. And so remember, pere substitution essentially means 14 substitution. Ah, and so there's our benzene. There's one chlorine, and there is a second chlorine in the pair of position. Okay, let's say B that was gonna be meta koro Phanom. So meta is gonna be 13 substitution. So here is a final, uh, and the carbon with the hydroxy group is always carbon one. So, uh, then we need to put a chlorine a carbon number three. And that's ah meta chloral phenomena. Okay, see, we're asked to draw para I Oto Anna, lean her in line. Ah, And so remember, aniline is gonna be a narrow Matic ring of benzene ring with on mean attached to it. So here's our mean Ah, and then pere again is like a 14 substitution pattern. So pere i Oto. So there's the iodine para to the, uh, meat. Okay, so then for d, we have Ortho bro. Mo nitro benzene. OK, so, as the name implies will start with benzene ring again. Um, and then we have poor so, bro mo nitro. Okay, so, uh, we need to put a nitro group on it and then orth o to that which again means, like, one to substitution eso or so that we will put a roomie. Okay. And so there's d for e were asked to draw to six die meth Oxy Tali. Helene. Okay. And so remember, tell you Lean is gonna be, ah, benzene ring with a meth old group substituted onto it so we can put the metal group here. So there's our tally lean, and that's gonna be carbon one, the carbon with metal group on it. And so now we need to meth oxy groups that carbons two and six. So there's one and there's the other. Okay, s so then for F Ah, we're asked draw to fennel one beauty mean, Um okay, so we need to draw to fennel. One pew teen. So in this case, we're naming this as a butane chain with double bond, innit? Um, and a fennel ring substituted onto it, and so the fennel ring's gonna be a carbon to and carbon one is gonna be, um there's gonna have a double bond. So I'm gonna draw that like this. 1234 Here's our double bond at carbon one. And then here is our gentle rain. Okay? And so that's Ah to fennel. One beauty in there. Ah. So then Fergie, whereas to draw to fennel to probe in one all Okay, uh, and so again, we're naming this, uh, as a propane chain s 01 to 3. There's our propane chain, uh, with an alcohol on carbon one. So let's put that there s so we're gonna number right to left here. Um, a double bond carbon to So there we go. Carbons 2 to 3. Ah, and then a fennel ring. Also in carbon too. So we need a federal ring right here. Okay. I'm so lastly for H. We're asked to draw trans one. Benzel three fennel, Psychlo, Pantene. Okay, so we need to draw the cycle plantain to start. So there's our psycho plantain on. Now we have trans one Benzel three central. So let's ah, let's put the ah three carbon right here. So this is gonna be our central ring. Okay, And then that would make this carbon three. And we need to put a benzo group here. Ah, and remember, the Benzo group is gonna be ah, centering with with an extra Corbett. So a fennel ray, um, separated by whatever it substituted onto separated from whatever it's substituted on to buy one carbon. And so this is the trans orientation here. We could draw it the other way if we wanted, you know, switch, which is wedged in, which is dashed. It doesn't the problem. Doesn't specify. It just specifies that they have to be trans to one another as they are here. Okay, so that's all of these, Um, and in orderto answer these, we just need to remember the rules for naming, um, molecules with bending rings in them on. Then apply those rules to these eight names that were given to come up with these structures. And that's the answer to Chapter 17. Problem number 29


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