- Introduction to proton NMR
- Nuclear shielding
- Chemical equivalence
- Chemical shift
- Electronegativity and chemical shift
- Diamagnetic anisotropy
- Spin-spin splitting (coupling)
- Multiplicity: n + 1 rule
- Coupling constant
- Complex splitting
- Hydrogen deficiency index
- Proton NMR practice 1
- Proton NMR practice 2
- Proton NMR practice 3
Using the n+1 rule to predict how many peaks to expect from spin-spin coupling. Created by Jay.
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- At9:44- I do not understand why splitting is seen in left molecule. The hydrogen atoms are in chemically equivalent environments. Would this not show as 1 peak?(21 votes)
- 9:44is where the video ends, so I'm assuming you're referring to the last example he gives, like at9:42. He is stating that these would split if we imagine that they are in chemically different environments, he just doesn't draw out the other atoms that would need to be attached to make us able to see that they are in different chemical environments.(18 votes)
- at8:45those protons look like they're in an almost identical environment, why and how are they different?(7 votes)
- I think he's saying "if" they were in different environments, as in if the carbons adjacent to them had different substituents. He just hasn't drawn them in.(6 votes)
- At2:17, you have pointed so that the proton next oxygen is at ~6.7 ppm, while proton next to the chlorine atom is ~7.3 ppm. Why is this? Wouldn't oxygen deshield the proton nearby more than chlorine would (since oxygen is more electronegative)?(4 votes)
- Resonance effects are more important here.
The OCH₃ group is strongly activating and ortho/para directing, while Cl is weakly deactivating.
The effect of OCH₃ is more important. Through resonance, it increases electron density on the ortho and para carbons.
The ortho H is therefore more shielded, and its signal moves upfield.(7 votes)
- From8:30~8:50, aren't the red and blue hydrogens chemically equivalent?(4 votes)
- It looks that way in his drawing.
What he is saying, though, is that chemically nonequivalent protons that are on adjacent carbons will split each other.
If they are separated by just one carbon atom, they will not split each other.(1 vote)
- Why do the equivalent protons not change the effect of the magnetic field?(2 votes)
- What does he mean by chemical equivalence? Also, the red and the blue protons at0:49are not equivalent since one is close to the oxygen and the other closer to the chlorine? Wouldn't the protons have to be sharing different environments to have an effect on one another in terms of coupling? Essentially, does chemical equivalence mean same environment?(2 votes)
- Chemically equivalent protons does mean they share the same environment, generally this means they are on the same carbon atom but that isn't always necessary.
When there are two (or more) protons next to one another and they are not in the same chemical environment you will see splitting (more than 1 peak)
Example: in benzene every single hydrogen is in the same chemical environment. In the 1H NMR we only see a single signal and one peak.
Example2: in phenol each hydrogen is no longer in the same chemical environment, as some are closer to the -OH than the others. We would expect to see three separate signals from the aromatic region and if a high enough resolution is used we will see many peaks.(2 votes)
- Woah, wait, hold on....the first two protons aren't chemically equivalent! We showed that in the last video!(2 votes)
- Is this proton NMR how you can tell triplet O2 and singlet O2 apart?(1 vote)
- no, proton NMR only applies to compounds containing bonds with HYDROGEN atom.the proton nmr deals with the proton (i.e., the hydrogen atom) attached with a bond.the concept of singlet and triplet O2 comes from Molecular orbital theory.and proton nmr is about spin of proton, but 'singlet and triplet O2' is about spin multiplicity (2S+1) of electrons in molecular orbital of O2.(3 votes)
- In the first example, why is the red proton at a higher position than the blue proton. The red proton is next to the oxygen which is more electronegative than chlorine so shouldn't the red proton be at a lower position because the more electronegative oxygen will de-shield it?(2 votes)
- So as I commented earlier, the oxygen is more electronegative. However, single bonded oxygen can donate electron pairs to the benzene rings offering stabilization while the chlorine is a weak benzene deactivating system (takes electrons away from the ring). Therefore the differences in the chemical shifts cannot be answered by electronegativity alone. Look up ring activating and ring deactivating functional groups for aromatic compounds.(1 vote)
- At0:24Jay says that the n+1 rule works only when protons are chemically equivalent. At2:57though he says that two red protons have one proton neighbour so we expect 2 peaks - but that neighbouring proton is not in the same environment, so why does it split those peaks? The same applies to when those two red protons split the peak for the blue proton into three.(1 vote)
- The neighbouring protons do not have to be in the same environment as the one that we are considering for the n+1 rule.
The red protons have 1 neighbouring proton, the blue proton. Since there's only one neighbour it doesn't matter here. We expect 2 peaks in the signal (n=1, 1+1=2) and we see 2 peaks.
The blue proton has two neighbouring protons, the red protons, and the two red protons are in the same environment as one another. We expect 3 peaks in the signal (n=2, 2+1=3) and we see 3 peaks.
That is what the n+1 rule refers to.(2 votes)
- The n plus one rule allows us to predict how many peaks we would expect to see for a signal in an NMR spectrum. So if we think about the signal for one proton, if that proton has n neighboring protons, we would expect to see n plus one peaks on the NMR spectrum. The n plus one rule only applies when the neighboring protons are chemically equivalent to each other. So in the last video we looked at this molecule and we focused in on this proton right there in red and we said that this signal right here was due to the proton in red and we also talked about this proton right here in blue and this signal's due to the proton in blue. Let's look at the proton in red. Alright, so let's see how many neighboring protons we have. So if you think about this carbon right? And you think about the carbon next door, so this carbon right here. So this proton is a neighbor right? So we have one neighbor so far. We go over here to this carbon, this next door carbon here. There are no protons on that carbon. So we have a total of only one neighboring proton for the red proton. So one neighboring proton, so n is equal to one. So we're going to see one plus one peak. So one plus one is equal to two. We expect to see two peaks for the signal for the red proton. So here's the signal and we see our two peaks. So this is called a doublet. So the signal for the red proton is split in to two peaks because of the presence of the neighboring blue proton. Alright, let's do the same thing for the blue proton. So we think about the signal for the blue proton. How many neighboring protons do we have? Well if we go to the carbon next door, so this carbon there are no protons on this carbon. And on this carbon, we of course, we have one proton. So only one neighbor. So n is equal to one. We're going to see n plus one peaks. So one plus one is equal to two, we expect to see two peaks. So we go over here to the signal for the blue proton and we see our two peaks. We get a doublet. One neighbor gives us a doublet here. What about the protons over here? These protons in magenta? So how many neighbors do we have for those protons? Go over here to the oxygen, no protons on that. No neighbors so n is equal to zero. N plus one peaks. So zero plus one is equal to one. We would expect only one peak for the signal for these three equivalent protons and of course, this is the signal. Only one peak and we call this a singlet. Alright, let's do another one. So we also saw this molecule in the last video. So over here and let's look at our protons. We expect one signal for the blue proton and we would expect one signal for these red protons here. Let's think about the red protons first. So how many neighboring protons do we have? We go to the carbon next door and then we have one proton. So one neighbor. So n is equal to one, so we expect one plus one peaks, so we expect two peaks. So the signal for the red protons needs to have two peaks. So this one right here, so there's one peak and there's the second peak so we get a doublet. What about the signal for the blue proton? So we get a signal for the blue proton. How many neighboring protons do we have? So we go to the carbon next door and we have one, two neighbors. So for the blue proton, we have two neighbors, n is equal to two. Expect n plus one peaks, so two plus one is equal to three. So three peaks for this signal. So this signal is right here and we get one, two and three, three peaks which is called a triplet. Let's go to the next one. So let's look at this one right here. We have Bromoethanes. So let's first draw in the protons. So on this carbon we have two protons and on this carbon we have three protons. Alright, so let's start for the signal for these protons right here. So these are equivalent, we would expect one signal for those protons. How many neighbors do those protons have? We go to the carbon next door and we have one, two. We have two neighbors. So n is equal to two. We would expect n plus one peaks. So two plus one is equal to three. We would expect a triplet for this signal and here is our triplet. One, two and three peaks for that one. Next, let's think about the signal for these two protons. How many neighbors do those two protons have? So we go to the carbon next door. One, two, three neighbors. So n is equal to three, so three plus one is equal to four. We would expect a signal with four peaks for the protons in blue and here's our signal with four peaks one, two, three and four. So we call this a quartet. So a four peak, we call it a quartet. Let's do another one. Let's look at this molecule right here. And let's first draw in the protons. So on this carbon there are three protons. So here are the three protons. On this carbon there are three protons and on this carbon, there would only be one proton, so here's this one. Let's think about how many signals we would expect to see. So not thinking about the n plus one rule or spin-spin splitting here. So a signal for these protons, these protons are in the same environment as these protons, so that's one signal for those. And then, we would expect a signal for this proton right here in a different environment. Let's think about the signal for the red protons first. How many neighboring protons do the red protons have? We go to the carbon next door and this is the carbon next door for both of those red protons and there's one neighbor. So one neighbor, n is equal to one. So one plus one is equal to two, we would expect a doublet for this signal and so that must be this. So this is kind of like a zoom in and it's not the exact same drawing. I hand drew all this stuff so it's not exactly perfect but you can see there are two peaks here. Two peaks, so this is a zoom in, this is supposed to represent a zoom in of those two peaks there. So that's that signal. What about the blue proton? So how many neighbors does the blue proton have? So we look at this carbon and we look at the next door carbon. So this is the carbon next door, one, two, three. This carbon is also next door to this carbon, so one, two, three. So a total of six, so n is equal to six. So expect n plus one peaks. So six plus one is equal to seven. So we'd expect seven peaks, called a septet. Let me go ahead and rewrite that here. So let me go ahead and write septet here and let's look at this signal. It must be this right here. This was pretty hard to draw and once again, this is a blow-up of what you're looking at. So we'd expect seven peaks. So one, two, three, four, five, six and seven. So that's the idea of the n plus one rule. And let's talk a little bit more about spin-spin splitting and when to expect a signal. So if you have chemically equivalent protons, they don't show spin-spin splitting. So if I look at these two protons right here. These two protons are in the same environment as these two protons, think about symmetry. So we'd expect only one signal. We'd expect only one signal on the NMR spectrum. And these protons are not going to split these protons even though they're next door because they're chemically equivalent to each other. So chemically equivalent protons do not show spin-spin splitting. Let's look at these examples right here. So what we've been talking about is next door protons. So this proton and this proton will split each others signal if they're in different environments so splitting is observed. We're talking about these next door protons here. But if you have this situation, so let's once again make this the red proton and make this the blue proton. There's an extra carbon in between. So splitting is generally not observed for this situation. So the red and the blue protons won't split each other generally, again there are exceptions but for simple NMR spectra, you really have to think about, you really have to think about the situation on the left here. So these protons are too far apart for them to feel any effect. And then finally, it is possible to have splitting from protons on the same carbons so if these are in different environments right? So if this proton is in a different environment from this proton, it's possible for splitting to be observed and we'll talk more about that in the next video.