Help with NMR spectroscopy please!

[BenSCBSc]

Hi everybody,

I'm struggling to understand how ¹H NMR works. I think it might be best to explain what I do know and maybe somebody can fill in the gaps for me and/or correct me?

 

The external magnetic field aligns the magnetic moments (spin) so that they are parallel/antiparallel (in a low energy state) to the magnetic field.

A radio frequency is pulsed (400 MHz in this case) and the nuclear magnetic spins flip (putting them in a high energy state).

So with the pulse, the spins are put through a process of relaxation & realignment, whilst doing this the nuclei emit a weak radio wave KHz which I can measure, but how exactly does this describe the molecular structure of whatever sample I have in there?

And where does the rest of the energy go? I put MHz in and get KHz out, due to the conservation of energy I've lost some somewhere?

 

A brief description will suffice, I just need to get my head around what is going on? Or if somebody can provide a link that explains it? I've looked and I have got a little information but not what I want.

 

I want to know how the radio radiation signal I'm recording describes the chemical compounds I'm supposed to be looking at?

 

Many thanks to anybody that can help me. Cheers Ben

[hypervalent_iodine]

From a chemical perspective, nuclei within a compound will experience different degrees of magnetic shielding, which come about from magnetic fields generated by other atoms / functional groups. This shielding will affect the energy required to flip a spin. Thus, we can differentiate protons from one another in a molecule since the energy required to flip their spins is different, causing the peaks to come up at different frequencies.

 

A wealth of experimental data has allowed us to describe frequency ranges for where we might expect certain nuclei to pop up. For example, we know where protons attached to a phenyl ring occur, and we can tell them apart from the protons on, say, a methyl group based on the fact that methyl protons appear at very different frequencies (chemical shifts).

There are other things too, such as anisotropic effects. These are explained here http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch13/ch13-nmr-3.html

 

Most of what it comes down too is the distribution of magnetic fields within a molecule.

[BenSCBSc]

Many thanks, I also found that the integral of the resonance peak increases with the concentration or number of molecules and since I am using CHCl₃ and CHBr₃, it means my hydrogen atoms increase by the same value. So you would know exactly how many hydrogen atoms are in your molecule. Damn that's clever!

Many thanks for the link, I haven't had chance to look yet but I'm sure I'll be equally amazed.

[hypervalent_iodine]

Yes, peak area correlates quite nicely to concentration, since it essentially measures the population of spins that have flipped. You can't really do the same for carbon, but I have performed quantitative experiments for 19F NMR. If you have the right standards and probes, then you can get some very good data on that front. You can also use the integration to work out the relative number of protons in one peak compared to another. This is useful for determining the number of protons in a particular function group when comparing peaks from the same molecule, or in working out the abundance of a particular contaminant / byproduct / reactant when comparing peaks from different molecules.

 

You can also do 2D NMR, which allows you to relate 1H peaks to the 13C peaks they are bonded to or near.

[BenSCBSc]

Wow more magnetic fields (anisotrophy), so any anisotrophic effects will inform you if the system is pi or not. Can you determine anything else about the compound from anisotrophic magnetic effects? It looks like there's an awful lot to learn here, I imagine it takes quite a bit of practice before you can look at the resonance lines of a variety of organic compounds and accurately describe their atomic structure.

 

I have been provided with chemical drawings for different molecular compounds and I have to be able to determine what their peaks would look like, also I am supposed to be able to describe what the compounds look like based on their names. I'm studying physics, so chemistry notation etc is not my forte. Looking at the scope of the subject, I'm not sure we have been given enough time (about 3 weeks). Perhaps I've misunderstood though, perhaps my tutor was referring to chemists or other users of NMR and not me, when she said 'alkane, alkene, alkyne, alcohol, amide, nitro & carboxylic acid are important? Or is it quite possible that I should be understanding the atomic structure of these function groups?

 

Thank you very much, you have been very helpful. Do you know of any papers, journal articles on basic NMR particularly to do with 1-iodopropane & 2-iodopropane or something similar. I have looked at a couple:

Unfortunately they either seem to be very complicated, far too complex for what I am doing or I can't get access with my athens?

[hypervalent_iodine]

Wow more magnetic fields (anisotrophy), so any anisotrophic effects will inform you if the system is pi or not. Can you determine anything else about the compound from anisotrophic magnetic effects? It looks like there's an awful lot to learn here, I imagine it takes quite a bit of practice before you can look at the resonance lines of a variety of organic compounds and accurately describe their atomic structure.

Sort of. When looking at a spectra of a pure compound, there are several things that you might look at:

• Chemical shift. This would give an idea of what sort of functional groups are involved, or allow you to identify your peaks if you already know (or have an idea of) the structure. Anisotropic effects play into this a lot, and we use this to explain the fact that, for example, aromatic protons occur at such a relatively high chemical shift. Tables such as this are very useful.
• Integration. We discussed this earlier.
• Splitting pattern. I didn't mention this before. Take the ¹H NMR spectrum of methyl propanoate:

From http://legacy.chemgym.net/as_a2/topics/nmr_intro/introduction.html

 

The simplest peak you will see in an NMR is represented by the methyl peak at ~ 3.7 ppm. We would call this a singlet peak, since it appears as only one peak. The remaining two peaks look a bit funny by comparison, but there is an explanation for this. The protons labelled A are all what we call magnetically equivalent; that is, they are in the same magnetic environment as one another and thus converge as the same peak (I say peak in the singular, but I am referring to the entire triplet). The protons on the neighbouring carbon, B, are not the same as A. They experience a different magnetic environment, and hence occur at a different chemical shift. My very basic understanding of how this plays into splitting is that because B can be either spin up or spin down, A will experience three different magnetic environments (which I guess comes from the 3 possible combinations of spin states on the two B protons), and therefore appear as three peaks (a triplet). Similarly, the protons on B experience 4 different environments, and is hence a quartet. The general rule is that a given proton will split into n+1 number of peaks, where n is the number of protons on neighbouring carbons (that is, protons attached to carbons directly next door). With this information in hand, we can construct an idea of which peaks are next door to each other based on this and the integration. If I have a triplet peak with an integration of 3 and a quartet peak with an integration of 2 (as in the above example), I could reasonably expect that these two proton sets are right next to each other in the molecule.

The ratios of the peak heights is also important, and can help identify the splitting pattern. This gets pretty complicated pretty quickly (you can, for example, get very complex splitting such as a doublet of doublet of doublets), and there are exceptions, but you should have no trouble if the compounds you are dealing with are as simple as the iodo compounds you mention. Certain systems, notably aromatics, can experience coupling through multiple bonds. You can also get fine coupling through space. One other thing to note is that you can also get coupling to other types of atoms, depending on their nuclei. Phosphorus is one example that comes to mind. For more: http://www.chemguide.co.uk/analysis/nmr/splitting.html and https://www.khanacademy.org/science/organic-chemistry/spectroscopy-jay/proton-nmr/v/complex-splitting

• Coupling constants, or J coupling. These are the distance, in hertz, between the split peaks in a signal. You calculate it by calculate the difference in chemical shift between two peaks, and multiply by the applied magnetic field power in MHz. The best way to visualise them is to construct a splitting tree, such as these:

From http://www.slideshare.net/Oatsmith/13-nuclear-magnetic-resonance-spectroscopy-wade-7th

For reference, both of these peaks are doublets of doublets, which arise when a proton (A) is coupled to two other sets of protons that are not magnetically equivalent and do not have the same coupling constant (B and C). B and C are different in this case because there is no rotation around the double bond, forcing B and C to exist in different environments.

J couplings are quite useful, and can tell you a lot of information if you know what you're looking at. You'll notice in the above example that the J_ab coupling in the peak for H^a is the same as the one for H^b. This is true for all such couplings, since the coupling that A experiences to B should always be equal in magnitude to the coupling experienced by B to A. We can therefore relate peaks to one another in this way by looking at their coupling constants, and begin to get an idea about which peaks are next to one another (with help from the splitting pattern).

The other thing you can glean from this comes about from this wonderful diagram:

From https://www.studyblue.com/notes/note/n/cyclic/deck/14307587

The dihedral angle between neighbouring protons has a huge effect on the magnitude of the coupling constant. As you can see in the example, this is extremely useful when you are discerning between non-equivalent protons on a single carbon such as those on a cyclohexane ring or a sugar. They are also very handy when looking at ​cis and trans double bonds, since the dihedral angles between the protons either side of the double bond are very different.

 

 

 

I have been provided with chemical drawings for different molecular compounds and I have to be able to determine what their peaks would look like, also I am supposed to be able to describe what the compounds look like based on their names. I'm studying physics, so chemistry notation etc is not my forte. Looking at the scope of the subject, I'm not sure we have been given enough time (about 3 weeks). Perhaps I've misunderstood though, perhaps my tutor was referring to chemists or other users of NMR and not me, when she said 'alkane, alkene, alkyne, alcohol, amide, nitro & carboxylic acid are important? Or is it quite possible that I should be understanding the atomic structure of these function groups?

 

Thank you very much, you have been very helpful. Do you know of any papers, journal articles on basic NMR particularly to do with 1-iodopropane & 2-iodopropane or something similar. I have looked at a couple:

Unfortunately they either seem to be very complicated, far too complex for what I am doing or I can't get access with my athens?

 

 

You should look into some basic organic chemistry text books. Organic Chemistry by McMurry was a great starting point for me as I recall, and there are heaps of examples and chapter questions there to help you learn. I assume you are at college, in which case I would go to your library and see what they have. Clayden is another great one, though I can't speak for how well they go into NMR. The basic compounds you describe are also well characterised. You could very easily google their NMR spectra, or look it up on SDBS (or Reaxys / SciFinder if you have access to those). I am also happy to go through any specific spectra you are looking at.

 

I couldn't comment on what is expected of you in terms of learning, but given the timeframe I'd say you're not going to be going into huge amounts of depth. Much of what I've described is well beyond the scope of what you would encounter when you initially come into NMR. In fact, most of it I didn't learn until the third year of my chemistry degree. I only bring it up in so much detail because I find it interesting.

[BenSCBSc]

Thank you very much, I'm very grateful.

There is quite a lot to get into there, but I have a couple of questions,

Do the protons that are double bonded not rotate, because they are double bonded?

 

Where you were talking about J-coupling (sentence below Splitting Tree diagram) and using protons A, B & C are these represented the first diagram? I'm not sure I understand how the number of peaks is representing the number of environments? I mean I'm not sure how these environments are counted, what determines a different environment? How is A different from C, I know C is bonded to an Oxygen atom and thus has different neighbours but how does this alter whatever it is your counting? I'm guessing because it has no hydrogen neighbours but if I take A that has 2 H neighbours how does that result in 3 peaks or is that 4 peaks because the central peak is twice as high (so it's integral is informing us that two of the protons are in the same environment)? I still don't know why it's 4 though and where you get 3 different spin states?

 

I have been given more material to work through so maybe J-coupling etc will come up.

 

I've kind of bubbled my way through half of the year, performing experiments including using a robotic telescope to image clusters and investigate stellar evolution & Compton scattering and had nobody to discuss any of it with so it's very refreshing to be able to ask questions about things I'm not sure about. Many thanks.

Hi again, I was just looking at my new material and found this for 2.2 dimethylpropane?

• propane indicates that the structure consists of a single chain of three carbon atoms: carbon–carbon–carbon,
• ‘2,2’ indicates that there are two substitutions on the middle (labelled 2) carbon atom,
• dimethyl indicates that both substituents are methyl groups,
• hydrogen atoms are attached to the ‘end carbon atoms’ to ensure that the fourfold valency of carbon is satisfied.

Can you tell me what is meant by 'substitutions' please? I haven't come across this term before and the diagram for 2.2 dimethylpropane doesn't show a chain of 3 carbon atoms? It has 3 x H bonded to 4 x C and those are attached to a central carbon. So it's symmetrical in 2 dimensions, the carbon atoms form a cross, not a chain?

[hypervalent_iodine]

Thank you very much, I'm very grateful.

There is quite a lot to get into there, but I have a couple of questions,

Do the protons that are double bonded not rotate, because they are double bonded?

Yes. Double bonds have no free rotation as single bonds do.

 

Where you were talking about J-coupling (sentence below Splitting Tree diagram) and using protons A, B & C are these represented the first diagram? I'm not sure I understand how the number of peaks is representing the number of environments? I mean I'm not sure how these environments are counted, what determines a different environment? How is A different from C, I know C is bonded to an Oxygen atom and thus has different neighbours but how does this alter whatever it is your counting? I'm guessing because it has no hydrogen neighbours but if I take A that has 2 H neighbours how does that result in 3 peaks or is that 4 peaks because the central peak is twice as high (so it's integral is informing us that two of the protons are in the same environment)? I still don't know why it's 4 though and where you get 3 different spin states?

Sorry, I wasn't clear. I was referring to Ha, Hb, and Hc, from the structure in the splitting tree diagram.

 

To be honest, I don't worry too much about what gives me the number of environments in so much detail. I generally just pay attention to the n+1 rules how how that gives rise to splitting. My understanding is this:

 

In the green picture, the protons labelled A are all equivalent. They experience the same environment as each other. They see the protons labelled B. In terms of spin states, they can either be both up, both down, or one up one down. Hence, three possible environments total, which causes the peak for A to split into three. The spin states of A can be either all up, all down, one up and two down, or one down and two up. Four possibilities, hence the peak for B is split into four.

 

The splitting tree diagram on is a little different, but the same concept. Ha sees Hb, which can be spin up or spin down, and is thus split into two. It also sees Hc, which can once again be spin up or spin down, and so each doublet peak arising from the Ha - Hb coupling is split into a further two. Another way of representing the tree might be this way (excuse the quality, I drew it with my finger...):

 

 

I have been given more material to work through so maybe J-coupling etc will come up.

 

I've kind of bubbled my way through half of the year, performing experiments including using a robotic telescope to image clusters and investigate stellar evolution & Compton scattering and had nobody to discuss any of it with so it's very refreshing to be able to ask questions about things I'm not sure about. Many thanks.

Hi again, I was just looking at my new material and found this for 2.2 dimethylpropane?

 

• propane indicates that the structure consists of a single chain of three carbon atoms: carboncarboncarbon,
• 2,2 indicates that there are two substitutions on the middle (labelled 2) carbon atom,
• dimethyl indicates that both substituents are methyl groups,
• hydrogen atoms are attached to the end carbon atoms to ensure that the fourfold valency of carbon is satisfied.

Can you tell me what is meant by 'substitutions' please? I haven't come across this term before and the diagram for 2.2 dimethylpropane doesn't show a chain of 3 carbon atoms? It has 3 x H bonded to 4 x C and those are attached to a central carbon. So it's symmetrical in 2 dimensions, the carbon atoms form a cross, not a chain?

Substituent is a general term for anything hanging off the parent chain of something. The parent chain being the longest continuous chain of carbons (in this case, the propyl chain).

[BenSCBSc]

I'm getting there, slowly. It looks like I do need to know all this stuff but I understand spin-spin coupling now and the intensity ratios are Pascals triangle. (I feel quite clever having recognised the pattern ) Actually just being able to understand what is going on is no mean feat! Thanks again for your help.

[hypervalent_iodine]

I'm getting there, slowly. It looks like I do need to know all this stuff but I understand spin-spin coupling now and the intensity ratios are Pascals triangle. (I feel quite clever having recognised the pattern ) Actually just being able to understand what is going on is no mean feat! Thanks again for your help.

 

 

 

It gets pretty complicated pretty quickly. Glad I could be of assistance!