NMR signals

[663697]

In trying to predict the NMR spectrum for the structure below, I was confused by 2 things:
1) Why is the blue H split into a septet? Wouldn't all the neighboring H's be chemically equivalent, producing a heightened doublet peak?

2) Why is the signal for the blue H more deshielded than that of the red H's? Aren't the methyl groups connected to the blue H considered electron-donating groups? I feel like my reasoning is off here.

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[rabbott1971]

Splitting uses the n+1 rule, so count neighbors and add one, that's why it shows 7. This also explains the number of peaks on the other hydrogens (red has no neighbors, so just one peak, and green has one neighbor, so two peaks).

As far as deshielding, I think it is because the blue H is on a more substituted carbon. So those other CH3 groups (green H's) are pulling away electron density. Seeing your comment about EDG, I believe that would apply to reasoning about substituent effects in electrophilic aromatic substitutions etc., but not as much for proton NMR. Carbon still pulls electron density. Others can correct me if I am offbase, but that's how I see it.

 

[663697]

Splitting uses the n+1 rule, so count neighbors and add one, that's why it shows 7. This also explains the number of peaks on the other hydrogens (red has no neighbors, so just one peak, and green has one neighbor, so two peaks).

As far as deshielding, I think it is because the blue H is on a more substituted carbon. So those other CH3 groups (green H's) are pulling away electron density. Seeing your comment about EDG, I believe that would apply to reasoning about substituent effects in electrophilic aromatic substitutions etc., but not as much for proton NMR. Carbon still pulls electron density. Others can correct me if I am offbase, but that's how I see it.

For the first part of the question, I think I know what I'm mixing up. Since there are 6 chemically equivalent hydrogens, it would integrate to A = 6, right? So the peak for those green H's would then be a tall doublet?

Also, when finding the signal for the blue H, why do we not take into account the carbonyl carbon like we do for the red H's?

 

[rabbott1971]

The splitting -- its saying -- for each group of equivalent hydrogens, whether there are six or just one, the signal will show up with ONLY ONE peak for the group of equivalent hydrogens. However, that signal will have additional peaks for each and every single neighboring hydrogen! So again with the blue one, its showing one peak for itself, and six peaks for each of its six neighbors. For the signal indicating the green hydrogens, there is one peak for all six equivalent green hydrogens, plus one additional peak for the blue neighbor. And for the red hydrogens, you have only one peak, because there are no neighboring hydrogens. Neighbors must be within one molecule, and the carbonyl carbon does not have any hydrogens.

For your second question, this is Proton NMR, so carbons are irrelevant. There is carbon NMR but this is not one of those, and I think proton NMR is much more common. You're right that H3C-COOH would show the H3C hydrogens as one peak (singlet), but that is due to there not being any neighbors (within a distance of one atom). The hydrogen on the carboxyl group is too far away, and it would also show up as a singlet, since it can't "see" the H3C group hydrogens. They just aren't in the same neighborhood!

 

[rabbott1971]

Integration-wise I think you're on track. That's going to try to tell you how many equivalent protons are in that group. This particular graph doesn't give much help with integration, because you can't really measure the area under those peaks. Some NMR will give you a little step-ladder type line that shows increments for assessing integration. But I'm not seeing one on this graph.

 

[aldol16]

1) Why is the blue H split into a septet? Wouldn't all the neighboring H's be chemically equivalent, producing a heightened doublet peak?

The physical basis of splitting is a bit beyond the scope of the MCAT but it does help explain the n+1 rule. For all MCAT purposes, use the n+1 rule to determine the multiplicity. So how this arises is as follows. Each proton in the vicinity of the proton in question also has a magnetic field. That field is what splits the signal of the relevant proton. So to see a concrete example, let's start with a simple system: F2HC-CH2F. I'm going to call the proton on the left Ha and the proton on the right Hb for simplicity's sake. Let's focus on Ha. Ha will "feel" the magnetic field of the instrument. That's the basis of NMR. But it will also feel the magnetic fields of the vicinal protons. Those fields will be randomly oriented - it's a statistical system. So the magnetic fields of the Hb protons (there are two of them) can be either both aligned with the field of the instrument, both opposed to the field of the instrument, or one aligned with the field and one not. So in other words, let's say that the field of the instrument points up. The fields associated with the two Hb protons can be either: up-up, up-down, down-up, or down-down. That's a statistical distribution. Up-down and down-up of course give you the same observable effect. When both fields are aligned with the instrument's magnetic field, you get either shielding or deshielding. Again, the physical basis of this is more advanced and I don't want to get into it but you can see the end result. You get a three separate peaks, corresponding to the up-up, up-down/down-up, and down-down fields, in a 1:2:1 ratio. This is a triplet and this is why two vicinal protons will split a proton like Ha into a triplet. Now you can extend this to six protons as you see in this case and see that the same statistical argument applies - it's just a larger statistical distribution now.

 

[aldol16]

2) Why is the signal for the blue H more deshielded than that of the red H's? Aren't the methyl groups connected to the blue H considered electron-donating groups? I feel like my reasoning is off here.

As far as deshielding, I think it is because the blue H is on a more substituted carbon. So those other CH3 groups (green H's) are pulling away electron density. Seeing your comment about EDG, I believe that would apply to reasoning about substituent effects in electrophilic aromatic substitutions etc., but not as much for proton NMR. Carbon still pulls electron density. Others can correct me if I am offbase, but that's how I see it.

So methine protons (-CHR2) will always be deshielded relative to methylene protons (-CH2R), which will be deshielded relative to methyl protons (-CH3). This is indeed due to the fact that no matter what R is, it's going to be more electronegative than hydrogen (yes, even carbon) and therefore the inductive effect will cause the deshielding.

There is also another effect called chemical shift anisotropy, but that's beyond the scope of the MCAT.

 

[rabbott1971]

NMR is one of those things that you can tell it is way more complicated than what is presented in undergrad organic chemistry. Just understanding the units of the x and y axis of the graph is more challenging than I thought it would be (and not particularly important for MCAT/Ochem purposes). I would love to learn more about it but I'm not trying to pull back the curtain on it right now, I'll just stick with my little basic dilettante principles and be happy. I don't think I've even seen an NMR question in any of my practice materials.

 

[Lawpy]

NMR is one of those things that you can tell it is way more complicated than what is presented in undergrad organic chemistry. Just understanding the units of the x and y axis of the graph is more challenging than I thought it would be (and not particularly important for MCAT/Ochem purposes). I would love to learn more about it but I'm not trying to pull back the curtain on it right now, I'll just stick with my little basic dilettante principles and be happy. I don't think I've even seen an NMR question in any of my practice materials.

NMR is a pretty complicated topic: it's basically my weakest subject in the sciences because it wasn't emphasized well in my ochem classes and labs, and the MCAT prep books + practice materials only briefly touched on it. The focus was much more heavily on IR and recognizing functional groups. I think NMR would matter more for analytical chemistry.

@aldol16 explained it well. I recommend Khan Academy for a good introduction and refresher on the topic.

 

[663697]

The physical basis of splitting is a bit beyond the scope of the MCAT but it does help explain the n+1 rule. For all MCAT purposes, use the n+1 rule to determine the multiplicity. So how this arises is as follows. Each proton in the vicinity of the proton in question also has a magnetic field. That field is what splits the signal of the relevant proton. So to see a concrete example, let's start with a simple system: F2HC-CH2F. I'm going to call the proton on the left Ha and the proton on the right Hb for simplicity's sake. Let's focus on Ha. Ha will "feel" the magnetic field of the instrument. That's the basis of NMR. But it will also feel the magnetic fields of the vicinal protons. Those fields will be randomly oriented - it's a statistical system. So the magnetic fields of the Hb protons (there are two of them) can be either both aligned with the field of the instrument, both opposed to the field of the instrument, or one aligned with the field and one not. So in other words, let's say that the field of the instrument points up. The fields associated with the two Hb protons can be either: up-up, up-down, down-up, or down-down. That's a statistical distribution. Up-down and down-up of course give you the same observable effect. When both fields are aligned with the instrument's magnetic field, you get either shielding or deshielding. Again, the physical basis of this is more advanced and I don't want to get into it but you can see the end result. You get a three separate peaks, corresponding to the up-up, up-down/down-up, and down-down fields, in a 1:2:1 ratio. This is a triplet and this is why two vicinal protons will split a proton like Ha into a triplet. Now you can extend this to six protons as you see in this case and see that the same statistical argument applies - it's just a larger statistical distribution now.

This makes total sense. I can see what I was mixing up now.

So just as an extra example, in the case below, since there are 2 H's on both of the carbons being pointed to, and they are chemically identical, they would be split into a quartet due to the -CH3, and would integrate to 4H, rather than 2H?

 

[663697]

NMR is a pretty complicated topic: it's basically my weakest subject in the sciences because it wasn't emphasized well in my ochem classes and labs, and the MCAT prep books + practice materials only briefly touched on it. The focus was much more heavily on IR and recognizing functional groups. I think NMR would matter more for analytical chemistry.

@aldol16 explained it well. I recommend Khan Academy for a good introduction and refresher on the topic.

I'm in the same boat when it comes to NMR. We only briefly touched on it the last 2 or 3 classes of my ochem course and after that I had forgotten almost everything.

 

[aldol16]

NMR is one of those things that you can tell it is way more complicated than what is presented in undergrad organic chemistry. Just understanding the units of the x and y axis of the graph is more challenging than I thought it would be (and not particularly important for MCAT/Ochem purposes). I would love to learn more about it but I'm not trying to pull back the curtain on it right now, I'll just stick with my little basic dilettante principles and be happy. I don't think I've even seen an NMR question in any of my practice materials.

One way they might touch on it on the exam is showing you some drug molecule or natural product and its NMR spectrum and asking you to identify a particular proton on there. That would require you to understand how the n+1 rule operates.

 

[aldol16]

NMR is a pretty complicated topic: it's basically my weakest subject in the sciences because it wasn't emphasized well in my ochem classes and labs, and the MCAT prep books + practice materials only briefly touched on it. The focus was much more heavily on IR and recognizing functional groups. I think NMR would matter more for analytical chemistry.

NMR is the single most important tool in all of chemistry. All fields use it - from organic to inorganic to organometallics. Of course, different fields use it in different ways, but NMR is one of the critical pieces of information you need to include in any paper. Reviewers would never let a paper slide through without NMR since it's the fingerprint of the molecule you said you made. IR can be misleading because it has less to do with overall structure of the molecule and more to do with individual parts of it, i.e. functional groups, since you're essentially measuring the "spring constant" of specific bonds.

 

[Lawpy]

NMR is the single most important tool in all of chemistry. All fields use it - from organic to inorganic to organometallics. Of course, different fields use it in different ways, but NMR is one of the critical pieces of information you need to include in any paper. Reviewers would never let a paper slide through without NMR since it's the fingerprint of the molecule you said you made. IR can be misleading because it has less to do with overall structure of the molecule and more to do with individual parts of it, i.e. functional groups, since you're essentially measuring the "spring constant" of specific bonds.

What references would you recommend for learning more about NMR uses in inorganic and organometallic chemistry? Because the only thing I'm familiar with is in organic chemistry and that too only mildly.

 

[aldol16]

So just as an extra example, in the case below, since there are 2 H's on both of the carbons being pointed to, and they are chemically identical, they would be split into a quartet due to the -CH3, and would integrate to 4H, rather than 2H?

Almost! Never confuse integration with multiplicity - completely different concepts. Integration always tells you how many protons gives you the signal. So in this case, since there are two protons in question, you should see a relative integration of 2 in your spectrum. The multiplicity will indeed be four due to the vicinal methyl group, since those protons can now be oriented up-up-up, up-up-down, up-down-up, down-up-up, up-down-down, down-down-up, down-up-down, and down-down-down. The bolded and italicized ones are identical to each other, so you get a quarter with intensity in a 1:3:3:1 ratio.

 

[663697]

Almost! Never confuse integration with multiplicity - completely different concepts. Integration always tells you how many protons gives you the signal. So in this case, since there are two protons in question, you should see a relative integration of 2 in your spectrum. The multiplicity will indeed be four due to the vicinal methyl group, since those protons can now be oriented up-up-up, up-up-down, up-down-up, down-up-up, up-down-down, down-down-up, down-up-down, and down-down-down. The bolded and italicized ones are identical to each other, so you get a quarter with intensity in a 1:3:3:1 ratio.

I thought the same as well. The reason I'm asking is because on a TPR practice passage it asked, "In the 1H NMR spectrum of Compound 2, what is the correct description of the peak appearing at 4.30 ppm?" I answered "a quartet integrating to 2H", but the answer according to them was "a quartet integrating to 4H" ; "Since Compound 2 is symmetrical, these signals will overlay each other exactly, meaning that the signal will integrate for four hydrogens, not two".

 

[aldol16]

What references would you recommend for learning more about NMR uses in inorganic and organometallic chemistry? Because the only thing I'm familiar with is in organic chemistry and that too only mildly.

Nothing better than reading the literature. So 1D NMR is what we teach undergraduates in undergraduate courses. There are also 2D NMR techniques such as COSY, which tells you which protons are coupled with each other, and NOESY, which tells you more about the local environment of the proton (are there eclipsing effects, etc.). So when you make an inorganic complex, you can tell exactly how the ligands are oriented around the metal. You can also use it to measure equilibrium constants, which is important for measuring the kinetics of reactions between inorganic complexes.

 

[aldol16]

I thought the same as well. The reason I'm asking is because on a TPR practice passage it asked, "In the 1H NMR spectrum of Compound 2, what is the correct description of the peak appearing at 4.30 ppm?" I answered "a quartet integrating to 2H", but the answer according to them was "a quartet integrating to 4H" ; "Since Compound 2 is symmetrical, these signals will overlay each other exactly, meaning that the signal will integrate for four hydrogens, not two".

Oh, I didn't even look at the other half of the molecule! My mistake. Yes, it should integrate to four. So here's how we usually integrate protons. There's a program called MNova that allows us basically draw a baseline below the peak and it calculates the peak area above it. But it needs a reference value, so the first peak you integrate is set automatically to one. So when you have a molecule with perfect symmetry (all protons have a mirror-image counterpart; in other words, the plane of symmetry doesn't lie on any one proton), you will actually get a reduced integration. The relative integration ratios will always reflect the absolute ratio of the protons. But in that case, the relative integrations would be some factor of the actual integrations. It's easier to see with an example. So let's use the simple example of propane. The NMR looks like this:

Two peaks, one should be a septet and one should be a triplet. There are 6 methyl protons and 2 methylene ones, so you know that the absolute ratio of integration between the methyl-to-methylene protons should be 6:2, or 3:1. But remember when you actually use the software to integrate, it doesn't give you an integration of 2 and 6. This is because you always must set something to 1 for reference. So in this case, you would integrate the septet first, which sets it to one. Then you integrate the triplet and you would get 3. Note that the relative ratio is still 3:1. But due to the plane of symmetry, there are actually 6 methyl protons and 2 methylene ones.

In your example above, one would set the proton at the 4-position in the pyridine ring to be one, since there's actually one proton there. That means that all the other integrations will now reflect the true proton count - that is, four for the protons in question since, due to the plane of symmetry, the methylene protons on the left and right would give the same signal.

 

[rabbott1971]

It is forbidden!!!