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Proton NMR practice 1

Practice determining the structure of a molecule from the molecular formula, hydrogen deficiency index, and proton NMR spectrum. Uses example of ethylbenzene. Created by Jay.

Video transcript

- [Voiceover] Let's say we're given this molecular formula. C5 H10 O and this Proton NMR spectrum. And we're asked to determine the structure of the molecule. The first thing you could do is calculate the Hydrogen Deficiency Index. And so if we have five Carbons here, the maximum number of Hydrogens we could have is two N plus two, where N is equal to five. So two times five plus two is equal to 12. So 12 is the maximum number of Hydrogens. Here we have only 10 Hydrogens so we're missing two Hydrogens. We're missing one pair of Hydrogens. So therefore, the Hydrogen Deficiency Index is equal to one. Immediately, that makes me think about a double bond might be present in this molecule or one ring. Let's go down to the integration value. For this signal, there's an integration value of 27. For this signal, the integration value is 40.2. For this signal, it's 28.4. And for this signal, it's 42.2. Remember what you do. You divide all four integration values by the lowest one. So the lowest one is, of course, 27. So 27 divided by 27 is, of course, one. 40.2 divided by 27 is pretty close to 1.5. 28.4 divided by 27, that's pretty close to one. And 42.2 divided by 27 is once again pretty close to 1.5. Remember, these are just the relative numbers of protons giving you these signals but you can't have 1.5 protons giving you a signal. We need a whole number. And we need to account for 10 total protons here. So if we multiply one by two, then we get two protons. So this signal represents two protons. If we multiply 1.5 by two, we get three. So this signal represents three protons. If we multiply one by two, we get two. And if we multiply 1.5 by two, we get three. So if we add all those up, two plus three plus two plus three, that's, of course, 10 protons. So now we have accounted for all 10 protons using our integration values. Next, let's look at each signal one by one. We'll start with this signal. So we have a CH2 since we have two protons. So this signal represents a CH2. How many neighboring protons do those CH2 protons have? All right, so we can figure that out by the number of peaks on the signal. So this signal here has one, two, three peaks. Now if you think about the N plus one Rule, if you have N neighbors, you get N plus one peaks. So if we have three peaks, all we have to do is subtract one to find out how many neighboring protons. So three minus one is equal to two. These CH2 protons have two neighboring protons. Let's think about the chemical shift for this signal. So the chemical shift for this signal is between two parts per million and 2.5. And that's in the region for a proton next to a carbonyl. Now it makes a lot of sense because we calculated an HDI of one indicating their (mumbles) double bond present and we need to account for an oxygen in our molecular formula. Let's go ahead and draw in a carbonyl. These CH2 protons are next to the carbonyl. We know that because of the shift, right? The carbonyl, the oxygen deshields these two protons a little bit and gives us a higher value for the chemical shift compared to something in Alkene type region. All right, let's color these protons in here magenta. So we're saying this signal is due to those two protons. Let's move on to the next signal. So we have three protons. That would be a methyl group, so CH3. How many neighboring protons for those methyl protons? Well, there's only one peak here. So one minus one is zero. So zero neighboring protons. What about the chemical shift? The chemical shift for this signal is once again past two parts per million. So this signal is deshielded, all right? So these protons must be deshielded a little bit. Those must be next to our carbonyl. We'll draw in our methyl protons, right? Being deshielded because they're next to the carbonyl. Let's make these blue. So these protons in blue here, these three protons are giving us this signal. And we should expect zero neighboring protons, all right? So those protons are on this carbon and the next door carbon has no protons on it. So zero neighboring protons makes sense. Move on to the next signals. This signal represents two protons, so that's a CH2. How many neighboring protons? All right, let's count how many peaks we have. One, two, three, four, five, six. So six peaks. We subtract one from that. Six minus one is five so we will expect five neighboring protons. Let's move on to the next signal. Three protons, all right? That's a methyl group, so CH3. How many neighbors? Well, we have one, two, three peaks. Three minus one is two. So we will expect two neighboring protons. Now these two signals, this signal here and this signal here, we're talking about under two parts per million now. So these must be the furthest away from the carbonyl, all right? They're not being deshielded as much as the two signals in this direction. So these two neighboring protons for the methyl group must be these two protons right here. Let's go ahead and draw that in. So we have our methyl protons right here. Let me make these protons red. So these protons right here in red are giving us this signal. From the signal, we know that these methyl protons are next to two neighbors and so we must have a CH2 next to that methyl. And so let's go ahead and draw in our CH2. This CH2 must be this signal right here. All right, so let's think about how many neighbors. We expected five neighbors for these CH2 protons. Let's count them up. One, two, three, four, five. So five neighboring protons, matches what we see on the NMR spectrum. Now, those protons, those magenta and red protons are actually in different environments. And so the simplified version of the N plus one Rule isn't quite true but it works for this example. It works for this example so we're gonna go with it because all we care about is getting the structure of our molecular here. All right, so let's think about the red protons again. So the red protons have two neighbors. How many neighbors of the red protons have? Here's one and here's two, so this makes sense. So that would make sense. This one makes sense right here. We just talked about that. Let's think about the magenta protons. So the magenta protons were supposed to have two neighbors. Let's look at the carbon next door to this one. So here's the carbon next door. Here we have a neighbor and here we have a neighbor. So this makes sense. And then, we already talked about the blue protons, right? Having zero neighbors. So everything seems to make sense here. To sum everything up, make sure to count all of your atoms and you will get, of course, five Carbons, ten Hydrogens, and one Oxygen when you do that. So this is the Dot Structure that we were trying to find. Let's take a look at another one here. So we have a molecular formula of C8 H10. Let's go ahead and calculate the Hydrogen Deficiency Index. So if we have eight Carbons, then we can have a maximum of two times eight plus two Hydrogens. So two times eight is 16, plus two is 18. For eight Carbons, 18 Hydrogens is the max. Here we have 10 Hydrogens. So we're missing eight Hydrogens or missing four pairs of Hydrogens. The Hydrogen Deficiency Index is equal to four. Remember, anytime your HDI is equal to four, you should think about a Benzene Ring. I'm going to go ahead and draw a Benzene Ring in here. Let's look at the integration. All right, so sometimes you'll see the integration given like this. This represents five protons for this very complex looking signal right here. This signal over here represents two protons and this signal represents three protons. Let's go back to these five protons with this complex signal, all right. This is in the Aromatic range, all right? So approximately 6.5 to eight. Those must be five Aromatic protons. We can go ahead and draw in five protons off of our Benzene Ring. Even though those protons are in slightly different environments, because this integration value is five here, we know a Benzene Ring is present, we're done, all right? We don't have to worry about the slightly different environments. We know that these five protons are giving us this complicated signal over here. All right, next, let's look at this signal. So two protons, so that must be a CH2. How many neighboring protons for the CH2? Well, there's one, two, three, four peaks. So if there's four peaks, four minus one is three. So these CH2 protons have three neighbors. All right, let's look at this next signal. So three protons. This must be a CH3. How many neighboring protons do we have for the CH3 protons? Well, we have one, two, three peaks. So three peaks. Three minus one is two, so two neighbors. Two neighboring protons, that must be these two neighboring protons. We'll go ahead and draw an Ethyl group, all right? So this is an Ethyl group pattern over here. There's a CH2 and there's your CH3. So let's draw in all of our protons. So we have these protons and we have these protons. Let's color coordinate here just to make sure everything makes sense. So these red protons right here, they must be giving us this signal, so these methyl protons, all right? We expect two neighbors because we have three peaks on our signal. And here are the two neighbors. One, two, so two neighboring protons. Let's look at these two protons next. So these two protons are giving us this signal. We'd expect three neighbors because we have four peaks. One, two, three, four. So three neighbors, so here's the next door carbon. So one, two, three, three protons. And then this is also a next door carbon but there are no protons on this carbon. So we see only these three neighbors. And so everything seems to make sense. Once again, count up all of your atoms and make sure you have accounted for everything. For example, you have eight carbons you have to worry about. So on your ring, we know there are six. One, two, three, four, five, six, and then we get seven and then we get eight. So the eight carbons is correct. And if you count all those up, we have our ten Hydrogens accounted for too. So this is the NMR spectrum for Ethylbenzene. All right, this one was a little bit easier than the previous example.