Subtitles section Play video Print subtitles [ Silence ] >> Alright, well, good morning again. So, today what I'd like to talk about is two of three really important concepts in mass spect. So, last time we introduced mass spect, we talked about how the technique worked. We introduced one big idea, and the big idea was that you had to have an ion. In the mass spect, we got that ion from kicking out an electron, and then we talked about CI and other soft ionization techniques like Mol D and electrospray ionization, and we said the big idea there is you get an ion by adding a proton or adding a sodium to the molecule. So, the three sort of big concepts in mass spect really aren't very hard, and in fact, nothing is hard about mass spect except we think differently than we've sort of gotten used to thinking as organic chemists. So, the first concept is exact mass. Ever since you were, well, taking high school chemistry, you got used to the molecular weights of compounds and calculating them based on the average mass, in other words, carbon S, carbon 12, and carbon 13, so you have an average mass of carbon. But here, each ion is being separated individually. In other words, you're separating each ion from each ion, which means you're going isotopomer from isotopomer, and so we need to get used to the idea of thinking about what the masses are of the individual isotopes, and we'll explore some of the implications of that. So, the second big idea which ties into that is isotopic abundances; the fact different elements have different abundances of different isotopes. Carbon, for example, has 1.1% of its carbons as C13, and the rest of them is carbon 12. Sulfur has a little bit of S34, in addition to S34, and we'll explore the implications of this, and then in the last lecture, we'll talk about fragmentation. So, this will be on Wednesday's lecture. We'll discuss fragmentation as it pertains to EI mass spect, which is really a subject or a special topic in its right. We are already in the CI in the electrospray ionization last time, so an example of fragmentation of a quasi molecular ion of a proteinated species. Remember the type of molecular ion you get from soft ionization chemistry, and that's essentially just protic chemistry. What we saw was a reaction that involving a leaving group of proteinated group of nitrogen. Leaving, that's chemistry that you've really been used to since your sophomore year, but when we talk about fragmentation and EI mass effect, we'll be talking about fragmentation of radical cat ions, and given the fact that in soft ionization techniques, you don't generally get a lot of fragmentation. Fragmentation isn't super important there, but in EI mass effect, where you have tons of fragmentation, you often can't even see your molecular ion. It's very important. So, I've given a couple of resources here, which you'll using in your homework assignments. These are also linked to the webpage for the course, so you don't have to type in all of this stuff into the computer. You can just click on the links that I've provided, and as you'd go through the homework, you'll get familiar with using these tools and choosing proper settings. What I'd like to do now and today, is to really talk about the concepts behind these first two issues over there, isotopic matches and isotopic abundances. And I thought maybe a place to explore this, since we were talking about mass spect techniques last time, and we talked about a variety of techniques, EI mass spect, ESI mass spect, CI mass spect, is to introduce a variant of any of those techniques, any of those ionization techniques that's often referred to HR mass spect or high resolution mass spect. In general, when you get a mass spectrum, the mass that you get is good to within a few tenths or a few hundreds of a mass units. In other words, if you get a number like in one particular exercise, I'll show you later on, the homework, you have number 1239.2. You might say, okay, that number 1239.2 is probably good to within a few 10ths of a mass unit. However, with specialized instrumentation and calibration, you can do far better. So, with better instrumentation and calibration, you can often get masses to very high precision with mass to charge ratios to, I'll say high precision, typically may 5 millimass units or 5 parts per million or better, and I'll show you what I mean by that, and in the case of a technique called ion cyclotron residence mass spectrometry, sometimes even one magnitude better than that. So for example, by five parts per million, I mean if I had a mass of 300.0000, then the mass of a small molecule may be like a steroid or something like that, we could get that within five parts in a million; in other words, within .0015 or what a mass spectrometrist would refer to 1.5 mmu or millimass units. And, in this particular case, you can start to distinguish among different species that have nominally, in other words, to unit mass, the same mass. So, the one thing, and I mentioned before is that your molecular weight is going to be dictated by your isotopes present. [ Pause ] So, let's take a moment to talk about isotopes and their masses. So, if I go, and I look at our periodic table over there, and I looked at carbon, and I look under atomic weight, I'll see that the atomic weight is 12.001115 depending on how many digits. I'm sorry .01115, depending on how many digits they give you. However, carbon is a mix of carbon 12 and carbon 13. It's 98.9% carbon 12, and 1.1% carbon 13. The mass of carbon 12 is set by definition as 12.00000, as many 0's as you wish to write. The mass of carbon 13, mass of carbon 13, is 13.00336. So, when you go and say I want to weigh out a moll of carbon or I want to weight out a moll of carbon-containing compound, what you really doing is taking this number, which is the weighted average of these two numbers. In other words, 98.9% of this number and 1.1% of that number. But, as I said you're separating molecules isotopomer from isotopomer, and so you're going to get a peak that corresponds to the isotopomer that's all C12. Other elements also have isotopes. Hydrogen, for example, if you look, your atomic weight in the periodic table is 1.00794, and yet hydrogen is a mix of hydrogen and deuterium, sometimes called heavy hydrogen. It's mostly, mostly, mostly hydrogen, 98.984% H1, and only 0.16% H2, and so when you're thinking about mass spectrometry, you want to use the mass of H1 and the mass of H1 is 1.00783. [ Pause ] Other common elements that you encounter in organic compounds; nitrogen, the atomic weight is 14.067,0067. This 00 there. And yet, nitrogen N is a mixture of N14 and N15. It's 99.62% N14 and 0.038% N15, and so the atomic mass of N14, the mass that you would use in thinking about mass spect is 14.00307. Oxygen. I guess I'm not writing out the elements. I'll just write 0. Again, I'll give you the mass that you're used; the atomic weight, it's 15.9994, and yet oxygen is a mixture of oxygen 16, oxygen 17, and oxygen 18. Oxygen 16 predominants at 99.76%. There's just a tiny smidgen of oxygen 17, .04% and just a little bit of O18, .20%. And, so the mass when you're thinking about mass spect is 15.99491 for oxygen. Alright, so we've gone through some common elements here. Let's take a moment to explore the implications of this. You can see why this is really valuable, why the concept of exact mass is really valuable. So, let's take two simple molecules. We'll take propane, CH3, CH2, CH3, and we'll take acid aldehyde; CH3, CH0. So, they're a nominal molecular weight of 42, of 44, pardon me. But, with a high resolution mass spectrometer, you can tell these molecules apart and more. So, let's take a look. If I have C12, H8; if I have C12, 3 and H1 8. In other words, the predominant isotopomer of propane, then we will see that it's exact mass is equal to 3 x 12.0000, as many zeros as I choose to write plus 8 x 1.00783, and when I tally that up, I get 44.0626. If we compare our propane to our acid aldehyde, C12, the main isotopomer is going to be C12 2, H1 4, O16, and the exact mass is equal to 2 x 12.0000, etc., plus 4.00783, plus 15.99491 is equal to 44.0262. So, the point is that there is a significant difference at the 100th place in mass between the two different molecules. That difference is 36.4 mmu, which actually is a lot, as far as high resolution mass spectrometry, and what's nice, is you can do more than just distinguish two molecules from each other. For a small molecule, you can have a computer give you all the molecular formulas that fit the exact mass. [ Pause ] So, for example if you go to John Greaves and give him a small molecule that you've synthesized, and you get a high resolution mass spect on it, and you get a particular exact mass, he can give you all the molecular formulas that fit within 5 or 10 millimass units. Or you can go to that website I mentioned and linked on the home page for the course or on the materials for the course, and you can go ahead and say I have an exact mass of 44.0262, and then you give it some parameters, and you say, give me all of the compounds that fit within five parts per million or within 10 parts per million; you don't want to give it too wide a range. You don't want to give half a molecular weight unit, or you'll get many compounds, and you say those compounds could have up to 10 carbons and up to, and you'd pull down from the menu say 40 hydrogens, and up to 5 oxygens, and up to 5 nitrogens, and you'd find out. And for a molecule this small, you'd find this is the only fit. Or you'd find maybe just a couple of fits, but then some of them would have absurd formulas like 5 nitrogens, 1 carbon, and 1 hydrogen. Alright. Thoughts or questions at this point. >> If you're doing positive ions, do you have to worry about adding the exact mass of sodium? >> Ah, beautiful question. Well, you're in ions. So, the question was, if you're doing positive ions, do you need to worry about adding the mass of sodium. So, of course, EI mass spect also gives you positive ions, but there, you don't add a sodium or a proton, but in the case of electrospray, you probably do have a sodium or a proton or a potassium on there, so absolutely, then you need to think about adding the exact mass of sodium or potassium or one more proton on there. And if you have a multiply charged species, say with two protons on there, so a mass to charge that's based on being dicad ion, then you'd add in two protons or two sodiums, or a sodium and one proton in. Great question. >> This is only if you're looking at the parent peak. >> This is only if you're looking at the parent peak. And with EI mass spect, sometimes, John Greaves will come back and say this is for a fragment, but it's a logical fragment of your molecule, where as with soft ionization you usually can get it. Now, typically in small molecule mass spectrometry, the journals basically say above a thousand molecular weight, don't even both to do high resolution mass spectrometry, because for small molecules like 300, you really will get unique formulas. But, for much bigger molecules, you may have many formulas that fit. Now, what's interesting is with ion cyclotron resonance techniques, you get about an order of magnitude higher precision, so I was reviewing a paper, and all of the exact masses fit to the last decimal point within a few 10ths of a millimass unit, and my first thought was, this guy must be cheating here. Somebody must be doing fraudulent research, and I called up John Greaves, and I said, what could be giving such precision. He said do they have an ion cyclotron residence mass spect, and I said they didn't list in it their materials and methods section, but I went to their website, and it's like, bam, sure enough they do. And, with that technique, because there you're spinning the ions around and the longer you spin them, the more precision you get, you can get that level of precision. >> Is that another name for an ion trap? >> That's a similar idea. Yeah, so ion cyclotron resonance was a detection method we didn't talk about, so an ion trap can go ahead and basically be used in mass spect, mass spect, where you'll first isolate the ions of a particular mass, and then you further collide them or fragment them. But an ICR is a detection technique. What you're doing is basically having a little cyclotron, a little magnetic field and a varying frequency, and only ions of a certain frequency will circulate in resonance, and so you tune your resonance, and you basically say okay because we're right on resonance, we're getting the ion at this mass. >> Was that in the US? >> This actually was in the UK. >> I had a class from a guy. It was like David [inaudible]. Do you know him? >> I have heard his name. >> He has a circular [inaudible] >> It's a pretty neat machine, and in fact, one of our professors who now retired to start his own company was actually developing ICR right here. He was on the third floor of this building. He had a superconducting magnet. One of my laboratories was on the fourth floor, and anytime I put a computer with a CRT monitor in that laboratory, the CRT monitor would get messed up by the magnetic field from his superconducting magnet. Making a better mass spectrometer or a new mass spectrometry technique becomes something that one does for a PHD thesis. So other thoughts or questions? These are good ones. Alright. So, the take home message is really simple. Think about the main isotopomer. Alright, let's now talk about the second concept, isotopic abundances. And again, it just takes a moment to wrap your head around this. It's nothing profound. If for example, I take methane, and I'm going to sketch out an EI mass spect of it, you'll see a line at 16, and then you'll see a second line at 17, and that line will be 1.1%, so if I give, I'm just going to give these, because I'll be drawing a number of these, I'll just give the relative intensities as 101.1. So, we call this peak and M+1 peak. And, of course, the M+1 peak comes because you have C12 and C13 in a hundred to 1.1 ratio. You have a miniscule amount of deuterium, but it's really negligible for almost anything, because you have hydrogen to deuterium in 100 to 0.016 ratio, so essentially, even if you had a molecule with a hundred hydrogens in it, they would only contribute 1.6% to the M+1 peak, but if you had a molecule with a hundred carbons in it, now your M+1 peak would be just about as big or maybe even a little bigger than your M peak. So, let's by comparison look at ethane CH3, CH3, so now you see a peak at 30, and that peaks relative intensity is 100. And you see a peak at 31, and it's relative intensity is 2.2, so that's your M+1 and its relative intensity is 2.2. So, that's your M+1 peak then has a probability of one C13. We'll just forget about heavy hydrogen, because it's essentially nil, but the probability of 1C13 in your molecule is approximately equal to; it's not exactly equal to, because this is not how you do statistics, but it's approximately equal to 2 x 1.1% or 2.2%. Now a miniscule number of your molecules have two C13's in them. So, your M+2 peak, you would expect to come from the probability of two C13's, which is going to be 1.1% of 1.1%, which is 0.012%. So, in other words for a two carbon molecule, you're not going to see an M +2 peak. If you went to a hundred carbon molecule, you'd have a significant probability of having two C13's. In other words, some molecules would have no C13's, some would have one, some would have two, some would even have three, and you'd get an isotope pattern reflecting it. Thoughts or comments at this point? >> So, the peak that's M+, is that always going to be the highest? >> No, okay, great question. The peak that you call M+ is going to be the lowest isotopomer peak. So, for example, if you had a compound with 200 carbons in it, you would see a pattern where you saw a peak, another peak, another peak, and I'm not giving you the exact pattern, but something like this, and this would still be your M+. This would be your M+1, M+2, and so forth. And of course, if you're dealing with a soft ionization technique, then your first peak is going to be M+H or M+ sodium, but basically the first isotopomer peak may not be the biggest in certain cases. Other thoughts or questions? Alright, let's take a look at some other elements with isotopes. [ Pause ] [ Writing ] So, we've already talked about carbon, about hydrogen, about nitrogen, about oxygen, all of these are common in organic compounds. Let's talk about silicon. Silicon consists of an isotopic mixture of silicon 28, silicon 29, silicon 30. The ratio is 100 to 5.10 to 3.35. Sulfur, another common element consists of a mixture of sulfur 32, sulfur 33, and sulfur 34 in 100 to 0.78 to 4.40. Now, what's interesting about sulfur and silicon, is if you look at them, they can whisper in your ear, because for a small molecule with just a few carbons in it, you're not going to see a significant M+2 peak. But if that molecule has a sulfur in it, or if it has a silicon in it, you'll be able to pick out a low M+2 peak and it will whisper in your ear, hey, there's something up here. Not all elements whisper, some of them scream [laughter]. Chlorine is one that screams. Chlorine consists of a mixture of chlorine 35 and chlorine 37 in a 100 the patient 32.5 ratio, so chlorine by having a very pronounced M+2 peak screams out at you, hey look at me. I've got a chlorine here. Bromine also screams at you with two isotopes, bromine 79 and bromine 81. In 100 to 98 ratio, you have a very substantial M+2 peak. Other common elements don't say anything to you. Fluorine, phosphorus, and iodine all have single isotopes. There are, of course, other isotopes of these elements. Fluorine has an isotope of fluorine 18, but it's radioactive, and is only generated transiently. You wouldn't find it in a naturally-occurring molecule. You would find it only if you put it as a radiolabel and immediately injected it into a patient, say for PET imaging. Phosphorus has phosphorus 32 and phosphorus 33, but those also are radioactive with short half lives, and are used just for imaging and medical and bioanalytical purposes. Ditto for iodine. Alright. [ Pause ] So, as I said, sulfur in SI gives small but noticeable M+2 peaks. [ Writing ] So, let's take a look. I'll make a little sketch here of a compound with a line at 64 of 100 relative intensity, a line at 65. I'll actually offset my drawing on this, and a line at 66. The line at 65 has 0.09. The line at 66 has 5.0. And, so the question is, this is a real sketch of a real EI mass spect, and so the question is what's the compound? [ Pause ] Okay, we have DMSO, what's the, so DMSO is going to 15, 15, sulfur and oxygen, so we're at 32, 16, 48, 30, that doesn't work out. Too big, and we'd expect our M+1 peak with two carbons to be a little bit bigger. Anyone else? [ Pause ] >> It's got a sulfur and silicon here. >> It's got a sulfur and a silicon. Oh. >> [inaudible]. Sulfur dioxide? >> Sulfur dioxide? So, what do people think? Mass works out right? 32, 16, 16, so that's 64. What about the isotope pattern. >> Pretty consistent. >> Pretty consistent. You have your sulfur giving rise to a little bit more here. Now I've given you the numbers based on the exact isotope pattern. Oxygen contributes a teeny, tiny smidgen of 017, and it contributes a teeny tiny more smidgen of 018. Right, we said that there's about 0.02%, 018, so that boosts us above the 4.4% for the sulfur. So, in reality, you probably, unless you were doing specific quantitative mass spect for isotopes, which is a whole field in itself, unless you were doing that, you'd probably just glance at this spectrum and say, hey, I see an M+2 peak, sums things up what fits that mass spect. I bet there's a sulfur or silicon, and then you'd be scratching your head and doing exactly what you did, and coming up with the formula. [ Pause ] >> Was that example of M+ [inaudible]. >> The M+1 would be from the little bit of sulfur 33 and the teeny tiny bit of 017, so I just erased the numbers, but it's 0.8% sulfur 33 and 0.04%, a miniscule amount of 017. Alright, chlorine compounds give a large M+2 peak. So, 3 to 1 ratio roughly for one chlorine. So, let me sketch out another mass spect, and we'll start here, and I'll try my best to give you the relative intensities and to represent them with some accuracy here. Alright, so 36 with 100, 37 with 4.1, 38 with 33, and 35 with 12 is the relative intensities. >> Can I make a comment? >> Yeah, who am I? Right, this is the game who am I? >> HCL. Got it. Now, explain it. [ Pause ] Alright, so let's start with big guy here. What's the big guy? >> CL >> HCL with chlorine 35 >> HCL with chlorine 35, and if you remember, I said that the way you generate a molecular ion; we'll talk more about fragmentation, but in the EI mass spect, the way you generate a molecular ion is to kick out an electron. Okay, what about this peak here. So, that's the 35 peak. That's the CL35 peak, so what about the peak at 38. That's the 37 one, so now, so far, so far so easy. Okay. Now, we've got this guy over here. CL what? CL35 what? >> Plus. >> CL35+. And again, to an organic chemist, it is much more disturbing than to a freshman to be writing these sorts of structures. We'll talk more about fragmentation, but with all the energy that's gone into the molecule, the bonds are vibrating with an electron knocked out. The bonds are weakened, and so the molecule flies apart, to give you H dot plus CL+, and you don't see H dot, because it has no charge. And then the peak at 37, CL37+. [ Pause ] So, of course, bromine compounds give very large M+2 peaks. [ Pause ] [ Writing ] So, you can say by ICM to M+2 is a 1:1 pattern for one bromine. Let's try one more example here. Again in EI mass spect. So, let's say we're at 94 with relative intensity of 100. You also have 96 with relative intensity of 98, and you've got a little guy at 95 with intensity of 1.1, and a little guy at 97, also with relative intensity of 1.1, give or take. Who am I? Methyl bromide. And of course, here your C12 BR79 isotopomer. Here's your C13, BR79 isotopomer. Here at 96 is your C12 BR81 isotopomer, and here at 97 is your C13 BR81 isotopomer. [ Pause ] So, if you have multiple CL's and BR's, you can have M+4, M+6, peaks, etc. So, for example, you will get different patterns if you have two chlorines in a molecule, you'll see a peak, and then another peak at M+2; that's about 2/3 of the height, and then a peak at 1.4, so I'll say M+2. We'll call this 9 to 6 to M+4. We'll call one. So much of spectral analysis is about pattern recognition. It's not about numerical analysis like a computer. It's about looking at something, and the human brain is fantastically good at picking out patterns. And, you'll look at something and you say, hey, this looks interesting. There's this really big M+2 peak that I wouldn't expect this really big M+4 peak, and then you think it's got to be some element with some predominant, with some substantial amount of isotopes. If you have two bromines, then you'll see a pattern that goes in about a 1:2 to 1 ratio, your M, your M+2, and your M+4. And so there's a caution here. The caution is that in polyhalogenated compounds, if M and M+2 or if M+2 was bigger than M, you might not even recognize M. So, I'll say caution, if M+2 is bigger than M, you might think it's M and get confused. [ Pause ] [ Writing ] And I'll just show you one more pattern. Three bromines would be a good example where you might look at this and say, oh, my goodness, what do I have here. And so, if you have three bromines in roughly a 1:3 to 3:1 ratio, you'll see M, M+2, M+4, and M+6. Going to take one moment to show you something from the homework, so just to get you to start to think about mass spect homework, though IR homework is probably more on your mind right now. [ Pause ] [ Background noise ] Oops. >> Thank you. >> You're welcome. This happens to have a problem from an old midterm exam, and it wasn't, it's a compound that was actually synthesized in my laboratory. It wasn't made to drive chem. 203 students crazy, but rather for purposes of a project where we hope to grow crystals of a compound for x-ray crystallography, and it's often advantageous to have bromines as a compound. And, so here's a mass spectrum. It's an ESI mass spect of a compound in which we have 168 carbon atoms, two bromine atoms, 38 nitrogen atoms, 42 oxygen atoms, and 4 sulfur atoms. And, this is the sort of pattern of isotopomers that might see. Now the first thing that one does in looking at this spectrum and saying something; something's interesting here. You look at this spectrum, and you say the lines aren't spaced by units. What's going on there? It's triply charged. The lines are spaced a third of a mass unit apart, 0.3 apart, and every now and then 0.4 apart. So, this is, this is a +3 charge, and assuming that no sodiums are on the molecule, assuming this is just charged by in the ESI, remember we said you can pick up protons or you can pick up sodiums, what does that call you about what we would call this species? How many protons is this picking up? Three. So, the species that we are seeing here is M+3H, 3+. Now we've talked about bromine. You can see that you're getting lots and lots of big peaks here. Bromine is going to be contributing to M+2, but tell me what this line over here, the second line is saying to you? The second line is bigger than the first line. What's that telling you in this 168 carbon compound? There's a lot of C13. By the time you get up to compounds that have about a hundred carbon atoms in this, the M+1 peak is going to as big or bigger as you get beyond a hundred, then the M peak, because the odds of having a molecule with; to put it another way, most molecules are going to have at least one C13. The odds of having a molecule with one C13 or two C13's is higher than the odds of having a molecule with no C13's. What other elements are going to be contributing, so you could have one C13. You can only have one unusual isotope for this one. You could have one C13. What other important contributors are there going to be to this 1239.5? Nitrogen and sulfur, and which will be a bigger contributor, nitrogen or sulfur? >> Sulfur? >> So nitrogen has, what did I say, 38 nitrogens, and so roughly, the percentage of molecules or the fraction of molecules with one N15 is going to be roughly 0.38% x 38. The percentage of molecules with one sulfur 33 is going to be what? [ Pause ] [ Background conversations ] So, nitrogen is actually going to be a bigger contributor to this peak. So, the biggest contributor is going to be 1C13. The next biggest contributor is going to be 1N15, and then down the line, you'll have maybe one sulfur 33. Alright, well, that will get you started on that last problem. It's all conceptual. It's all something that involves really exploring this idea of what exact masses and isotopic abundances are. [ Pause ] >> It's more like an anamar multiplex. >> [Laugher] And for those of you. Yes, James, it looks more like an anamar multiplex, and for those of you who go on. Absolutely fascinating patterns, from ruthenium compounds and osmium compounds and compounds with multiple rutheniums or multiple osmiums. ------------------------------f87a215109ba--
B2 peak sulfur ion molecule carbon pause Chem 203. Organic Spectroscopy. Lecture 05. Isotopic Masses, Abundances, and Mass Spectrometry 38 1 Cheng-Hong Liu posted on 2015/01/25 More Share Save Report Video vocabulary