ochem week 8 part 2

ochem week 8 part 2

peaks on number of hydrogens and carbon placement

4 and 5 are on the same carbon with the same type (methyl) so they will combine

peak combinations come from symmetry, but top and bottom of arene will be its own idea

usually focus on isopropyl group

with an alkene you'll usually have two peaks together

Hydrogen NMR

1 proton one 1electron (1 neutral would make it deuteriom)

arene usually shows up betweem 6.5-8, then between 6.5-5.4 is vinyliv or alkene with 1 hydrogen

hydrogen attatched to a carbon double bonding to another carbon= vinylic Y=ON, Halogine Allylic=hydrogen attatched to Csp3, which is attatched to a carbon double bonded to somehting somewhere

homotopic- two moleculues in the same peak

hydrogens are more downfield the closer they are to larger bond (double triple etc)

its allylic when attached to a carbon thats attatched to another in a double bond

its vinylic when its directly attatched to a carbon directly in a bond

its more downfield when it can have a double bond AND is closer to something more electronegativ

this would be the integration value for each atom in that molecule

intensity of peak makes for type, or +1 hydrogen from what u have

numbered integration, letter is splitting

types of peaks- singlet, doublet, triplet, quartet, pentet, OR multiplet

triplet ratio 1:2:1, quartet 1:3:3:1, pentet 1:4:6:4:1

same peak with integrate to a higher value

ochem week 8 part 2

Transcript

Now, if our carbon has the exact same electronic environment as another carbon in our molecule, our NR can't distinguish between.

So if we have two carbons that see everything else in the molecule, there's some symmetry there that see everything else the exact same way.

Those two carbons will have the same heat. And because our carbon NMR isn't the most, it's gives us a limited amount of information.

We can't look at a peak and immediately be able to deduce if there is two or three or four carbons associated with them.

Essentially, the peaks are different heights for multiple reasons, not just correlated to how many carbons.

So if we have these overlapping carbons, these carbons are thinking in the same kind of space.

We have something P, we're going to encourage need to be able to look at a molecule, kind of parse that before we can just immediately know what we've got.

So we're gonna do a little bit of practice with that.

So here are three molecules. All these carbons are numbered. Which of these carbons would share peaks? Right. Would be in the exact same spot. So if we look at our nmr. Right. So for this first one, in theory, we should see seven peaks.

However, we don't see seven. We see less than that. So which of these carbons are overlap? All right, so if we're looking at our first compound up here, we'll just go left.

Right. The numbering on chem drawn to be somewhat arbitrary sometimes.

I'm not sure why we're starting at 4 and going to 1 carbon 4.

Are there any other carbons that will show up in the same place as carbon?

There are. What is that? Carbon. Carbon five. Carbon five. Right. Both of these guys. Okay. Both of these guys are going to show from the same place.

Right. Whenever we have that isopropyl group or a TR butyl group to do the same thing, all of those carbons are in the same electronic environment because they're connected to the same carbon that's been connected to everything.

Okay, so those two guys will show up in the same spot.

Carbon 1. Does it have anybody saying no? What about 2, 3, 4, 5? We already did 4 and 5. 6, 7. Nope. So just those two. Right. So we would see. In our overall NMR, we would expect to see seven peaks.

We would only see six peaks because two of them are the same.

Hope that makes sense. All right, this is the next one. Does carbon one have a bunny? Yes. What is it? Nine. Nine. Right. So these two are paired. Does carbon two have a bunny? Yes. Which One Is it five. Does anything else have a paired carbon? No, it does not. Right. So carbon three and four, although they're both methine, they're both different distances away from the carbon.

Right. And then carbon 7, 8 are both methyls, but they're connected to different carbons.

Right. Carbon 8 is a metal connected to a methylene. Carbon 7 is a metal connected to method. Because they're going to give us degrees, not dramatically.

Right? Right. One of them has PV of 28 and the other one is 29. But that's two different signals. We can see the difference. So that's what we're. What we're looking for. All right. And then. And specifically, too. Right. These two are paired, and those are paired. And then lastly, we have our one free dichlorobenzene carbon bond.

Does that have a buddy? Yes. What is it? Carbon 5. Carbon 2. Does that have a buddy? Yes. Which one is it? Three. Two has a buddy with four. Right. And then three and six. Are those paired together or are they each their own thing?

Their own thing. They're each gonna be their own thing. Right. So six will give us a peek. Three will give us a peek. And then we will have two other peaks. One of them will be both 1 and 6, and the other will be both 2 and 4.

Any questions? All right, so we'll do a square cap. So here is a carbon NMR. You can see that there's five peaks on this carbon NMR.

There's two that are in that 120 to 130 region, one of them around 60, and two of them around that 0 to 40.

So the code is SCD5MQ. So there's multiple possible answers. The N of R only and N of R of one thing. But we don't have a ton of inverters, so we can only draw so many conclusions from it.

So of the options presented down here. Right. Which of these could be reflected by this NMR's method?

There's also another question in there, which is just my usual.

How would you rate the test on a scale of 1 to 10? If you put a billion, it will be counted as wrong. Please put a number between 1 and 10. Also, this room is a little empty. Easier than usual, which is fine. Spring break is going up, so you're getting a few more points than usual for your sprinters.

Oh, let's go. You showed up. Congratulations. You can get some extra grip for that. Let's go. Oh, wait, you're right. It might be the CSC allergy crap down Down. All right, so there's a couple of things that we can look at based on these structures right off the bat that'll help us to narrow some stuff down.

Right, so how many peaks do we have overall in the section?

So we need to have five unique carbons in our compound.

That doesn't mean we can't have a compound with six carbons.

Right. If two of those carbons are unique. Right, so let's look at A. For example, we've got 1, 2, 3, 4. And then these last 2, 5 and 6 are both CH3 groups. They're both connected to that same carbon. Right. So they're the same distance away from every other atom in this molecule.

So are they going to share a peak in our spectrum? Yeah, they would share a peak. So even though we have six carbons, we would only expect five peaks.

Right. So could A still be an option even though we only have five?

All right, so we don't before we are even looking at chemical shift or things like that.

Right, so B and C, let's look at B. Right, we have 1, 2, 3, 4, 5, 6 total carbons. Are any of those carbons going to share peak? No, they're not. Right. Everything is a different distance away from something else.

So we would need six peaks and we only have five. So is D an answer? No. Nope. All right, D has five. F. 1, 2, 3, 4, 5. We've got six. Do any of them share a peak? Yes, these last two here. So that's a total of five. So that's still in the running. G is going to look very similar to F. Right. Where we have these two that share a peak and now we're lost.

So G would only be four peaks. Right. So 1, 2, 3, and then those are the same. That's four. That's not enough. So G is out and then H, we've got five carbons and they're all different.

Right. All right, so just counting peaks, we can rule out two.

We can also come in and look at the chemical shift values.

Right. What is this region right here that these two are in?

What's that usually associated with? Does anybody remember? Oh, the carbon is further, a little further. Right. So normally carbon needles are on that end, but because we haven't made 11, 200, the inspector cuts it off.

Yeah, yeah, but that's. It brings up a good point a lot of times on these spectra.

If there isn't anything beyond a certain point, rather than showing you the empty spectra, it cuts it off.

So just make sure to. To keep an eye on that. I try and just give you the whole 200 range for that.

So you can see there's nothing there. But I can't always make the spectrum do what is too far.

But yes, it's the outkase, the double bond region.

Right. So if we know we have to have a double bond because we have carbons in that double bond region, we can go ahead and rule out C and F as.

Right. Because those don't have V double bonds. How many carbons do we have in the double bond? We only have two. So could we do E? E would have to have four carbons in the double top region.

Right. So we can also get rid of E because there's only two.

And then the last thing we have between A, B and H is this carbon with a halogen on it for these two guys.

Right. So if we have a carbon that's bound to a more electronegated atom or a polarizable atom that is going to shift one of those peaks a little further down, it's going to pull some of the electrons away from the carbon.

So we should see if we have one of these halogenated, we should see something in this 60 region, which we do.

Right. We have one carbon in there. So if we were going to choose between AH and B, is there one we could rule out or say.

That is much less likely to be the case. We probably rule out B because there isn't. Essentially, B would look like this. Right. We would have three in that region. Right. One of them being further away means that we don't have B.

And so now A and H. And here's a hint, since it's just like all the line, we've only got two answers left.

Those are your answers because they're not going to tell you, select all that apply and only give you one answer, no questions.

But also, if we're looking at A and H, they both have an alkene carbon.

Carbon to the bond. A carbon that's singly bound to a more electronegative or polarizable atom and the right number of carbons for the number of peaks.

Any questions? Yeah. So when you have an alkene, you're always going to have the.

These two in the. Yeah. With. Yeah, yeah. At least two. You could see one if it was symmetrical in a way where it's shared.

Right. So the way we've seen it can actually. That would just be ethylene, so maybe that's less of a concern.

Well, if it's a perfectly symmetrical alpha one There.

So, for example, if we had like this compound here, we would only see three total peaks.

Actually, we only see two total peaks because all of these are the same and then those.

So we'd only see two, even though there's six carbons.

But that's also a pretty rare incident. Think that's going to happen? Because it would have to. All four things would have to be symmetrical. You'll more often see where you have like an area ring, but you only see four points instead of six.

But you can't ever just have one carbon with the double bond.

So even if you only see one peak there, there has to be two.

You're just sharing. Any other questions? Any questions on carbon nmr? Since this is a carbon nmr, there's no way to tell whether it's pulling or bromine.

Not like mass spec. Not like mass spec. We could run a mass spec or we could spend a long time fiddling with our instrument and get it to tell us run a chlorine in a mar or something.

That would be more trouble than it's worth. The mastect is way easier. Any other questions? So that's carbon and ore. It's going to give us some more tools in our toolbox to try and take summary of molecule we don't know anything about.

Look at these spectra and put it together. The last tool that we're going to get is hydrogen number H, which is going to be dependent on our odd number of protons or neutrons.

Hydrogens. Right. So the majority of our hydrogens, the vast, vast majority, have one proton and that's.

And one electron, but doesn't contribute to electricity.

It doesn't have a neutron. If it has a neutron, we call that a deuterium. We've seen that before a couple of times. We'll see a few more kind of niche uses as we keep going.

But it's used very commonly in nmr because if we run a hydrogen or a carbon nmr, the deuterium turns it invisible.

So we use deuterated solvents to kind of hide the solvent in the NMR.

We don't need to see that. I have 90% ethanol with my product dissolving it.

That's not helpful. I know that the ethanol is there, so we stick the deuterium atoms on there instead.

And that takes care of that. They also tend to be more. Because it's harder to get. They tend to be more expensive than just deuterated solvents.

We tend to use them in smaller amounts and they are kind of niche.

Oh, yes. So when we are looking at our hydrogen nmr, the spectrum is going to look very similar to a carbon bar.

We're still going to use PPS as our x axis. We're still going to be looking at that chemical shift.

The more those electrons get pulled away from the hydrogen atom, the lower downfield they will be.

Our relative scale is going to be different. So on carbon, we're looking at 0 to 200. On hydrogen, we tend to be around 0 to 12. So the relative scale shifts. But because there are so many more hydrogens, we can get more specific data that we can look at that is more reliable for certain pieces of information.

So we get two more pieces of information. One of them is integration, and the other one is splitting.

Splitting is a little bit more complex. Integration. Is everybody in here taken some version of. Is everybody vaguely familiar with some of the ideas in calculus?

I'm not going to make you do calculus. But what effectively, integration is an integral is the area under a curve.

So what happens is when we run the spectra, our peaks that show up, the computer does the math, but you take an integration value of that area, and that tells you the space under that peak that you're looking at, and that correlates to the number of hydrogens associated with that peak.

So in a carbon nmr, we don't know how many carbons are associated with each peak.

In a hydrogen nmr, we can take the integration value and that will tell us how many are there.

So if there's two or three or four, we do a little April and we see, oh, there's three there.

So it just tells us. Which is nice because it's usually more hydrogens than there are carbons.

Now, what's technically happening is it's just a ratio of the height of that to everything else.

But for the purposes of this class, it's just going to tell you how many hydrogens are there so we can get more information.

Again, more information adds also complexity. The more we have to look at, the more we have to interpret.

All right, well, we'll start with chemical shift.

So for our hydrogens, our chemical shift is going to be more based on what that hydrogen is attached to.

Right. So generally it's a carbon. Every now and then, it might be an oxygen or a nitrogen.

And since we're now looking at the carbon that the hydrogen is attached to, the same way that the carbon is impacted in a carbon NMR tends to reflect on how a hydrogen is impacted.

Hydrogen. Right. So if our hydrogen is attached to a carbon in A double bond, it gets shifted down here the same way.

If our carbon is in a double bond, it's shifted down.

Right. The main difference is going to be that the hydrogen is more impacted by an aromatic system than our carbon.

So I mentioned our aromatics and our regular double bonds.

In a carbon bar, they have a lot of overlap, but they may tend to shift more downfield.

And in our hydrogen nmr, there is a distinct difference.

If you have an arene that shows up between six and a half and eight.

If you have an alkene that is not an Arene, that is between like 4 and a half and 6 ish, usually it's around 5.

Right. There's a very distinct cutoff between those. Which also means that if you look at your hnmr, you can usually pretty clearly tell if you have an arene, which is helpful since sometimes it can be hard to look at your IR and see if that carbon carbon double bond is really there.

If it's some noise or if you have those aromatic teeth or if those got covered up.

Right. If you have hydrogens in your aromatic region of your hnmr, you have an ear.

There's no other reason that those are going to show up.

I mentioned a little bit ago we had to start talking about naming stuff, the terms bimylic and allylic.

We didn't go to use them a ton, mostly because this is where I talk about them.

So sorry about that. But I talk about them here because they're going to be.

They're going to refer to the hydrogens as they would show up on our hny.

So if our hydrogen is attached to a carbon in a double bond, that will shift it further downfield than if our hydrogen is attached to a carbon that is attached to a carbon in a double bond.

But even that one is still further downfield than just our hydrogen attached to a carbon with no double bond anywhere nearby.

So our just normal ch sp3 hybridized, no, nothing else fancy going on, tends to be between like zero and one and a half.

If our hydrogen is allylic, it gets shifted further down.

If it is vinylic, it can shift it even further down.

If it's aromatic, it can shift it even further down.

The last two things that we have here, it is similar to on our carbon R, that hydrogen is attached to a carbon with an O or an N or a chlorine or chlorine, something like that, it will get shifted further downfield, not as far as that double bond.

So it'll be in between single all Single bonds and double bonds.

We have this heteroatom area. A couple of things. Okay, I did say that. So somewhere between 4 and 6 can tend to be where your NH and OH show up.

Because our hydrogen that's in. We're looking at all of our hydrogens. If our hydrogen is attached to a nitrogen or an oxygen, sulfur or phosphorus or something other than a carbon, it does still show up in the hydrogen.

A mar, it shows up weirder. The NH and the OH is the ones that we'll tend to see.

They are way more impacted by other kind of random stuff because our resonance structures tend to involve nitrogens and oxygens a lot.

So they. I say four to six. I've seen them as far as. Well, they just move wherever they feel like moving to.

They're more unpredictable. They're kind of usually not something I rely on looking for.

They also broaden. So remember how in our ir, our OH group gets really broad.

Our carbon gasses basically eats the front half of our IR spectrum.

That also will happen in an nmr. If you have an oh. That hydrogen bonding takes these nice sharp peaks that we're used to looking at and actually kind of makes them really wide.

I've seen them also where they essentially get so wide that they just fade into the baseline and they're basically invisible.

They kind of just disappear. So we can run into some issues trying to find those guys.

I tend to put that locating that NH and that OH as like the last thing I look for.

Right. If you look at an IR and you know you have an oh, but you can't find it in the nmr.

You have the oh, it's just being weird in your nmr.

Don't worry about it. If it's on a carbon, it will be here. If it's on an oxygen or nitrogen, it should be here, but the IR will be more reliant.

And then the other one, which is for some reason just not in your textbook, which really should be, is that aldehydes and carboxylic acids.

So aldehydes tend to show up first, and then carboxylic acids tend to show up here.

So this is like 12 to 11. This is like. Like 10 to 9. That tends to be where you see those. And again, the carboxylic acid ones can disappear the same way that our alcohol ones can disappear, but they're so far downfield that it can be hard to see them anyway.

Those aldehyde ones are pretty much always there.

Those aldehyde ones are very reliable. So Again, if you're looking at your IR and you're trying to figure out if you have an aldehyde or.

Right. The difference there is that, you know those little vampire things next to the CSP3 hydrogen, carbon hydrogen bond.

Sometimes it can be hard to tell if that's really what you're looking at or not.

Well, if you have an aldehyde, it's going to have to be in your hnmr, because that's a hydrogen and it's going to be around 9 or 10 and it's the only thing that's there.

So it's very obvious from the HNMR if to have an L from the ir, it may be a little harder to parse out.

Does that make sense? So with a lot of the spectra, there are some things that.

Right. So our IR will jump up and scream at you. There's an oh here. The NMR will kind of vaguely hint that maybe there's an oh there.

But the aldehyde does the other way around. Right. Your HNMR will screen there's an aldehyde, whereas your IR may be harder to predict.

So for some of these functional groups, the IR and the hmr, they both tell you it's there, but some of them tell you more loudly.

That's why we use all of them a lot. Any questions so far on chemical shift? Oh, yeah. So the only thing important here is this term homotopic.

That just means this is a specific term for when we have two hydrogens that are in the same.

Right? So the same way that our carbon, if we have two carbons that are in the same electronic environment, they show up in the same place.

If we have two hydrogens that are in the same environment, they also show up in the same place.

So generally this will correlate to hydrogens on the same carbon.

Right. So if we have like ethanol, for example, there's two H's here and three H's here, we would see two peaks in our H NMR for those two guys.

And then we would see that oh, maybe. Right. So these two same and these three are also the same.

So we would call those three hydrogens on that methyl homotopic to each other.

If we ever have carbons that are also going to share a peak, the hydrogens on those carbons are also Right.

So if we had isopropanol, Right. Those two carbons would share a peak in our carbon nmr.

So all six hydrogens share a peak in the hydrogen and they would integrate to six.

So can we put six hydrogens on a carbon or. Yeah, we cannot. Right. We cannot put six hydrogens on a carbon. So if we take an integration value and we get a value that's 4 or larger, what that tells you is there's multiple hydrogens or multiple carbons all sharing a heat together.

Right. So if we took the E, our integration would tell us that we've got one hydrogen and six hydrogens.

What it's really saying is we have one and two sets of that are chemically equivalent.

So we'll talk more about the integration of that.

But that's what that hollow topic is referring to.

If our carbons share a peak, the hydrogens on that carbon also shear.

So let's practice. Right, so, oh, this is chemical shift. So if we're looking at these parts, these hydrogens, which of these is going to be shifted the most downfield?

So there's that chart that I just showed you. I don't expect you to necessarily have it memorized immediately.

Right. But some of this should be a little bit more obvious.

Right. So if we're looking at hydrogen one and two in our first one, Hydrogen one or hydrogen two, which of those is closest to the thing that is electronically damaged?

1. 1. One is next to the double bond. Two is further away from the double bond. So which do we think is going to have be shifted more down here?

One. One. Right. Because it's the nearest thing that's sucking up the electrons.

Right. This one hydrogen is in the allylic position. Right. It's on the carbon that's on the carbon that's in the double bond.

So that'll be shifted a little further down. If we're looking at our second compound, we've got a hydrogen that's in that double bond, right.

On the carbon of the double bond that. And then we've got a hydrogen that's on a carbon that's attached to an O.

So which we think is going to be more impact, there's a hydrogen that's directly connected to a carbon in the bond, or a hydrogen that's connected to a carbon that's connected to something electronegative, one or two.

So one is going to be further downfield. This is our vinylic hydrogen. Right. All right. And lastly we have an aldehyde hydrogen and an aromatic hydrogen.

Which of those is furthest downfield? Aldehyde. It's gonna be the aldhe. So our aromatic would be further down the field than anything else we've seen so far.

But the aldehyde is even further Right. The same way that in our carbon nmr, our carbonyls are the furthest downfield in our H and M R, M R hydrogen is on a carbonyl carbon.

That's the furthest that field. There's just only one way for a hydrogen to be on the carbonyl carbon, and that's an alpha.

They're all one or they split it. Any questions? All right, all these answers were one that was not on purpose.

It also. Okay. Any questions on the chemical shift in general? Yeah. Could you repeat why it was one for the aldehyde again?

Yes. So the aldehyde shifts just in this region and the Aryan shifts in this region.

So in part, it's just those numbers. But. But the rationale too is essentially that carbon double bonded to an oxygen is going to pull the electron density more away from Rh than the carbon double bonded to another carbon.

Essentially we have the effect of the double bond in both cases, but in one way we also have that electron to get out and it matches our carbon NMR pattern.

Our carbon NMR goes for carbonyls and then double bonds and then other stuff.

Here we have a hydrogen on a carbonyl, a hydrogen on a double bond, a hydrogen on other stuff.

It's the same. Any other questions? All right, so the other two things are integration, splitting.

I already kind of talked about integration. It's a pretty straightforward concept because it's literally just how many hydrogens are associated with this.

Right. We saw in our carbon large, we can have multiple carbons associated with pica.

There's no way to know how many. For hydrogen, we can know how many. It's the integration. So this example is. Yeah, here's our isopropyl group. So we can see that our integration values would correlate up to one and six for each of these guys.

Right. And again, we can't put six carbons on a hydrogen or six hydrogens on a carbon.

So it's going to be at least two. Could be up to three, depending on what the deal is.

Okay, so these calculations are technically just relative numbers.

Right. So our actual compound were. Say it had. We knew, we did a mass spec and we knew it had to have 10 hydrogens.

And we add up our integrations and we got 2 and 3 for our integration values.

Well, 2 and 3 does not add up to 10, but it adds up to 5.

So if that 2 was actually a 4 and that 3 was actually a 6, that would add up to 10.

So there just might be some symmetry in this there where four hydrogens show up in the same place, and six hydrogen show up in the same place.

And all we can look at is the relative ratio of those numbers to each other.

So it's the simplest fraction that we end up actually seeing.

A lot of the times it actually doesn't. That doesn't cause us a problem because like we can see here, we have one hydrogen on one of our carbons.

Well, that means that the simplest fraction of 1 to anything is the exact random numbers.

That's why we see. So a lot of the times, and generally what I'll give you in this class, I will just give you the correct number of hydrogens.

But you may encounter at some point where you're.

You're trying to do something and the numbers don't add up.

They're exactly half or exactly a third of what you expect them to be.

That's probably because what you're really just looking at is a ratio.

You need to kind of multiply that up based on some symmetry.

Hey, a. A common example is. So I mentioned that if we have 10 hydrogens, this guy here is perfectly symmetrical.

But because it's 2 and 3 and 2 and 3, we only see that 2 to 3 ratio, even though it's actually 4 and 6.

So if I gave you an NMR of this compound, I would say 4 and 6.

But if you stuck it in the instrument, the instrument would spit out 2 and 3.

That's the most complicated part of integrations, is that ratio bit.

Otherwise, it is really just counting. This can also help you identify the correct molecular formula for your mass spec.

Right, because your mass spec will tell you the number of carbons and hydrogens.

But there's oxygen, or if there's a nitrogen, or if there's a carbon, chlorine, or bromine or degrees of saturation, some of that can get complicated.

But if you also have integration values, you know, there should be 10 hydrogens.

If you do your math, 10 hydrogens, everything works out right.

We know that that's the correct molecular problem.

Again, this, these, all these spectroscopy techniques.

Talk to each other. All right, Any questions? Let's do some integration practice. So here's a molecule. It's one we looked at earlier, but this time, tell me what the integration values you would expect from the peaks are.

Keep in mind, right, that carbon four and five, we've already decided they're chemically equivalent.

What's the best way to do this? Do we just say, like, how many hydrogens would be associated with these?

Okay, it's. It. How many hydrogens are on carbon 7. 3, 3. Does carbon 7 share a peak with anybody else? No. So the peak that's associated associated with those hydrogens would integrate to what, the three?

That's the answer, right? So we have a peak integrates to 3. For carbon 6. How many hydrogens? 2. Does it share its peak with anybody? So what's our integration value? 2. What about 3? 2, 2. What about 2? There's nothing. Right. There's no hydrogens at all in 2. So we just don't even see see that in our H in a bond.

We would see that on carbon, but we don't see it in the hydrogen.

How many hydrogens on carbon? 1. 1, 1. And then on carbon 4. How many hydrogens you have? 3. 3. But does it share its peak with anybody? Yes. Yes, it shares with 5. So how many total would we see? 6. 6. So we see a total of 6 of that peak. So we'd see 5 peaks in our NMR peaks. The peaks associated with 1 and 3 would be further downfield.

If they're close to that C double up O, the other guys would be a little further up field with their length, and they would have relative integration values of 1, 2, and then 2, 3, 6 as well.

Any questions? All right, so the last little thing we'll talk about today, last little bit we'll get into, is going to be splitting patterns in H&MR.

Splitting is perhaps the most complicated of the things we can determine from an hnmr.

If you notice, when we looked at our examples of an hnmr.

I'll go back in a second. The peaks here are not one line, they are a bunch of lines, right?

There's a whole bunch of things which is part of why we have an integration value.

There's more stuff curve that we can take the area under.

But what those number of peaks that we see under each actual peak tells us is about how many hydrogens are near the hydrogen we're looking at.

So essentially, when we look at a hydrogen on a carbon, right?

So if we look at that one that we had before, actually, let's look at an easier one first.

We will look at this guy right there. So that guy right there has two hydrogens on it, and that guy has three.

What our spinning atoms do is they take this peak here that would integrate to two, and it says, how many hydrogens do I have on the adjacent carbons?

Because as these hydrogens are spinning and the hydrogens next to them are spinning, they can see each other, right?

They, they are have A some degree of coupling is the term that we use.

There's a coupling constant that is associated with those two hydrogens that are both spinning and both being spun in the same direction in this magnetic field.

And so they have some degree of essentially being the same that they're in space.

And so what they will do is they will split for every one hydrogen they see.

So if there are zero hydrogens nearby, we get one peak.

And then for every additional hydrogen nearby, we split that peak.

Right? So if there's one, we see two peaks. If there's two, we see three peaks, Right? So this methyl group here would actually split into three different heaps.

That's what we would observe because it's next to two hydrogens.

So it's the number of hydrogens. Plus one way to remember that is if there are zero hydrogens, you still see some zero is A.

And then subsequently, since our methyl can see two hydrogens, those two hydrogens can also see the methyl.

If I can see, you can see me, that's LV's hydrogen.

All right? So these three, see those two splits into three separate peaks.

So the next peak over is going to look back and see those three hydrogens.

So it will split into four different peaks. It looks something like that. So when we start split these peaks into a certain number of 1 or 2 or 3 or 4, we refer to that splitting pattern as a singlet, a doublet, a triplet, a quartet.

So on a pentet, a sextet, a septet, octet, octet, that all is number one.

Because once we get past cortex, generally it becomes unreal.

They become very, very, very small little peaks on the sides, they tend to overlap with other peaks, right?

If, if we have two things right on our carbon nmr, if we had a chemical shift like 28 and 29, that's just two straight lines next to each other.

If we had these kind of broad peaks at 20, 29 and Asian or same, we'll say 2.82 point.

They're going to be on top of each other, which is going to make all that splitting impossible to read.

So in those cases, we just generally say multiplit.

There is something happening here. It is happening a lot. So whatever it is seeing is complicated. And believe it at that splitting can get incredibly complicated and very difficult to read.

Part of the reason that the really nice instruments that are millions of dollars to high resolution everything are so nice is that they pull these splitting patterns closer together.

And in doing so, this overlap issue that you can get gets really minimized instead of having a peak that spreads out across like, you know, 0.2.3 PPMS and then can overlap with a bunch of stuff, it gets trunk.

And that means that, that there's less overlap. And we can really specifically look at that one time.

We can always zoom in on these things. Singlet, doublet, triplet, quartet, those are the main ones to know in terms of the way that we will describe our compounds.

We can always look at a compound and predict what we should see.

It's just in reality, looking at a spectra, once we get past four, it can be hard to actually see five or six or seven and kind of distinguish them between each other.

So I can look at something and predict that it would have an octet.

The reality of observing an octet and actually being able to describe that is much less.

So we can see here, right? This is our. Is this. This is an ethyl bromide alcohol. But an ethanol compound will look very similar to this right here.

And so we have our. That ethyl group right where this integrates to two, this integrates to three.

And we can see that splitting pattern of a quartet and a triplet.

So there's a number and then there's a letter, which is usually how these get written.

Just to be explicit. Right. We can look at this and count to four for that quartet.

That's the one. The part that's zoomed in. The person that's not zoomed in is a bit harder to see.

But we look at count of four. So we call that a quartet. It's usually abbreviated with a Q. And the number that you see that number is the integration value.

The number is. Those are the two pieces of information you're getting.

Numbers integration letter is split. Just keep that in mind. The recitation that you're going to be working on, that's like not next week, but the week after.

Really. When I give you an nmr, it will give you a number and a letter.

That's what this is or the number of integration.

The letter is the split. There is a distinct ratio that we see with our splitting patterns.

You can notice in this quartet, the middle are larger than two on the side.

You can see the same in the triplet, except it's just one in the middle are larger than the two on the side.

So that relative size differential, it follows a pattern.

I Forget if it's PA or something like that. There's like a mathematical pattern for it, but it tends to allow us to at least distinguish between odd and even numbers.

If there are Two things in the center that are tall or one thing in the center that is tall.

Whether or not we have a triplet or a quartet, distinguishing between, say a quartet and a sextet can be more complicated, but at least two versus just one.

We can distinguish between them, but that ratio is roughly 1 to 2 to 1 for a triplet, 1 to 3 to 3 to 1 for a quartet, 1 to 4, 4 to 6, 4 to 1.

You can read, right? And the distance between these is referred to as its j value or coupling constant.

What is improving when we have higher resolution in Mars, because that coupling constant, that j value, is getting smaller.

There is a way that you can mathematically calculate that and calculate the PPM from the megahertz of the instrument.

It's probably in your textbook. I'm not going to make you do it because you don't need to do it.

And if you ever need to, you can look up the calculation.

It's not the most complicated, but it's kind of an extra thing to make you do for, in my opinion, in this class, no real reason, so we're not going to worry about that.

But it is a calculation that you can do. Another thing that we do often run into with nmr. So remember how for ir, some of the questions that you saw just gave you a list of numbers and you had to know what those numbers were instead of actually looking at.

In IR spectra itself, people will do the same thing with N.

So they will give you the location of the peak, the integration of the peak, and the splitting pattern, if that's relevant.

Right? So that S for singlet, D for doublet, T for triplet, Q for quartet, M for multiplet, they'll just give you those.

And you don't actually get to look at the thing. You just have that information. So you do need to memorize those regions because you may just be given a number and says, this is the ppm.

What do we have? What are we looking at? All right, but for the last little bit, here is a compound we've seen twice now.

But this time we're going to predict these splitting patterns.

So again, if we have the same peak, right? So if we have some overlapping in the peak, those will integrate to a higher value.

The splitting will still be dependent on what they see.

Generally, if it's the same peak, they're going to have to see the same thing, right?

So we don't count it twice. We only count them for one time because we only see the one thing.

But by definition, if they're sharing that peak, if they're homotopic, they see the exact same thing.

So they normally slit the exact same. It's all right, so carbons 4 and 5 are sharing a peak.

Right. The hydrogens on those carbons are sharing peaks as well.

If we just looked at four, how many hydrogens are on a carbon next to carbon four?

Just one. Right. So what would that split carbon four into two? It'd be two peaks. Would it be all that doublet? A doublet. Right. If we look at Carbon 5, Carbon 5 has the exact same situation because it is also attached to carbon one.

It only sees one hydrogen. It would also split into a double. So what we observe when we look at these two peaks is one double and it integrates to six.

Right. So that's what we would see in our nmr. Now, instead of looking at these same peaks, looking back, we look at carbon 1.

How many hydrogens are on carbon 4? 3. How many hydrogens are on Carbon 5? 3. 3. How many hydrogens are on carbon 2? 0. 0. So that's a total of 6. Right. So how many peaks would we expect to see for carbon 1?

7. So even though carbons 4 and 5, they see carbon the same, one singular hydrogen.

So we only split once. When we're looking from carbon one's perspective back, we see everything.

Right. So we see all six. We split the seven. So if we have something that looks like this on our spectra, that is probably an isopropyl group.

Something integrates to one and six. That's a great indication that we have this isopropyl group.

It's a very common recurring pattern that we see, similar to how that ethyl group was a 2, 3 integration with a quartet triplet ratio.

That's also indicative of the methyl. There's some patterns that we see kind of pop up a lot, but.

All right, so that's our first two. And then if we look at carbon three, how would we expect carbon three to split?

What would we call it? Triplet? It would split into a triplet. Right. Carbon six has two hydrogens. So if we're looking here, we've got two here, we've got zero here.

So there's only a total of two. So that's a triplet. It splits three, two plus one. So if three can see six, six also has to see three. Right. So if three gets to see this way, when we come over here, we also see backwards.

But now six can see both three and seven. Right. So how many total hydrogens can we see from carbon6?

We can see a total of how many are on three? Two. How many are on seven? Three, three total. Right. So five splits into what? Six sex. So we would say that this three is triplet. This is a sex step. We might just write multiplet because it might be pretty complicated.

And then if 6C 7, 7 sees 6. So how would 7 split triple? Another triplet. Right. So 7 and 3 are both triggers. However, 3 is going to have a chemical shift that's further downfield because it's closer to the or.

Yeah, further downfield because it's closer to the O.

It's also going to integrate to a higher value. It'll. Or to a lower value, integrate to 2. So more downfield, lower integration, 3 more up field, higher integration.

Or 3. So we can still distinguish between them even though they're both.

Any questions on splitting? All right, so this is basically the extent of what.

What you need to know to do. That should be uploaded now. You'll have two weeks to work 60. What that representation primarily is going to be.

There's a decent chunk of it which is just reading in a box.

Tell me what the OR says. It is also going to contain several problems that say, here is a mass spec, an ir, a carbon NMR and a hydrogen nmr.

What do I have? And you have no other information than those vectors.

That is the kind of standard way that we do these questions the most.

Come back after spring break on Tuesday. We will do a lot of practice with. When I say we, I mean we, because I forgot what I put on these slides last semester.

So we're all going to figure out what these compounds are together, but we will work on that piggyback right now.

Have a good free rake.

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