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Structural Isomers by NMR
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0:00
Welcome to Chem Help ASAP. This video provides practice problems on distinguishing one structural isomer from another based on a proton NMR spectrum. We will emphasize four features of the spectrum – the number of signals, chemical shift, multiplicity, and integration. The video description contains links to printable copies of these problems and their solutions. Furthermore, links during the video will connect you to other potentially helpful videos.
0:29
Before we jump into the problems, let’s review the key features of an NMR spectrum that we will use to distinguish one isomer from another. First, the number of different signals is very useful. At most, each hydrogen can generate one signal, but symmetry within the molecule means that the number of signals is smaller as some signals will likely overlap. The second trait is the chemical shift.
0:58
The chemical shift refers to the position of a signal along the x-axis and is determined by the electronic environment around the hydrogen. Terms like upfield, downfield, shielded, and deshielded are common when discussing a signal’s chemical shift. The third trait is multiplicity, which is the pattern of the signal – singlet, doublet, triplet, etc. – and is determined by the number of neighboring (vicinal) hydrogens on adjacent carbons. Finally, we have integration.
1:34
The more hydrogens that contribute to a signal, the larger that signal is based on the area under the peak. The area under the peak determines the integration. Of all these features, I would strongly recommend spending some time to memorize the representative chemical shift values – aromatic C-Hs, aldehydes, alcohols, etc. Memorizing these values takes some time, but knowing typical chemical shifts is extremely helpful for analyzing NMR spectra.
2:06
The different spectra in this video were generated by the website nmrdb.org – a fantastic resource. Alright, as we work through these problems, I am not going to follow a specific set of steps. I tend to do these problems by feel. If you are looking for the best steps for interpreting spectra, then I don’t have that. Hopefully, if you watch all these problems and solutions, you will see multiple different approaches to answering these questions.
2:36
You can then rely more on a specific method if that one works better for you. OK. Let’s go.
2:44
Here's the first problem. Now when I see a spectrum, I first look at the spectrum. I'm not going to look at the structures below the spectrum. Let's just get a feel for what the spectrum is showing us. So what stands out to me is, I see three signals. Now it's possible that this really isn't three signals -- that some of the signals are different yet overlapping in a strange way. But to me, these these signals
3:10
look fairly clean. That's a nice clean doublet. That looks like a nice clean doublet. This is probably three signals. So three different types of hydrogens in the molecule. So it's a fairly simple molecule. Then I'll look at the integration. This looks like two hydrogens, one, and six. And six, that's a pretty big number. So there are a lot of identical hydrogens. Six is two times three. Anytime you see threes show up, that often means methyl groups. So this six looks
3:39
like two different methyl groups, or two methyl groups that happen to show up in the same place. Okay, so now let's look at the molecules we have and see if this can clarify anything. Our first structure on the left... This molecule should have at this CH2. Maybe that's signal A, B, C, D. So on first glance this molecule, this leftmost molecule, looks like it should have four signals.
4:09
It's not guaranteed, but that's what it looks like at first glance. I don't think this spectrum matches this first molecule because our spectrum has three signals. So let's tentatively eliminate this structure. Okay. Our next structure... Let's see, there's a CH2. That should have a signal. This CH between the methyl groups should have a signal. And then we have a methyl and another methyl. These two methyl groups
4:36
are identical by symmetry. Okay, so this molecule should have... We would expect
4:46
three signals. So that's a possibility. Now let's keep working to the right.
4:51
We have this molecule. This molecule has three methyl groups, and all three of these methyl groups by symmetry are identical. So this molecule should have just one signal. Now it's going to be one big signal because all three of those methyl groups. That's a total of nine hydrogens, and it's going to be a big singlet because they don't have any neighbors. But it's one signal. That doesn't match the spectrum that we have. So it's looking pretty good for molecule B. Now let's
5:18
look at this last one. So that methyl, B, C, and D. Now notice we have two methyl groups, A and D, and I've said they're different. I've said they're different because I gave them different letters. And you might say, "Hey, methyl groups. They're all the same. Well, no they're not all the same. You know that this methyl group is attached to a carbon, just a CH2. This methyl group is attached to a CH with a chlorine. That's going to change its electronic environment, and probably
5:50
(not certainly) but probably means A and D will not be in the same place. Now is it impossible? No, but A and D will likely be different. But let's just leave a question mark here and leave that option open. And now we'll let the spectrum give us more information to help us distinguish... Do we have the second compound or the fourth compound?
6:12
Okay, so let's look more at our spectrum. We have... Let's focus back on this big, this big signal -- six hydrogens. So we have these six hydrogens together. This could be
6:27
a CH3, which is a singlet -- not signalet but a singlet -- and another three hydrogen singlet that happen to be right next to each other. So you know this could be two separate methyl groups. Oh, that kind of sounds like this... two separate methyl groups. But they would need to be both singlets, meaning they have no neighbors. Well that doesn't fit this compound because
6:56
methyl group A has a neighbor so methyl group A would be a doublet. Furthermore
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methyl group D would be next to the two hydrogens on C, so that would actually be a triplet. These singlets don't match. Okay instead, what signal (this six hydrogen
7:20
doublet) looks like is it looks like a doublet. And I just said, it this looks like our two CH3s with one neighbor, and so that's a doublet. That's a doublet by symmetry. They land on top each other. That's six hydrogens all together. There's a doublet. So I think we can eliminate this last compound. And we're definitely... Here, let's finish the spectrum and just see what we have. So we (I) mentioned, we have this hydrogen. This hydrogen has a ton of neighbors.
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It has three here, three here, and two on carbon... on position A. So that's a total of eight neighbors. It's gonna have all kinds of multiplicity. It's going to be split like crazy, and that's pretty much split like crazy. And it's also one hydrogen. That's a good match. Now notice these two hydrogens. That's a CH2. There's two hydrogens. We're saying it's a doublet, or the signal, the spectrum says it's a doublet. It does have one neighbor. So those should be a doublet,
8:24
and furthermore, they're further downfield because this chlorine is next to... next door. And chlorines and oxygens typically, based on their electronegativity, you know, they're doing an inductive effect that causes deshielding of the carbon and those CHs. And that moves these down to typically between three and four. That's the typical range. So everything looks good. This is almost certainly our second compound. That's our answer, which on the question handout would be B.
8:54
All right, problem two. Let's see if we can work through problem two a little faster than we did problem number one. So again, let's look at the spectrum. So we see what seems to be four different signals. These four hydrogens -- and I'm looking at the integrations, those red numbers -- four hydrogens, two, three, and three. And I always think when I see threes, very often those are methyl groups. So we're going to have a molecule that probably has two methyl
9:22
groups, and since they have a different chemical shift, they're going to be distinct methyl groups. Now let's look down here at this four hydrogen signal, and this may not be one signal. Maybe it's something overlapping, but notice it's at about 7 ppm, between this 6 and 8 window. That's an aromatic ring. And, of course, if we look below the spectrum, we see lots of aromatic rings. Now the integration can be extremely helpful. So what this tells us is, we have
9:51
an aromatic ring (because that's what the... where the chemical shift is, and we have four hydrogens. You only get four hydrogens on an aromatic ring if you have a benzene ring that has two groups on the benzene ring, which means there are only four spots left over for hydrogen. So let's immediately look at our structures at the bottom. This first structure has a benzene ring, and that benzene ring only has one substituent on it, meaning it's going to have
10:24
five hydrogens on the ring. Five hydrogens would not be four hydrogens, so we can eliminate this compound immediately just based on the integration of the aromatic region.
10:36
Let's look at our next structure. Here's a benzene ring that is monosubstituted. This benzene ring will have five hydrogens on it. We can knock that out of consideration. This is getting great. Okay. Now this next structure, this is a tri-substituted benzene ring, meaning it's going to have... it has three substituents, meaning it has three hydrogen substituents to fill out the rest of the ring.
11:00
This would have an integration of three. That is not four. It can't be that one. Now let's look at our last structure. We have a benzene ring, and it is disubstituted, meaning we have one, two, three, four. We must be this last answer. So that's answer D. Now let's look at what else we see. So I said, "Oh, here are these two peaks that have integrations of three. That's very often (not always but we'd expect to see) two methyl groups." And that's
11:37
exactly what we have. We have a methyl group here. Note that this methyl has no neighbors. There are no hydrogens on that neighboring carbon. So this should be a singlet.
11:49
And that's exactly what we see. This is a three hydrogen singlet. Okay. That makes sense. And here is a CH3 group. There's our integration of three, and that signal has two neighbors. This is the n plus one rule. 2 plus 1 is 3. So this CH3 group ought to be a triplet, and that is a triplet. So that looks good. And finally, let's address these two hydrogens right here. Notice we have 2 left. That's exactly what our peak is. It's two hydrogens. These two hydrogens
12:26
have three neighbors based on that methyl group. Three plus one is four. So these two hydrogens should be a quartet, and indeed they are. Okay. So this... Our correct answer for this problem is D.
12:41
Okay, before we do problem three, I need to make an admission. I have doctored this spectrum. So NMRdb -- and I love the website -- however NMRdb does not make a prediction for OH and NH groups. And you can identify those typically in in an actual spectrum (not a simulated spectrum) as being broad peaks. This broad peak right here is from an NH or OH, and if you look to the bottom of the screen, you can see it's going to be an NH.
13:15
So this signal is from an NH. NMRdb does not generate this, so instead I went into
13:23
Paint and I stuck that peak in there. And it has an integration of two. But, if you were to take the final answer from this question, put it into NMRdb.org and generate a spectrum, you would not see this signal because NMRdb.org does not do OHs and NHs. It only shows CHs. So that's a... one disclaimer. Now let's look at this spectrum, and I've already highlighted what I think is the most interesting signal. Here we see a bunch of signals that have really nice, clear
13:54
multiplicity, which is what you typically see on CH peaks. And then we kind of see this blob, which is what you tend to see for OHs and NHs. Now that depends on what solvent you use, but for most spectra in introductory organic chemistry, you use solvents that give NHs and OHs as sort of as broad peaks. So we have a broad peak. That's an NH or an OH, and it has an integration of two.
14:21
Okay. That's very helpful information, looking at our structures at the bottom of the screen. Because if we look at our first structure, that only has one NH. That wouldn't give an integration of two. So we can't be this signal, but only based on this one peak, we can already eliminate that structure. And this is why I tend to look at the spectrum, and I look for information that's going to be as helpful as possible. Let's take as many compounds
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or possibilities off the table as quickly as we can, and as it turns out, based on the structures we're given, that's really helpful information. Now again, this is an NH with two hydrogens.
15:03
We look at our next structure. Oh, this has an NH2. That is a possibility. We haven't looked at the rest of spectrum, but that's... that at least belongs under consideration. Our next structure also has an nh2 so that's a possibility. Our last structure doesn't have any NHs, so we can eliminate that from the get-go. So now we're down to two possibilities, and we need to look at the other peaks. So let's... You know, I tend to gravitate towards these methyl groups if I can. So
15:32
here we see there are two different methyl groups in this structure. So if we look at this second structure (and which is answer B on the sheet), we can see that, well okay, these... we should... these NHs should show up. There's a CH2. There's a CH. That methyl and that other methyl are the same by symmetry. They're going to overlap. So we shouldn't see two methyl groups for this second structure. We should just see
16:04
one six hydrogen signal for both of those methyl groups. That is not what we see. We see distinctly two methyl groups. So let's... Before we eliminate that structure, let's go to the other one and make sure that the other one is going to give us, the, this multiple methyl groups. So let's go over here. Let's just label these A B C D. The nh2 is E, so so this would, if we're correct, this is almost certainly going to be signal E. That's our NH2. Now we do have two methyl groups.
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If... These don't look identical or we wouldn't predict they'd be identical by symmetry, so seeing two different ones it's completely reasonable. The signal for A has one neighbor,
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that hydrogen on B. Therefore A should be a three hydrogen doublet. And we have that right there, so
17:03
that's reassuring. Now let's make sure the other methyl group also lines up. So for D, we're going to have two hydrogens on C, two plus one is three. So we'd expect signal D to be a triplet, and it is. So that's reassuring. Okay. Our methyl groups make sense, and I think we can safely eliminate the second structure. Now let's go ahead and assign the rest of the spectrum.
17:30
Peak A, we've already assigned that, so let's try peak B. Peak B, that's a hydrogen that has neighbors on A and C. That's six neighbors... I'm sorry. It's five neighbors -- 2 on C, 3 on A. 5 plus 1 is 6. So we see this high multiplicity signal, and it is one hydrogen. That looks good. And then let's look at these two CHs. So they have three, four neighbors, and we see a multiplicity of five for that peak. And again, the integration works. Now
18:08
one last comment on NHs and OHs. When I was counting hydrogen neighbors for B,
18:16
I only counted the carbons. I did not count the nitrogen.
18:21
It depends on how you run your NMR spectrum. It depends on the solvent. NMRdb assumes you're using a solvent where you won't see multiplicity that is contributed to by the NHs and OHs. Depending on the solvent, you might see it. So I'm saying this has... Signal B has three plus two neighbors -- five neighbors total. It's possible, you could get up to seven. It depends on the solvent.
18:52
Okay. Here's problem number four. Just looking over the spectrum, what stands out? Not a whole lot, I will say. This signal is pretty far down field, but I'm... I like to focus on the threes if I can. So there's a three, and there's a three. Looks like we have two different methyl groups in this compound. And you know that's not a whole lot to work with, but let's go ahead and see what we have. So we look at the first structure. A. Now B - B doesn't have anything. There's B
19:22
and C. So again, this carbon doesn't have any hydrogens so it won't show anything. But we do have a methyl and a methyl. Hey, at least by our criteria, this could possibly match. it's not unreasonable. Let's go to our next compound. Next compound... We have a methyl, and by symmetry this other methyl would overlap. So these two methyl groups would have the same chemical shift and they'd overlap. They'd both be singlets. That is not what we see. Furthermore, aldehyde hydrogens
19:54
show up at about 10 ppm, and the spectrum only goes to 5, which implies there's nothing at 10. So we can eliminate this structure. Now let's look at the next structure. We would have a signal for that CH2, that CH2, and that CH3. We only have one methyl group in this third structure,
20:14
so that's not a possibility. Now let's go to our last structure. A -- there's a methyl. B -- that's a CH2. C -- that's a CH2. That doesn't have two methyl groups, so that is not a possibility. That doesn't match the spectrum. Now let's go back to our original structure. We think we know the answer. The answer is going to be A, but let's go ahead and assign it. So
20:39
signal A -- that's a CH3. There are no neighbors, so that should be a singlet, and we do have a three hydrogen singlet. Let's go to the other methyl group. Here's this CH3. It has one neighboring hydrogen right there. That should be a doublet. It is a three hydrogen doublet. Fantastic. That's signal C. And now let's look at this last signal. B -- this has...
21:08
Let's put in the CH3. CH3. This... That hydrogen has three neighbors, so it should have a multiplicity of four -- integration of one, multiplicity of four. That is
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B. Okay, so everything lines up. The answer for problem number four is A.
21:29
Okay. Problem number five. Again, let's just take a look at our spectrum and see if there are any obvious clues that we can use to to eliminate some of the possibilities. I see a three. This is a very downfield three, but that looks like a methyl group. And here we have a six. It's two times three. That's very likely going to be two methyl groups. And then we have this singleton hydrogen here, around 2.5. Okay,
21:55
So let's see if we can find a structure here that has three methyl groups. So our first structure
22:03
has... Let's label the number of different signals we'll have. That's a CH3, a CH2, a CH2, and a CH3. We only have two CH3s, so it looks like we are not... This has two methyls.
22:22
This doesn't look that encouraging. Okay. Let's look at the next structure. We should have A, B, C, D, and there's a methyl, and there's a methyl. Otherwise, we just have CH2s. This looks like two methyls. This is probably not our structure. Let's look at our third one. There's A. That CH is B. C. And oh! I was going to label that D, but notice, by symmetry, these two should be identical. So that doesn't... That's not a D. That's a C. Okay. That has three
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methyls. That could be what our spectrum is. Now let's look at our next one. We have A. There's B. Oops. Okay. This methyl group is by symmetry the same as that one, so we'll label this A as well. And then that makes that C. So we have three methyls here too. These are both possibilities. So let's look at our... Let's go back to our first one. We have these two methyl groups,
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and they have one neighbor, so that should be six hydrogens together as a doublet. And we have a six hydrogen doublet. Okay. And furthermore, we have this CH3 by itself. There are no neighbors from this oxygen so that should be a singlet. Okay that's, that's a match. Let's see if the other compound can match that description. Here are two methyl groups. They both have
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one neighbor, so that should be six hydrogens as a doublet. And that's what we see. Okay, and then we have this CH3 and it has no neighbors, so that should be a singlet.
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Okay. These are not obviously different, but what is different about these is the position of, you know... The one that stands out to me is this one. Look at this. This is way down field. How did this CH3 get all the way down past 3.5 ppm? Well, it must be next to, directly attached to, a very electronegative atom. And if we compare two structures,
24:38
this one has the CH3 that is, that fits, that is the singlet next to an oxygen. This CH3 is not directly attached to an oxygen. Now I know you look at this CH3 in the last compound and say, "Well, it's not next to an oxygen, directly next to, but look at these oxygens over here." They really don't have a big impact. It's... What is the atom that's directly next door? For this CH3 it's just a carbon. For this CH3 it's an oxygen. The chemical shift
25:11
of this signal much... is a much better match for our third structure, answer C. So let's... and let's... Let's look at this a little more closely. So how about for problem C, where would this peak show up? Now I just said this hydrogen, it's on a carbon, and it's next to another carbon. Not very, not very fancy, not very deshielding. And notice that signal shows up around two and a half or so.
25:47
If our spectrum matched our rightmost structure, then this CH is right next to an oxygen. It ought to be way down field, like past at three and maybe into four. But notice, that peak is a little bit more up field. So the position of this peak also supports structure C, not structure D. So we're going to eliminate D and go with structure C. But just by looking at the methyl groups, we could... we could get the discussion down to C or D. So just
26:21
looking at some key features can help narrow the structures that you're going to have to consider.
26:28
Okay. Here's our last problem. I'm going to play the same trick that I did on our previous structure that had aromatic rings. So if you notice, this spectrum has signals down, you know, past six and a half, between six and a half and eight. This is typically the aromatic region of a spectrum. And notice that we have a total integration of five. We have multiple signals down here, but in aggregate, in total, we have five hydrogens. That means our structure
26:56
needs to have a benzene ring that only has one non-hydrogen group coming off of it. So let's use that as our first pass to filter some structures. So as we look at structure A,
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we have five hydrogens on our ring. That's a possibility. As we look at structure B, we have five hydrogens on the ring. Okay, that's still a possibility. Structure C? It has two substituents on the ring. That means it has four hydrogens. We can eliminate this because we need a molecule that has five total benzene ring hydrogens. And as we go to our last structure, we again have a monosubstituted benzene ring.
27:38
This is a valid candidate. So we've only eliminated one structure. Now let's... You know, I love looking at methyl groups. So here we are, this structure looks like it has two threes in it. Two methyl singlets. So let's see which of the remaining compounds actually have two methyl groups in there -- two methyl singlets. So now we look at our first structure. We have a methyl and a methyl. This is a possibility still. We look at our next structure. We have a methyl,
28:08
and we don't have any other methyl groups. This is less likely. And notice with... This has an NH. Those NHs tend to be broad. We don't have that in the spectrum. This is not looking promising. Okay. Now let's look at this last structure. This last structure should have a methyl,
28:27
and there's no other methyl. Furthermore, it should have an NH, which we don't see in this spectrum. Therefore we can likely eliminate this. Now let's just make sure everything makes sense based on what we've said. So here's our benzene ring. Those are our five aromatic hydrogens in the roughly seven to eight region. That looks good. We have two methyl groups.
28:50
Let's label these as A and B. Now notice A is next to a hydro... Or, I'm sorry, a nitrogen. That's a more electronegative atom. That's going to move it further downfield. And so this methyl singlet that's further downfield is likely A because it's attached to the more electronegative nitrogen that is deshielding and causes the peak to show up at a higher chemical shift. B, however, is a methyl
29:19
group, and it's only attached to a carbon. It is less deshielded and shows up at a lower chemical shift. And that is this signal right here. So everything makes sense as we look at the rest of the data, and everything is consistent with this first structure. So this would be answer A. I hope you enjoyed these problems and that they help you better understand how you can use NMR spectroscopy. Please subscribe to the channel, like this video, or leave a comment. Thank you
29:50
for watching. Remember that links to these problem sets are in the video description. Take care.
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