# Proton nmrWatch

#1
When looking at a proton nmr thing, when you are trying to find the splitting pattern am i right in thinking it is the number of hydrogens on the adjacent carbon plus one. But what if there are carbons on both sides?
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10 years ago
#2
it's the hydrogens that are attached to the carbons...
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10 years ago
#3
start at the RHS, work to the left
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#4
Ok i have never heard that rule before but thanks
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10 years ago
#5
The splitting pattern depends on the number of equivalent protons that you're coupling to. So, for example, if you have a molecule such as
R1-CH2-C*H2-CH2-R2
where R1 and R2 are different, and you're looking at the NMR spectrum of the proton on the carbon at the centre marked with a star, you will see coupling to both CH2 group protons SEPARATELY. You're thus expecting a coupling pattern, where is the number of protons in each environment (2 in each case) and , the nuclear spin quantum number for hydrogen.
You're therefore expecting to see 9 peaks in the spectrum - but not that some of them may be superimposed on each other, and they may not be in the weak coupling regime (i.e. roofing), so you may not be able to distinguish all the peaks.

If however R1 and R2 were the same, then CH2 groups on either side of the proton in question are chemically and magnetically equivalent, and you will simply see (2nI+1) peaks, for n=4, i.e. 5 peaks only, with a binomial intensity distribution.

The easiest thing you can do to rationalise these is to draw stick diagrams. I'm attaching a drawing I made just now, which unfortunately isn't the prettiest stick diagram around, but it should hopefully explain why the coupling pattern is like it is.
I've exaggerated the difference between the coupling constant for protons 1/2 and 3/4 - in reality the peaks would likely coincide if the environments are reasonably similar. Oh, a word regarding my 'numbering' - protons 1 and 2 refer to the two protons on one of the CH2 groups, and these protons are equivalent (hence the coupling constants are the same, J1 = J2); likewise for protons 3 and 4 (J3=J4).
The diagram shows that you're expecting an intensity ratio of 11411 in the ideal case.
Now, hopefully you can draw a stick diagram for the case where R1=R2, i.e. where J1=J2=J3=J4, to confirm that you do get binomial peak intensities.
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10 years ago
#6
(Original post by Sinuhe)
The splitting pattern depends on the number of equivalent protons that you're coupling to. So, for example, if you have a molecule such as
R1-CH2-CH(Cl)(Cl)-CH2-R2
where R1 and R2 are different, and you're looking at the NMR spectrum of the proton on the carbon at the centre (the one with two Cl groups attached, assuming no coupling to Cl is seen), you will see coupling to both CH2 group protons SEPARATELY. You're thus expecting a coupling pattern, where is the number of protons in each environment (2 in each case) and , the nuclear spin quantum number for hydrogen.
You're therefore expecting to see 9 peaks in the spectrum - but not that some of them may be superimposed on each other, and they may be in the weak coupling regime (i.e. roofing), so you may not be able to distinguish all the peaks.

If however R1 and R2 were the same, then CH2 groups on either side of the proton in question are chemically and magnetically equivalent, and you will simply see (2nI+1) peaks, for n=4, i.e. 5 peaks only, with a binomial intensity distribution.

The easiest thing you can do to rationalise these is to draw stick diagrams. I'm attaching a drawing I made just now, which unfortunately isn't the prettiest stick diagram around, but it should hopefully explain why the coupling pattern is like it is.
I've exaggerated the difference between the coupling constant for protons 1/2 and 3/4 - in reality the peaks would likely coincide if the environments are reasonably similar. Oh, a word regarding my 'numbering' - protons 1 and 2 refer to the two protons on one of the CH2 groups, and these protons are equivalent (hence the coupling constants are the same, J1 = J2); likewise for protons 3 and 4 (J3=J4).
The diagram shows that you're expecting an intensity ratio of 11411 in the ideal case.
Now, hopefully you can draw a stick diagram for the case where R1=R2, i.e. where J1=J2=J3=J4, to confirm that you do get binomial peak intensities.
explain all of that again please, since I didn't understand one word. Simplify if possible. Also the molcule you showed doesn't exist: the centre C is bonded to 5 other atoms it seems.
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10 years ago
#7
If your exam board is AQA you don't need to know the splitting pattern for something that has two different proton environments on either side of it. For the rest you can just stick to the n+1 rule, where n is the number of hydrogens on the adjacent element.

So for example, would have 3 peaks - a singlet (O-H), the will be a triplet (due to the adjacent - 2+1=3) and the will be a quartet.
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10 years ago
#8
(Original post by Wenzel)
If your exam board is AQA you don't need to know the splitting pattern for something that has two different proton environments on either side of it. For the rest you can just stick to the n+1 rule. where n is the number of hydrogens on the adjacent element.

So for example, would have 3 peaks - a singlet (O-H), the will be a triplet (due to the adjacent - 2+1=3) and the will be a quartet.
cheers

Just wondering, WHY does, say, the CH3 split into a triplet? What do the two hydrogens do to make it "split" in such a way? (presumably it is something to do with the spins of the protons, since that's the only way a binomial distribution and the "n+1" idea comes in)
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10 years ago
#9
(Original post by aster100)
explain all of that again please, since I didn't understand one word. Simplify if possible. Also the molcule you showed doesn't exist: the centre C is bonded to 5 other atoms it seems.
Eh, reality messing with my explanation, how nasty of it.
I just wanted to differentiate the central carbon so it would be easy to see which one it was, without making it diastereotopic. Is it better now?

I'm not sure which bit you don't understand. A proton will couple to its neighbours - I don't think you want to hear the physical reasoning behind this? But yes, if you have two protons, Ha and Hb, then the energy of the system depends not only on the isolated spins of Ha and Hb, but also on the relative spins with respect to each other. These energy changes occur for all energy states in a similar, and consistent, fashion, so you get changes of +J/2 and -J/2 to the energy of each state (in the weak coupling regime), and overall for each such coupling you get TWO peaks, one at frequency+J and the other at frequency-J, where 'frequency' is the Larmor frequency of the proton you're observing in a magnetic field - the frequency its magnetisation precesses around the static homogeneous magnetic field.
This coupling is a through-bond effect (through-space (dipolar) effects are averaged to zero when molecules tumble rapidly in solution), so it depends on the number of bonds between the coupled protons - typically in organic NMR you're considering coupling to protons three bonds away (so-called 'vicinal' coupling), although other couplings can also be important.

So, to simplify, let's rephrase all this in the form of a statement: two protons will couple to each other if they are close enough in a molecule.

When you have more than just two protons, several protons can couple to the proton you're interested in. You can show (by simple combinatorics and an energy level diagram) that the number of permissible transitions in an NMR experiment, and hence peaks in the NMR spectrum, is given by the formula 2nI+1, where n = number of equivalent protons; I = nuclear spin quantum number (I=1/2 for protons, but for other elements and even deuterium it can be different).
(This formula basically means that you have (n+1) peaks in simple 1H-NMR spectra, as the OP noted.)

But if you have inequivalent protons that you're coupling to, EACH OF YOUR PEAKS WILL AGAIN BE SPLIT by the new, different proton. Since the coupling constant is different, you can't simply add this new proton to your count of 'n', but need to consider splittings proper. If you split (2nI+1) peaks into a further (2I+1) peaks, you'll get (2nI+1)*(2I+1) peaks. If you couple to TWO equivalent nuclei, but which are different from your original coupling, you'll get (2nI+1)*(2*2*I+1) peaks.

Does this make more or less sense now?
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10 years ago
#10
(Original post by Sinuhe)
Eh, reality messing with my explanation, how nasty of it.
I just wanted to differentiate the central carbon so it would be easy to see which one it was, without making it diastereotopic. Is it better now?

I'm not sure which bit you don't understand. A proton will couple to its neighbours - I don't think you want to hear the physical reasoning behind this? But yes, if you have two protons, Ha and Hb, then the energy of the system depends not only on the isolated spins of Ha and Hb, but also on the relative spins with respect to each other. These energy changes occur for all energy states in a similar, and consistent, fashion, so you get changes of +J/2 and -J/2 to the energy of each state (in the weak coupling regime), and overall for each such coupling you get TWO peaks, one at frequency+J and the other at frequency-J, where 'frequency' is the Larmor frequency of the proton you're observing in a magnetic field - the frequency its magnetisation precesses around the static homogeneous magnetic field.
This coupling is a through-bond effect (through-space (dipolar) effects are averaged to zero when molecules tumble rapidly in solution), so it depends on the number of bonds between the coupled protons - typically in organic NMR you're considering coupling to protons three bonds away (so-called 'vicinal' coupling), although other couplings can also be important.

So, to simplify, let's rephrase all this in the form of a statement: two protons will couple to each other if they are close enough in a molecule.

When you have more than just two protons, several protons can couple to the proton you're interested in. You can show (by simple combinatorics and an energy level diagram) that the number of permissible transitions in an NMR experiment, and hence peaks in the NMR spectrum, is given by the formula 2nI+1, where n = number of equivalent protons; I = nuclear spin quantum number (I=1/2 for protons, but for other elements and even deuterium it can be different).
(This formula basically means that you have (n+1) peaks in simple 1H-NMR spectra, as the OP noted.)

But if you have inequivalent protons that you're coupling to, EACH OF YOUR PEAKS WILL AGAIN BE SPLIT by the new, different proton. Since the coupling constant is different, you can't simply add this new proton to your count of 'n', but need to consider splittings proper. If you split (2nI+1) peaks into a further (2I+1) peaks, you'll get (2nI+1)*(2I+1) peaks. If you couple to TWO equivalent nuclei, but which are different from your original coupling, you'll get (2nI+1)*(2*2*I+1) peaks.

Does this make more or less sense now?
Bit more lol, except the 2nI + 1... no idea how that comes about. I also got a bit confused only on the last paragraph

But very good.

Show off
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10 years ago
#11
The 2I + 1 term just comes from quantum mechanics, don't worry about how

In the last paragraph sinuhe is explaining that if you have two different adjacent protons then they will not split in the same way, so the simple n+1 rule you learn at A-level fails.

For example R2CH - CH2 - CH2N2CH3 the CH2 in the middle wouldn't have a quartet (4 peaks) signal although its coupling to three protons (ignoring J4/5 couplings), you would likely see a doublet of triplets (or a triplet of doublets) which is 6 peaks...bottom line, you don't need to know this for A-level
10 years ago
#12
(Original post by aster100)
Bit more lol, except the 2nI + 1... no idea how that comes about. I also got a bit confused only on the last paragraph

But very good.

Show off
Hehe.

Ok, so, the total energy of a system in a magnetic field, measured relative to that without a magnetic field, is given by

where
m is the magnetic quantum number (m = -I, -I=1, ..., I-1, I -- just like the quantum numbers l and m in atomic orbitals you're familiar with)
is the gyromagnetic ratio (a number, essentially a constant of proportionality)
is the reduced Planck constant (i.e. ) (to keep the units in order)
B = magnetic field strength in the z-direction

The energy DIFFERENCE between two states differing by (that's the selection rule for NMR transitions) then is
, or
(this is the Larmor frequency I mentioned earlier)

When spin-spin coupling is included, the energy of two coupled nuclei, A and X, is

where is the spin-spin coupling constant. You can rationalise this by considering the scalar product of the vector operators of nuclear spin magnetic moments - but that's quite complicated and you'd need some knowledge of quantum mechanics for it. So let's just assume it's true.
Then the Larmor frequency of nucleus A is simply going to be

So, consider the coupling of a proton with FOUR equivalent protons (for protons, m=+1/2 or m=-1/2, since I=1/2):

In other words, you have 5 lines, with intensities 161 (as shown by the number of times you get a particular frequency out). This is EXACTLY analogous to the stick diagram from my first post, except that you do it with numbers. But the principle is the same.

More generally, for an AXn system, where all X are equivalent spin I=1/2 nuclei, you have (n+1) lines in the spectrum, and the amplitude of the i-th line, for i=0,1,2,...,n, is given simply by the number of ways you can put i spins up and the remainder down (or 1/2 and -1/2, respectively), so

The 2nI+1 formula follows directly when you consider a spin with I>1/2.

As for the last paragraph of the previous post, consider the coupling of proton A to TWO protons, X1 and X2, which are inequivalent. Let's construct the same table as above:

Since J_1 and J_2 are DIFFERENT, each line will give us a different frequency, and you get FOUR lines. If the two coupling constants were the same, then the second and third line would be the same, and you'd get a 11 triplet (as expected).

So, having inequivalent protons means that lines will SPLIT. If you considered each proton X1 and X2 in turn, you'd see that X1 gives us a doublet (n+1 = 2), and so does X2, but overall we get a 11:1 quartet - i.e. 2*2. That's the formula I gave in the last paragraph.

Hope that clarifies things somewhat. I think if you're only doing A-levels, this is a bit of an overkill; but if you find it interesting, you can find a decent and relatively straightforward explanation of simple NMR in the Oxford Chemistry Primer 'Nuclear Magnetic Resonance' by Peter J Hore (OCP #32).
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10 years ago
#13
You really do like NMR don't you sinuhe! So what book would you reccomend to someone with an intermediate level of NMR theory?
10 years ago
#14
(Original post by EierVonSatan)
You really do like NMR don't you sinuhe! So what book would you reccomend to someone with an intermediate level of NMR theory?
Haha, I don't know if I really like NMR, it just seems to crop up quite a lot on here (and evidently I'm very bored today - it's far too hot to go out!). Was it an NMR question where you told me off last time?

I don't know much NMR myself; the only other 'conceptual' (as opposed to organic/inorganic NMR) book that I have is NMR: The Toolkit (OCP #92 - incidentally my copy is a bit weird, with one of the central signatures of the book printed upside down; it's very funny!). That one is a bit more challenging because you have the full-on quantum approach, but I think that also makes it simpler in many ways: if you do things from a quantum viewpoint, you end up doing just maths, which is far nicer than explaining things with handwavy 'physical reasoning'.

How about Spin Choreography by Ray Freeman? It looks quite cool (well, it's got the best title of any chemistry textbook I know), but I haven't looked at it in detail yet. Have you?
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#15
I think this is a bit beyond the A level syllabus and as i am cramming at the moment i dont really have time to go beyond the syllabus. Does anyone know any websites or worksheets with spectroscopy questions for alevel because i have done the ones in the textbook.
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10 years ago
#16
(Original post by Sinuhe)
Haha, I don't know if I really like NMR, it just seems to crop up quite a lot on here (and evidently I'm very bored today - it's far too hot to go out!). Was it an NMR question where you told me off last time?

I don't know much NMR myself; the only other 'conceptual' (as opposed to organic/inorganic NMR) book that I have is NMR: The Toolkit (OCP #92 - incidentally my copy is a bit weird, with one of the central signatures of the book printed upside down; it's very funny!). That one is a bit more challenging because you have the full-on quantum approach, but I think that also makes it simpler in many ways: if you do things from a quantum viewpoint, you end up doing just maths, which is far nicer than explaining things with handwavy 'physical reasoning'.

How about Spin Choreography by Ray Freeman? It looks quite cool (well, it's got the best title of any chemistry textbook I know), but I haven't looked at it in detail yet. Have you?
Yes it was an NMR question last time I was shocking never really taught NMR theory properly only the basics - the courses were far more directed towards application - the text for those course where williams and flemmings 'spectroscopic methods in organic synthesis' which has only a basic theory section. So I'm trying to get around to learning a bit more about things like quadrapoles, nOe's, instead of just knowing how to use them etc
10 years ago
#17
(Original post by EierVonSatan)
Yes it was an NMR question last time I was shocking never really taught NMR theory properly only the basics - the courses were far more directed towards application - the text for those course where williams and flemmings 'spectroscopic methods in organic synthesis' which has only a basic theory section. So I'm trying to get around to learning a bit more about things like quadrapoles, nOe's, instead of just knowing how to use them etc
That doesn't seem like an organic chemist talking ...

Quadrupole relaxation and the nuclear Overhauser effect are both quite well explained in OCP #32 (from a physical standpoint), so that'd be a good place to start I reckon. If you're well keen I can send you some of our NMR tutorial problems if you want to have a go at them.

(Maybe I should stop answering NMR/quantum questions on TSR, as my answers appear invariably to confuse rather than clarify things! )
0
10 years ago
#18
(Original post by Sinuhe)
Hehe.

Ok, so, the total energy of a system in a magnetic field, measured relative to that without a magnetic field, is given by

where
m is the magnetic quantum number (m = -I, -I=1, ..., I-1, I -- just like the quantum numbers l and m in atomic orbitals you're familiar with)
is the gyromagnetic ratio (a number, essentially a constant of proportionality)
is the reduced Planck constant (i.e. ) (to keep the units in order)
B = magnetic field strength in the z-direction

The energy DIFFERENCE between two states differing by (that's the selection rule for NMR transitions) then is
, or
(this is the Larmor frequency I mentioned earlier)

When spin-spin coupling is included, the energy of two coupled nuclei, A and X, is

where is the spin-spin coupling constant. You can rationalise this by considering the scalar product of the vector operators of nuclear spin magnetic moments - but that's quite complicated and you'd need some knowledge of quantum mechanics for it. So let's just assume it's true.
Then the Larmor frequency of nucleus A is simply going to be

So, consider the coupling of a proton with FOUR equivalent protons (for protons, m=+1/2 or m=-1/2, since I=1/2):

In other words, you have 5 lines, with intensities 161 (as shown by the number of times you get a particular frequency out). This is EXACTLY analogous to the stick diagram from my first post, except that you do it with numbers. But the principle is the same.

More generally, for an AXn system, where all X are equivalent spin I=1/2 nuclei, you have (n+1) lines in the spectrum, and the amplitude of the i-th line, for i=0,1,2,...,n, is given simply by the number of ways you can put i spins up and the remainder down (or 1/2 and -1/2, respectively), so

The 2nI+1 formula follows directly when you consider a spin with I>1/2.

As for the last paragraph of the previous post, consider the coupling of proton A to TWO protons, X1 and X2, which are inequivalent. Let's construct the same table as above:

Since J_1 and J_2 are DIFFERENT, each line will give us a different frequency, and you get FOUR lines. If the two coupling constants were the same, then the second and third line would be the same, and you'd get a 11 triplet (as expected).

So, having inequivalent protons means that lines will SPLIT. If you considered each proton X1 and X2 in turn, you'd see that X1 gives us a doublet (n+1 = 2), and so does X2, but overall we get a 11:1 quartet - i.e. 2*2. That's the formula I gave in the last paragraph.

Hope that clarifies things somewhat. I think if you're only doing A-levels, this is a bit of an overkill; but if you find it interesting, you can find a decent and relatively straightforward explanation of simple NMR in the Oxford Chemistry Primer 'Nuclear Magnetic Resonance' by Peter J Hore (OCP #32).
From first principles, although it's harder to follow, it does make a lot more sense. I did get most of the maths. Cheers!! Excellent stuff.

I bet you get lots of sex with knowledge like that
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#19
Does anyone have any slightly more indepth explanations i think what is present so far is fairly obvious. Would anyone care to expand lol
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10 years ago
#20
(Original post by aster100)
From first principles, although it's harder to follow, it does make a lot more sense. I did get most of the maths. Cheers!! Excellent stuff.
Good, glad to hear it's making sense.

I bet you get lots of sex with knowledge like that
Why of course - knowledge of magnetic resonance is clearly at the very top of everyone's list of things to look for in others.
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