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  • [ 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--

[ Silence ]

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