NMR problem

Okay. Nuclear magnetic resonance. Hydrogen nuclei exhibit a characteristic known as nuclear spin. We don't actually mean that they're spinning, but it means that the hydrogen nucleus, as it has an odd number of nucleons, produces a magnetic field. Normally, this doesn't have any effect on anything whatsoever. But when we put the hydrogen nucleus into an external magnetic field, it means that it can adopt one of two states. It can either produce a magnetic field that's in line with the external magnetic field, or it can produce a magnetic field that's opposed to the external magnetic field. Both of these states have different energy. In order to change the state from low energy to high energy, we can put in electromagnetic radiation of a specific frequency. By measuring the absorption of electromagnetic radiation by hydrogen atoms, we can find out the chemical environment of those hydrogen atoms. The chemical environment. Is affected by atoms that are nearby, which also have their own magnetic fields. So they can draw electrons away from the hydrogen or they can not draw electrons away from the hydrogen. Electronegative elements will draw electrons away from the hydrogen nucleus and expose it more to the external magnetic field, meaning that more energy will be needed to change the energy states from low spin to high spin. So by examining the energy absorbed in nuclear magnetic resonance, it gives us an idea of the chemical environment all the hydrogen protons are in. And this allows us to deduce quite a few things about the structure. Now there, I'm going to move on a little bit faster. This is about tetramethylsilane. This is an external reference which allows us to standardize all of the NMR spectra by comparing them all to a particular chemical, which tends to produce a signal. It produces a single signal, a very strong signal, because there are 12 hydrogen atoms and they're all in the same environment, and they're all attached to a relatively electropositive element, silicon. So what we get, we get absorptions appearing on a spectrum reading from right to left 0 to usually to around about 9 or 10, although we can have values up to about 16. These are called parts per million. It's parts per million of the energy required to resonate the tetramethylsilane hydrogen nuclei. So the signal that moves towards the left appears at 1 part per million, or delta 1. This would be said to be shifted by 1 delta value and around about 1. This is a typical region where absorptions occur due to methyl groups or alkyl groups. Let's move on a bit. Again, these are the integrals. The integral is the area under the curve of the absorption. And this tells us the relative number of protons that cause the absorption. Now, the interesting thing about NMR is that we can find out information about the neighboring protons, in other words, protons on a carbon that's adjacent to the carbon with the hydrogens that are producing the signal. That's a bit of a mouthful to say, but imagine we've got two carbons next to each other. Then one will have some hydrogens, the other will have some hydrogens, and they're both neighboring hydrogens. One set of hydrogens will produce magnetic fields which cause splitting in the neighbours. And the splitting pattern is due to the two different spinning protons will produce magnetic fields which are. Or two different magnetic fields. Possible magnetic fields. Now that means that if you one hydrogen, it can either produce one magnetic field or another. So one hydrogen produces two magnetic fields. Those two magnetic fields will be felt by the neighboring protons and they will then be split into two separate signals. This is sometimes known as the N plus one rule. And it follows a pattern like Pascal's triangle. So one hydrogen which isn't split will have one signal. A hydrogen that's split by one neighboring hydrogen will split into two signals of equal intensity. A hydrogen that's split by two neighbouring hydrogens will split into a three. A signal with three peaks with intensity 1 to 2 to 1. And if a hydrogen that's adjacent to three hydrogens will split into four peaks with an intensity 1 to 3 to 3 to 1. These are. This pattern appears. It's called Pascal's triangle. So this gives us something called the n +1 rule. So 1 proton splits the neighbour into 2, 2 protons splits the neighbour into 3, 3 protons splits the neighbor into 4, etc. Okay, so we can see an example from this is ethanol. It's rather a crude spectrum as I drew it and looks pretty horrible, but never mind, it'll do. You can see here there's a signal on the right hand side. This signal here has got 1, 2, 3 peaks in the one absorption. So it's been split into a triplet. So this information tells us that there are two neighboring protons. This next absorption here is split into four. Well, that tells us it must have a neighbor that's got three protons. So we can see this pattern. A triplet and a quartet is typical of Ch3 Ch2. So the Ch3 is split into three by the Ch2 and the Ch2 is split into four by the Ch3. Okay, I think that's enough information. It's been very quick, but let's go to the example. The example we're given here is a molecule with the formula C5H10O. All right, so what have we got? Well, in our spectrum, you can see the spectrum here. There are one, two, three signals. That tells us that there are three environments. Come down here. We'll just write three environments. What can I do? There we are three environments. Okay, so that means those three environments must house all ten hydrogens. So the three environments with a total ten hydrogen atoms. Now, the first environment we can see is around about delta value 1. Now, that's typical of an alkyl group that's fairly far away from, quite far away from any electronegative element. All right, now we know it's been split into two. It's been split into a doublet. And that tells us that it must have a neighbor with one, a neighboring carbon with only one hydrogen atom. So what can we draw? Well, let's have a look down here. Let's. Let's just have a little, a little draw of the possibility. Well, we can see we must have the CH like that. There must be a CH group and it must be next to a C with lots of hydrogens. Well, not lots, but it must be next to C with some hydrogen. We don't know how many hydrogens. Okay, so let's have a look around. What else have we got? So this, this. That's all we've got for a moment. We've just got the fact that there is one hydrogen splitting these C x. And if I color these CX here, let's color them, and these appear at delta about delta 1.1, the Cx. Now, that tells us that over here, next to these CX here, we're not going to have any, we're not going to have any electronegative elements. It can't be CH2, because if it was CH2, we would see a triplet. With this CH would be split into a triplet. What have we got? We've got a singlet. So that means that this CH cannot be this singlet. Okay, so let's have a look. So let's continue. So the singlet at 2.2 must be due to CHX. Again, we don't know how many hydrogens next to carbons, carbons that have no hydrogen atoms. Now, how is this possible? Well, it's possible if we've got our formula there. We know there's an O in there. So what could we have? Now it's, it's also interesting that it's 2.2. That. That's typical of a CH3 group. That's. Or CH groups that are attached to a carbonyl group. And we can see that in our. In our list of absorptions. Okay, so it's looking like something like chx, C double bond O. All right. Okay, so now up here, let's, let's continue. And we've got here a signal at 2.6 around. So signal at 2.6. Now let's have a look. It's been expanded. Now you see this signal 2.6 is 1, 2, 3, 4, 5, 6. There are seven. It's split into a septet, split into a September. Now that means it must have. Must have 6 hydrogen equivalent. Hydrogens as neighbor, as a neighbour. Well, the only way that this can be possible is if we have CH3, 2. Okay, CH3 2 gives us six hydrogens as a neighbour font. Let's make this into a subscript so it looks better. Let's make this into a subscript. Okay, so if we put this, this idea down here with this one here, we can see this CHX must be due to the chx. Must be this whole group here because it's coming next to a CH group. So we put idea one here with idea three. And you'll see we've got Ch3, 2, Ch. Now what else have we got? Well, we've got a carbonyl co attached to a chx. Well, let's count up the hydrogens. We've used 3, 6, 7. So chx. So x must be 3. And does that fit? Well, let's have a look at our spectrum. Well, the signal at 2.2 must be due to this. It must be due to the Ch3.3 color that. Okay, I can't color it up here because it's a. It's an image, this ch, we'll color it differently. This is a different environment. This must be due to. This has been split into a septet. So this CH is due to the signal at one point, sorry, 2.06. And this CH32 is a doublet at 1.1. We just color those in as well, just to make it nice. And there we have the structure.