Ch. 15 - Analytical Techniques: IR, NMR, Mass SpectWorksheetSee all chapters
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Ch. 1 - A Review of General Chemistry
Ch. 2 - Molecular Representations
Ch. 3 - Acids and Bases
Ch. 4 - Alkanes and Cycloalkanes
Ch. 5 - Chirality
Ch. 6 - Thermodynamics and Kinetics
Ch. 7 - Substitution Reactions
Ch. 8 - Elimination Reactions
Ch. 9 - Alkenes and Alkynes
Ch. 10 - Addition Reactions
Ch. 11 - Radical Reactions
Ch. 12 - Alcohols, Ethers, Epoxides and Thiols
Ch. 13 - Alcohols and Carbonyl Compounds
Ch. 14 - Synthetic Techniques
Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect
Ch. 16 - Conjugated Systems
Ch. 17 - Aromaticity
Ch. 18 - Reactions of Aromatics: EAS and Beyond
Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition
Ch. 20 - Carboxylic Acid Derivatives: NAS
Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon
Ch. 22 - Condensation Chemistry
Ch. 23 - Amines
Ch. 24 - Carbohydrates
Ch. 25 - Phenols
Ch. 26 - Amino Acids, Peptides, and Proteins

Concept #1: Splitting without J-values

Transcript

Now we're going to move onto the third important piece of information that you can get from proton NMR and that's going to be what we call the spin split.
First of all, it's just important to recognize all the different terms that this could be called and how they're all the same concept. If you ever hear of spin-spin coupling, or J coupling or a term I mentioned earlier called multiplicity, these are all the same exact concept. Spin splitting it just has to do with the concept of neighboring protons interfering with each other and this will reveal to us the distances between different protons.
Now I have to throw out a big side note here, which is that this topic can be taught either in a really simple way or a really, really complex way. Since I have no way of knowing exactly the way your professor taught it this semester, I'm going to teach it to you both ways. I've actually made two different videos so you can pick the easy one or pick the hard one depending on how your professor explains it and I'm going to help you determine which one to watch. Obviously, if you just want to be thorough, you can definitely watch both of them.
But for right now I'm going to teach this topic without J values. J values are the things that complicates the spin splitting concept a lot. Just so you know I've been teaching organic chemistry for a long time and the explanation without J values, which is the simplest explanation works for about 9 out of 10 of my classrooms. Most likely you're in one of those classrooms where you don't really need to learn a rigorous explanation of J values. That being said, then this session, this one video would really just cover you for spin splitting.
Now if your professor goes deep into J values and starts talking about how to draw a tree diagram, that's where you're going to want to watch the second video. If you suspect that you do need to learn that, go ahead and click on the second video after this one and I'll get more into the specifics of how you know that it's important for your class. Awesome. Let's get back to the lesson.
Basically, if you're not learning J values, then this is a very simple rule. All it says is that adjacent non-equivalent protons will split each other's magnetic response to NMR. Now there's a really simple rule that we use to predict what these splits would look like and that's what we call the n plus one rule. The n plus one rule just basically says that n stands for the number of – so I'm just going to write this here. N is going to stand for the number of non-equivalent – the dyslexia is coming out. Equivalent adjacent protons. Basically, n is going to be equal to this definition here.
It turns out that Pascal's triangle is actually going to do a really good job of predicting what these different splits will look like. Now I know it might have been a really long time, you might have not seen Pascal's triangle since grade school. So I'm just going to go over this really quick just to remind you.
Basically, Pascal's triangle is this weird kind of mathematical revolutionary phenomenon. I don't know if that's the right way to put it. Where it basically is a pattern of numbers that if you add up the two numbers above, then what you're going to get is the number below. 1 plus 1 equals 2.
As you go to multiple layers you start getting bigger and bigger sums. What we can see is if we go down in one of these directions where you basically say the two numbers on the top here, two plus one would equal to three or three plus three would equal to six. So you basically take the sum of the two numbers and that would be the number on the bottom. Then you just so on and so forth.
It turns out that these values are really good at predicting the height of splits. This actually has to do with split heights. As n gets bigger you're going to get more splits that have varying complexity. For example, if you do n plus one and your n is actually equal to zero, basically meaning that you don't have any protons around that are splitting, then what we're going to call that is a singlet. That's going to be the name of that and you're just going to get one peak.
Now let's look at a more complicated example. Let's say that n in your example is actually equal to two. That means that you have two non-equivalent adjacent protons next to your target protons. That's going to be called a triplet. What a triplet predicts, according to Pascal's triangle, is that since we're on the third level of the triangle, you're going to get a single split of size one, a second split of size two and a third split again of size one. That's exactly what is represented right here. Notice that I basically have a one, two, one pattern.
Basically, you can use Pascal's triangle to model what your splits should look like going all the way - let's just do one more example, going all the way to a quintet or a pentet, really, technically, it should be called a quintet. Quintet is the better way to say it. If you have quintet, notice that your peaks – you're going to have five different splits and they're going to be of the size one, four, six, four, one and that's exactly what we see here. That is when n would equal to 4. So we'd have four plus one which equals five, which would be our quintet.
I hope that's making sense in terms of the shapes. Now you just have to actually apply it because I know that it's all a little confusing. Let's do the first problem as a worked example. We'll kind of all do it together. Then the second one I'll have you guys do on your own.
Let's go ahead and analyze this carbon right here. I'm going to make that one my red one. What I'm wondering, first of all, is how many adjacent non-equivalent protons it has next to it. What we find is first of all, adjacent means it's within one space. So if I got to the left, there's nothing there. If I go to the right, I do have a carbon here. Does that carbon have any protons on it?
No. It turns out that in this case if you go to the left there's nothing, if you go to the right, there's nothing. That means for red, n is equal to zero. Now using the n plus one rule, that means that zero plus one equals one, which means that I'm going to get a singlet. I'm going to get just a single peak for the red hydrogens. Does that make sense so far? So I would expect it just to look like a single peak. Awesome.
So let's keep going. Now let's look at the blue hydrogens over here. Now, I drew the same thing. I said how many adjacent, non-equivalent protons does it have next to it. Well, if I go to the left, again nothing. I'm next to a carbon with nothing. But if I go to the right, do I have any hydrogens. Actually, yes. How many do I have? Three. I have three to the right.
That means that this would mean that n is equal to three. If I use the n plus one rule three plus one equals four. Four gives me a quartet I would expect that I would get a quartet from those protons. Those protons would be split into a quartet because I've got the three protons splitting plus one.
Now let's look at green. Green is over here If I go to the right, nothing. If I go to the left, how many protons do I have? Two. So that means that for green, n is equal to two. According to n plus one, it would be two plus one equals three, which would be what we call a triplet. That means that here I have a singlet, a quartet, and a triplet and that's what we mean by splitting, that every proton has its own unique shape based on the number of protons it's next to.
On top of that, if you wanted to predict the shapes of these singlets, quartets, triplets, it's easy because you could just use Pascal's triangle. You could say that a singlet – by the way, I'm not drawing these in any particular order. I'm not drawing chemical shift here. I'm just drawing shapes. The singlet would look like this. A quartet would look like this with one, two, two, one.
Actually, that's not what it says. That's wrong. Let's look at Pascal's triangle. Pascal's triangle actually says for a quartet, that should be one, three, three, one. I drew that all wrong. Let's do it again. So it should be one, three, three, one. Then my triplet, as we said earlier is one, two, one. Then the triplet should look like this.
Now. by the way, don't worry about the heights here. It doesn't have to be a specific height or whatever. I'm just trying to show the ratios between the different splits. Now you know what a singlet, a quartet, and a triplet would look like based on Pascal's triangle.
Now I'm going to have you guys do (b) on your own. Just so you know, (b) actually comes with one special instruction before you can solve it, which is that it turns out that heteroatoms have a special rule. Heteroatoms. Do you guys remember what a heteroatom is? It's important for the rest of this course. Heteroatoms are just non-carbon atoms. That would be nitrogen, sulfur, oxygen, phosphorous, etcetera
Heteroatoms, this is a big deal, do not split. You can't split through a heteroatom. Think of it like a wall. You can't split through it. That means that think of this as being a complete wall blocking off one side from the other. When I go to analyze how many different adjacent hydrogens are next to that, well if I go to the right, obviously there's nothing. But if I go to the left, there's also nothing because I hit a wall. That wall is the heteroatom. Same thing when I go to analyze this one. When I go to the right, I'm going to hit that wall.
Just think of that. That's a special rule that I just wanted to include that you can't split through a heteroatom. Other than that everything else is still the same. So go ahead and try to solve for the rest of (b) to try to figure out how many splits you'd get for a proton type one, proton type two and proton type three. Now the rest is up to you. Go for it. 

Example #1: Proton Splitting

Transcript

All right, so proton type one. If I go to the left, nothing. If I go to the right, do I have any hydrogens? Yes, I do. I have two. That has nothing to do with the number two. It just happened to be two hydrogens there. That means that n is equal to two meaning that two plus one equals three, meaning that one proton type one should be a triplet. Hopefully all you guys got that.
Now red gets a little bit more complicated. I'm going to draw an arrow here to show what I'm talking about, but I have to move it a little bit away for more space. To the left, I actually have three protons. Three H. To the right, I have two H's. Correct? That means that I've got three on one side, two on the other. That means that n is actually equal to five. Total, right? Meaning that I have five plus one, which equals six. And is is going to be a sextet.
Since that's one we haven't drawn yet, just to show you, you could predict what a sextet would look like by Pascal's triangle. It would be a one, five, ten, ten, five, one split. I'll quickly try to do that. It's going to be tiny. It's going to be tiny, something like this. Something like that. We've got the tiny ones on the edges, then the five's and then the tens. So obviously, it might not be perfect, but that's pretty good. Notice that Pascal's triangle helped me to know that shape.
Then, finally, we've got proton type three. I'll draw another arrow. For proton type three, to the right I've got nothing. We just said that's a wall. To the left, I have two hydrogens. That means that n is going to be equal to two, which means it's going to be two plus one equals to three, which means it's going to be a triplet.
In this case, I've got triplet, sextet, triplet and do you guys know what this H is going to be on the alcohol? A singlet. Because I told you guys n is equal to zero all the time for heteroatoms, so it's always just going to be one peak by itself.
So anyway, this is the explanation of splitting that like I said is going to suffice for almost all of you out there. Now go ahead and watch the introduction to the next video just to confirm if you have to watch it or not, but just so you guys know, have confidence that if you practice this and get good at it, this is probably all you're going to need for your upcoming exam. Cool. Let's move on. 

Practice: Predict the splitting pattern (multiplicity) for the following molecule: 

Practice: Predict the splitting pattern (multiplicity) for the following molecule:

Practice: Which of the following compounds gives a 1H NMR spectrum consisting of only a singlet, a triplet, and a pentet? 

Draw schematically the signals that will appear in the  1H NMR spectrum for the protons labeled with an asterisk. 
What is the splitting of each of the protons on the following molecule? a. _____________________________________ b. _____________________________________ c. _____________________________________ d. _____________________________________ e. _____________________________________ f.  _____________________________________ g. _____________________________________ h. _____________________________________
Which of these compounds will show two triplets (among other signals) in their   1H NMR spectra? A) I, II, III B) I only C) I, III only D) II only
Indicate the splitting patterns expected for all of the equivalent sets of hydrogens in the molecule below. Assume that you observe the maximum possible number of peaks in each resonance. Also, assume that the splitting is negligible beyond 3 bonds.
Sketch the  1H NMR spectrum of methyl propionate. Match the peaks with the protons in the structure. Mark the splitting patterns (e.g. singlet, doublet, triplet, or quartet) and number of protons for each signal (e.g. 1H, 2H, or 3H).
Predict the spiltting pattern for the structure below in order of protons a, b, c:a. doublet, doublet, singlet b. doublet, triplet, triplet c. triplet, triplet, triplet d. triplet, triplet, singlet e. doublet, doublet, doublet
Splitting: For the structure below, indicate the splitting for each signal. Use these abbreviations for the splitting : s=singlet; d=doublet; t=triplet; q=quartet; etc. If there is complex splitting, indicate the total # of peaks and how the splitting is produced. e.g. d of t
Splitting: For the structure below, indicate the splitting for each signal. Use these abbreviations for the splitting : s=singlet; d=doublet; t=triplet; q=quartet; etc. If there is complex splitting, indicate the total # of peaks and how the splitting is produced. e.g. d of t
One of the following compounds will show a doublet as part of its  1H NMR spectrum. Which one? 
What splitting pattern in 1H NMR spectrum would you expect for the hydrogen atoms colored in red. s singlet t triplet m multiple t d doublet q quartet.CH3CH2COCH2CH3
What splitting pattern in 1H NMR spectrum would you expect for the hydrogen atoms on the methoxy group? s singlet t triplet m multiple t d doublet q quartet