- 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
Predicting the splitting pattern when a proton has two different kinds of neighboring protons using a splitting tree. Created by Jay.
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- Does it matter if you split with the higher or lower frequency first?(38 votes)
- So, for the last example with Jab = 12hz and Jac = 7 Hz (arbitrary letters), what would the ratio of the peak heights look like?
I imagine that it'd be like 1:2:3:2:1 or something similar due to overlapping coupling, but I'm not sure. Could someone clarify this for me?(4 votes)
- If the peaks are sharp and well-separated, the six peaks should be in the ratio 1:2:1:1:2:1, with peaks 1, 2, and 4 belonging to H(b) and peaks 3, 5, and 6 belonging to H(a).
However, the two middle peaks might overlap and look like a single peak. Then the signal would look like five peaks in the ratio 1:2:2:2:1.(4 votes)
- Shouldn't the second part of the splitting tree diagram be split 3.5 Hz both way rather than 7? The way it is shown is for 14 Hz and 14 Hz again, correct?(3 votes)
- No, J = 7 Hz, so the distance between each component of the triplet is 7 Hz. The distance between the two outer peaks of the triplet will then be 14 Hz.(4 votes)
- origin of NMR signal?(2 votes)
- The protons behave as if they are little magnets.
They can line up either parallel or antiparallel to external magnetic field, with the antiparallel arrangement being higher in energy.
When you irradiate the protons with a radiofrequency of the right energy (frequency), the protons will absorb energy and be excited to the higher energy level.
When the protons drop back to the lower energy level, they emit the energy at a characteristic frequency, which is detected by an internal receiver.
The signal is amplified and is displayed as a peak on a graph.(6 votes)
- General Question--> If there are two different neighbouring environments, and you have to do (n+1)(n+1), how do you know which environment to start with for example it could be a doublet of triplets or a triplet of doublets.(2 votes)
- It doesn't really matter.
Usually, you start with the larger J value.
If the doublet has the larger J value, we call it a doublet of triplets.
If the triplet has the larger J value, we call it a triplet of doublets.(3 votes)
- What happens if you have an H and an OH on the same carbon. Does the H on the OH count as a neighbor to the other hydrogen (since it is attached to the O and not the C)? How does it effect the other H on the same carbon?
For reference, I am trying to construct the H-NMR spectrum of Borneol.(3 votes)
- (neq 1)*(neq2) = #of splitts when f1 does not equal f2/n?(2 votes)
- How can we distinguish a double of triplet vs a triplet a doublet if the J values or the Hz values aren't given?(2 votes)
- More complex splitting occurs when a proton has two different kinds of neighbors. And a good example of this is the blue proton that I circled in Cinnamaldehyde. So the blue proton has a signal with a chemical shift about 6.7 parts per million. So down here is a zoomed in view of the signal for the blue proton. Let's look at neighboring protons. So the blue proton is on this carbon, and we have a carbon next door right here with one proton, so there's one neighboring proton. Here's another carbon next door with one proton, so we have two neighboring protons. So let's try to apply the n plus one rule here. So if n is equal to two, we have two neighboring protons, we would expect n plus one peaks. So two plus one is equal to three, so a signal with three peaks or a triplet. But that's not what we see for the signal for the blue proton. We see one, two, three, four lines here. So the n plus one rule doesn't work in this case. And that's because the n plus one rule works when the neighboring protons are equivalent and here the two neighboring protons are not equivalent. So we need a new way to explain the signal for the blue proton. And we're going to use what's called a splitting tree. So we're going to start with the signal for the blue proton, so here's the signal for the blue proton. It's going to be split by the proton next to it, this proton in red. And the coupling constant between the red and the blue proton is 12 hertz. So let's go ahead and show, let's show this signal is split into a doublet, so let me go ahead and draw in the blue lines here. So we get the signals split into a doublet, the coupling constant is 12 hertz, so this distance here represents 12 hertz. So why can we think about it being split into a doublet? Well you can think about a variation of the n plus one rule here, right? So we're talking about one neighboring proton, so n is equal to one, so one plus one is equal to two, so we split the signal into a doublet with two lines. Alright, now let's think about what happens with the other protons, so this one. Ok well that's going to take each line of the doublet that we just drew and split it into another doublet. So this line gets split into a doublet, and this line gets split into a doublet. The coupling constant this time is six hertz. So this distance here represents six hertz, and this distance represents six hertz. So why was each line split into two? Well once again we can use a modification of the n plus one rule here. We're talking about one neighbor here, so n is equal to one, so one plus one is equal to two, so each line is split into two. So each line is split into a doublet. For this line and for this line. This line is split into a doublet, too. So we get four lines for the signal of the blue proton. If we go over here we see those four lines, one, two, three and four. And we call this a doublet of doublets, or a double doublet. What would happen if the coupling constants were the same? So let's just pretend like they're both 12 hertz, so let's go ahead and draw what we would see for the signal. So we have our blue proton's signal, is split into a doublet by the red protons, so let's go ahead and draw in our doublet here. And let's say this distance represents 12 hertz, so 12 hertz here, and let's say the coupling constant over here was also 12 hertz, so instead of six hertz. So each line of the doublet we just drew is split into another doublet from our other neighbor. And this time I'm going to show a coupling constant of 12 hertz. So each line of the doublet in blue is split into another doublet. So the line in blue on the left is split into a doublet, and the line in blue on the right is split into a doublet, because of our one neighboring proton. Well I changed the coupling constant so now I just have to pretend like now we're dealing with 12 here, so the coupling constants are the same. And notice what this gives us, this gives us a triplet, right? So here is one line and then we're going to get this peak, and then we're going to get this peak here. So if the coupling constants are the same, you get a triplet, you get what the n plus one rule predicted. So that's just something to think about with the origin of the n plus one rule. Let's do another example. So let's go down here and let's look at this molecule. So we're going to focus in on this proton here in blue. And we have neighboring protons right? So this neighboring proton, the coupling constant between those two is 12 hertz, and then over here, we have two neighboring protons, and the coupling constant between the magenta protons and the blue is seven hertz. So let's think about the signal for the blue proton here, so here we have the signal for the blue proton, which isn't split, let's think about what the red proton is going to do. We have one neighbor, so one plus one is equal to two. So we're going to split the signal for the blue proton in two, so we get a doublet here, so let me go ahead and draw in the doublet. The coupling constant was twelve hertz, so this distance right here is 12 hertz. Now you have a situation where we're thinking about the magenta protons, we have two of them. So n is equal to two, so two plus one is equal to three. So the magenta protons are going to split each line of the doublet that we just drew into a triplet. So we're going to get a triplet here, let me go ahead and draw that in. So for a triplet, and our coupling constant is seven hertz, so let me see if I can draw that here. So that distance is supposed to represent seven hertz, So let me draw that in. So this distance is seven hertz, this distance corresponds to seven hertz, and one line of our doublet is split into a triplet now. Same thing happens for the other line, right? Same thing happens for this line right here, we need to split that into a triplet because of these magenta protons. So we're going to split that into a triplet, once again, we need to think about a distance of seven here, seven hertz. So we have on both sides, we have seven hertz, let me go ahead and draw that in, so this is talking about seven hertz, and this is talking about seven hertz. So that one line is split into a triplet, so let's draw that in here, one, two and three. So finally, how many peaks would we expect for the signal for this blue proton? So one, two, three, four, five, six. So we would expect six peaks for this signal, for this blue proton, because of the neighboring protons which are on different types of environments. So this is complex splitting.