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Redraw the structure and abel the carbon atoms; Assign each carbon atom to peak (for aliphatic peaks it is not necessary to figure out which is which exactly:) (6pt...

Question

Redraw the structure and abel the carbon atoms; Assign each carbon atom to peak (for aliphatic peaks it is not necessary to figure out which is which exactly:) (6pts)180160140120100 PPM

Redraw the structure and abel the carbon atoms; Assign each carbon atom to peak (for aliphatic peaks it is not necessary to figure out which is which exactly:) (6pts) 180 160 140 120 100 PPM



Answers

Identify the carbon atoms that give rise to the signals in the $^{13}C$ NMR spectrum of each compound.
a, $CH_3CH_2CH_2CH_2OH$; $^{13}C$ NMR: 14, 19, 35, and 62 ppm
b. $(CH_3)_2CHCHO$; $^{13}C$ NMR: 16, 41, and 205 ppm
c. $CH_2 =CHCH(OH)CH_3$; $^{13}C$ NMR: 23, 69, 113, and 143 ppm

So here we just got another animal example. So we're going to be assigning our structure. So I've drawn our structure and full on the screen so you can see all of our bonding and all of our protons. And as you can see, I've already made my assignments on the screen. You can see that I've identified all of my protons here.

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

So we have two carbon atoms present in our structure. So for the first carbon Adam, there are four bonding groups around since we have three hydrogen sze and then the following carbon. So we have four bonding groups with no lone pairs. This is going to be a tetra. He'd roll and tetra Hydra lt's usually have a hybridization of S P three for the second carbon. Our second carbon has two bonding groups. So we have the first we have the nitrogen in the second carbon in the first carbon, actually. So we have two bonding groups in zero lone pairs. Then this is going to be a linear molecular shape and linear molecular shapes have a hybridization of S t.

Hillary won. Today we're doing Chapter 19 problem eight, and this room asked to identify the structure of a compound with a molecular formula of C four. It's 802 and with each anymore, spectra of a triple it. That's your 0.9 at double its or multiply, representing two. Hydrogen is at 1.6 of triple representing, two agents at 2.3 and a single. It's representing 111.8. So how we should go about solving this is first we need to turn the degrees on saturation. So how we do that is, well, the degrees on saturation or the number of double bonds equals the number of carbons minus number of hydrogen and highlights over too. So it's eight or to my X number of nitrogen is over to. That's your over two plus one. So you're four minus four equals zero plus one equals once we one degree oven saturation. Sometimes you know we have one double bond in this molecule, and if you look at the beach and more data, we see that we have one proton that's going to be a single it so it's not neighboring anything that can ah couple or split the signal at 11.8. So right off the back, we can say that the only way that we can have a proton that is not neighboring any other protons is that if it is on this oxygen, that is next to a cardinal group. So essentially, we have a carbon so acid. So you know that this proton is extremely acidic, mean that is going to fall off very fast. And it's also highly de shielded by the left arm. A drying groups, uh, adjacent to it, one being this option one being the carbonnel auction. So we can say that because it's gonna be highly dif dif. You listen to be significant on the left side of special spectrum and have a high ppm value such as the one that says 11.8. So we know that this proton is gonna have the highest PPL PBM value. So we know that that proton is taken care of and now we have this one carbon here and this should be a second carbon here. And then let's draw the other two cartons to satisfy the molecular formula with straw straight chains and see if this is true. So let's now work on the most shielded proton. So we should have three protons here. They're going to be the most shielded furthest away from these electronic drawing groups. And these three protons are neighboring two protons here. So we should see triplets. Or we should see these three protons being represented by two plus one peaks that equals triplet for three peaks. So three proton should be representing three peaks that should have the lowest ppm value. And that's actually what we see. We see that a 0.95 ppm. We have three protons that correspond to a triplet. So now what about these two protons here? Well, their neighboring three protons All right, but they're also neighboring these two protons on the left. So these two protons and blue, these two protons and blue are gonna be represented by a multiplex. So remember, you have your n plus one times you're n plus one. So you're gonna have four times three is. You might have 12 peaks in this one maximal, and this should have an intermediate value of P p. M. Somewhere between the Black Protons summer between the green proton. So these blue protons, if you look in between. So we have 0.991 point 65 and 2.3. So we should be interested in 1.65 And in fact, yes. This indeed is a multi planet representing two protons. So this supports this potential structural we drew here. Now, what about the green protons on the left? There's no neighboring. Protons are right. There's two. So these two protons to be represented by a triplet and they should have the highest value compared to other than 11.8. And that's going to be actually 2.3 triplets representing two protons. And that's exactly what we see in our questions them. So this is the proposed structure with given the data from this question.


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