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## Proton NMR

# Multiplicity: n + 1 rule

## Video transcript

- 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.