Concept: Ionization Potentials5m
Hey guys. In this video we're going to discuss common fragmentation patterns found within mass spectrometry. So, basically we're going to discuss the step of ionization and everything that follows after that. So, before we can really understand why some fragments are going to be more favored than others we need to grasp this concept called ionization potential and what ionization potential tells us is how likely an electron is to be knocked off by this electron beam, it turns out guys that not all electrons are made equal, some of them are going to be held very tightly by molecules. So, they're more difficult to knock off and some of them are going to be held very loosely so that means they're the first ones to go when they hit this electron beam, these patterns, these ionization potentials can help us help us to predict what's going to be the major cation that's formed or what's going to is a major radical cations that form after ionization has taken place. So, guys here are peas made a very simple trend that I'd look for you guys to memorize and all it is is this, that basically one of the easiest ones to knock off is the lone pair of a nitrogen and guys this is for the same reason that we've kind of always thought of the nitrogen is having a very reactive lone pair, it's very loosely held, it's very easy to ionize that lone pair, okay? So, if you have a nitrogen with a lone pair on your sample that lone pair is the most susceptible to getting knocked off, one of those electrons is likely you're going to get knocked off during ionization, during the ionization stage, okay? Now, in terms of this kind of spectrum, we're just going to kind of go down one by one the different types of compounds that get a little bit harder to ionize with every step, so the next one would actually be aromatics, okay? So, aromatics, these are the general category of benzene and benzene like molecules, okay? And it turns out that any of the single bonds directly attached to an aromatic are actually relatively easy to ionize, okay? So, this is going to be something else that we're going to see, aromatics tend to give a radical cation that doesn't involve breaking the ring that ring is very stable. So, you want to keep that ring intact, we wind up just breaking off one of the ends one of the single bonds attached to it that's coming off of it.
So, in terms of ionization, we wouldn't ionize the actual bonds of the ring, we would ionize maybe one of the hydrogen's or something that's attached to the benzene ring, for similar reasons double bonds come next. So, basically, this is what we call vinyl, this is a vinyl position, a vinyl position means to rectly special bond, it's a little bit harder than benzene but it's still not that bad, we would prefer to knock off a vinyl position. Something that's on a double bond rather than something that's not on a double bond. So, if you're a directly to a double bond that's the one that you are likely going to ionize, then we get some lone pairs attached to oxygen, for the similar reasons as nitrogen just less reactive, less acceptable. So, this one would have a higher ionization potential and then finally the hardest one guys, kind of the ion of last resort would simply be a single bond that's attached to other single bonds, okay? So, in this case. Notice that I'm not attached to a double bond, I'm not attached to a ring, I'm not attached to a benzene, I'm just attached to something that is an alkane, okay? So, I'm just going to put here an alkane. Now, you might be wondering, Johnny does it matter that it's a ring? no, I'm just using rings here to keep everything consistent because what I'm trying to show you is that it's not the ring that matters it's really the stability of that ring, whether it's a benzene whether it's a double bond or whether it's just an alkane. So, an alkane similar to the methane that we used in our intro video, that would be one of the more difficult ones to ionize because there's really nothing, there's nothing helping those radicals to get loose, that's just going to take brute force to remove one of those electrons and make the radical cation, okay? There's no extra stabilizing factors for an alkane, alright? Awesome guys, so that's ionization potentials. Now, let's move on to simple fragmentation mechanisms.
Concept: Simple Fragmentation Mechanisms13m
So guys for the purposes of talking about fragmentation let's keep working with methane for a second since it's the one that we used in our intro and recall that the radical cation or the molecular ion that wasn't the only peak on the mass spectrum remember that there were other peaks present so how do we get those other peaks? Well guys that happens because a lot of times your radical cation is going to fragment into more stable ions, OK? So it's going to basically fall apart and deconstruct itself blow up in a sense so that it can become more stable and even though there's a lot of different ways that let's say a molecule like methane could break apart there is one overarching theme which says that the cation fragment usually determines the relative amounts of the fragments found in the sample, OK? So that means that every radical cation or molecular ion can separate into a radical and a cation and to determine which side is going to get the cation you would think about carbocation stability similar to the carbocation stability we've always used in organic chemistry so let me show you guys exactly what I mean, here is our molecular ion that I'm bringing over from our intro video, right? The same exact molecule we know this has an MZ equal to 16, OK? But it turns out that there's 2 different ways that this thing can fragment if it wants to, it could just stay as the radical cation which means it hasn't fragmented yet but it could also choose to fragment, now the way fragmentation would work is that right now there's a radical in between 2 atoms but that radical would choose to go to one or the other so it would either choose to go to the H or it would either go to the carbon let's see what happens if the radical goes to the H, well you would draw an arrow that has basically a fish hook arrow because only 1 electron is moving and what that means is that you would get an H radical because the radical moved to the H and now my carbon is missing an electron so you would get a cation fragment, OK? Now notice that not both of these are going to be detected by the mass spectrometer, right? We said what types of Ions are detected by mass spectrometry? Only positive charges so that means that the M to Z ratio that I'm going to see for this molecule is going to be 15, it's going to be 15 because we lost a hydrogen this one is not observed in my mass spectrum because it's not positively charged cool awesome but it turns out that there is another alternative mechanism that could have happened which is that instead of going to the H what happens if the radical goes to the carbon instead? Totally fair this could also happen, well I'm still going to get a radical and a cation separately now because it fragmented I'm going to get my radical that has basically it's a methyl radical and I'm going to get a positively charged H so I means that now my M to Z because it's only detecting the positive charge is going to be equal to 1, OK? So these are two possibilities that could happen for a single if a single electron is removed and if a single arrow is used for the radical in terms of the radical mechanism. Now do you think that both of these fragments both the M to Z 15 and the M Z of 1 are going to be of equal abundance, do you think it's going to be like a 50 50 ratio? Guys not at all it turns out that one of them is going to be very very common, OK? The MZ of 15 is very very common and the other one of the H of MZ 1 is almost not observed at all so why is that guys? Well it has to do with the fact that a positive charge and a carbon is going to be more stable than a positive charge and a hydrogen so this has to do with in terms of predicting fragments we would always predict the fragment that's going to give the more stable carbocation, alright? Now by the way just to point out, our methyl carbocation is very stable? Not at all this is not the most stable carbocation ever but it's better than a hydrogen, OK? You may also recall that some of your professors are still in some of your homework were supposed to avoid methyl carbocations and primary carbocations because they're not the best but guys keep in mind this is happening for very short periods of time this is happening on the level of like nanoseconds or even less so what that means that it is possible to get these carbocations because they're so it's such a fast process that by the time it hits the detector it's already gone, OK? So just just.... I know that it doesn't really jive with what you've learned about carbocations before but keep in mind this is all very extreme process that we're putting it under and it's over within a very very short period of time.
So guys now that we understand how to kind of predict which of the fragments is going to be more abundant let's talk about common splitting fragments that are seen on different molecules and what I'm going to be doing is I'm going to be showing you not the common radicals......I mean sorry not the common carbocations but the common radicals because it turns out that remember that we said that if you lose a hydrogen that's called an M-1, OK? Well there are other radicals that are formed that you see very commonly and these would be nicknames according to the a molecular weight that you're losing so for example if a CH3 gets a radical and just chops right off of your molecule this is what we call an M-15 peak because that means that whatever your M is I don't know what your M could be due to whatever size it is let's say it's a huge molecule it's very likely that if there's a methyl group present you're going to get M-15, M-15 meaning whatever your molecular weight is subtracted 15 from that and there's likely going to be a peak there, OK? We could say the same thing OH for example, OH remember that we said that it's very easy to take the radical off of the O so it's also very common to get an O on M-17 if an OH is present, by the way the way we get 17 is that the oxygen is 16 and hydrogen is 1, alright? And you're going to see that like these other splits that I'm showing you just have to do with basic arithmetic adding up the atoms and these are very common so for example we talked about a methyl group what about an ethyl group? An Ethyl group would be -29 so that M-29 is also very common because of the fact that it's very easy to lose an ethyl group so guys these aren't here for you to memorize as much as just be familiar with I want you to come away from this video saying, "Hey I have a pretty good idea of what types of fragments are more common and which types are less common" Alright? And you can just keep going we could lose a methoxy group, a chlorine, an ethoxy and a bromine these are very very common these are ones you should just be on the lookout for, OK? Now there is one that kind of stands out that I just need to point out really quick which is water, OK? This H2O, why does it stand out? Because notice that it's the only one I didn't draw with a radical, did I make a mistake? Did I forget to put the radical there? No it turns out guys that due to a very interesting mechanism that can occur this is not a simple mechanism there's a more complicated mechanism that involves several arrows but it turns out that water can be lost all on its own without a radical and this one would be due would be on M-18, so it's important for you guys to know that if you have an alcohol present on your compound it could either be lost as an M-17 which means that the OH gets chopped off and that's it or the alcohol could be lost as a water because on its way out it's going to grab one of the hydrogen and go with it so just keep in mind you don't have to draw that mechanism right now but just keep in mind that you would actually see both an M-17 and an M-18 if an alcohol is present.
Awesome guy so I'm just going to end off with the fragmentation of butane, this is the actual mass spectrum and I want to show you guys how it relates to these common splitting fragments that we're talking about so just so you guys know the M to Z ratio of my molecular ion should be 58 according to the molecular weight of butane but guys look when we look at our mass spectrum and I'm going to take myself out of the screen in a second but I want to point out how tall is my 58? Tiny, remember that I told you that your molecular ion isn't always your base peak in this case the base peak is actually one of the fragments so we're going to look into that but just keep in mind that your M your molecular ion is tiny because it's very rare to just knock off an electron here usually it's going to fragment because the fragments are more stable. So I'll go ahead and take myself out of the screen now and we'll keep talking so guys noticed that what is my base peak? My base peak has a M to Z of 43, how does that make sense? Why do you think my base peak would have an M to Z of 43? Well do you notice any common splits that could make that? Totally guys, the reason that my base peak is 43 is because that's the peak that forms after I lose one of methyl groups so after I chop through this molecule and lose a methyl group what I'm going to get is this cation, OK? And this is carbocation happens to be more likely than the one that forms without it fragmenting might say Johnny why they're both primary? And in that case, I'm going to bring myself back really quick, you'll say Johnny but they both look primary so what's the big deal? Well just think about it guys this is a very high energy process so all this is saying is that it's very likely that as you pass an electron beam through this molecule that you're going to break off a methyl, it's not saying that it's more stable it's just way more likely that it's just going to either the right side is going to fall off or the left side is going to fall off it's very difficult to keep this molecule all in one piece, OK? By the way, this is going to be a common trend where the bigger the molecule is the smaller than molecular ion it's going to be and the reason this could be very difficult to keep it together, OK? If you have a huge huge molecule and you run an electron beam through it you're going to fragment that thing you're going to shatter it into pieces whereas if it's a small molecule it's more likely to stay together, OK? So I just want to point out how that's a very common fragment that's the base peak now let's look at the next one I'll take myself out of the screen again so another really common peak another really big one that was 43 another with really big one happens at 29 why do you think something happens at 29? Well guys that happens to be the fragment that forms when you lose an ethyl group so when you lose an ethyl group now you get a cation that looks like this, again is that cation way more stable? Probably not a lot more stable but it's just very likely that it forms because it's very difficult to keep this thing all in one piece when you're running that many electrons through it like railroading it's going to break apart, OK? And guys there's a whole lot of other peaks here that I'm not going to explain to you and that's because it's not important you don't need to know how the 50 or how of the 27 formed or the 26 you can just imagine that means that I lost 2 hydrogens or you know 2 methyls and a hydrogen whatever the biggest point here is that I want you to figure out and I want you to be able to understand how to determine when you would have a common fragment like O this makes sense that I would have a very high peak here because that's losing a methyl or that's losing a water or that's losing an ethoxy something like that, cool? Awesome guys so we're done with splitting fragments let's go ahead and do some practice so that we can solidify everything that we learn on this page.
Concept: Clarification on the Base Peak2m
Really quick, I want to make a clarification before you guys start typing this into the question box, a few of you guys might be saying, Johnny I get why 43 is a big peak and I get why 29 is a big peak, it has to do with those fragments that we spoke about, but how would you predict ahead of time that 43 is going to be bigger than 29, and the answer is that question is you wouldn't, it takes very complicated math and physics to model what happens inside one of these machines and really the only way to know the size of these exact Peaks is to run it through the mass spectrometer, so the point of this exercise wasn't that I expect you to know that 43 is going to be the biggest, I just want you to know that 43 is going to be big and also 29 should be big as well because both of these have very common fragments that are going to, radicals that are going to come off and then make cations of this size, okay? Now, there may be some practice questions in your textbook where it's very obvious that a certain fragment should form, for example, if you can form a tertiary carbocation then, you know, maybe that has a very high chance of being a common fragment. So, then you should be able to predict that, but in this case all of these carbocations were of equal, similar stability, so there's no way that you could predict that 43 was going to be taller than 29 except just to know that both of them are going to be highly present inside of your mass spectrum. Awesome. So, let's go ahead and flip the page.
Problem: Draw the most likely ion fragment of the molecule5m
Problem: Draw the most likely ion fragment of the molecule2m
Problem: What would be the value of the base peak (m/z)?5m
(a) Show how fragmentation occurs to give the base peak at m/z 58 in the mass spectrum of ethyl propyl amine (N-ethylpropan-1- amine), shown below. (b) Show how a similar cleavage in the ethyl group gives an ion of / z2.(c) Explain why the peak at m/z 72 is much weaker than the one at m/z 58.
Use equations to show the fragmentation leading to each numbered peak in the mass spectrum of octan-2-one.
Propose a fragmentation to account for each numbered peak in the mass spectrum of n-butyl isopropyl ether.
The mass spectrum of n-octane shows a prominent molecular ion peak (m z 114). There is also a large peak at m z 57, but it is not the base peak. The mass spectrum of 3,4-dimethylhexane shows a smaller molecular ion, and the peak at mass 57 is the base peak. Explain these trends in abundance of the molecular ions and the ions at mass 57, and predict the intensities of the peaks at masses 57 and 114 in the spectrum of 2,2,3,3-tetramethylbutane.
Predict the masses and the structures of the most abundant fragments observed in the mass spectra of the following compounds.
(e) cyclohexyl isopropyl ether [ cyclohexyl-O-CH(CH3)2]
(f) CH3CH2CH2NHC(CH3)3 tert-butyl propyl amine
Account for the peaks at m>z 87, 111, and 126 in the mass spectrum of 2,6-dimethylheptan-4-ol.
Problem-solving Hint In general, you should be able to propose favorable fragmentations for two or three of the largest peaks in a spectrum. Also, the spectrum should contain large peaks corresponding to the most favorable fragmentations of your proposed structure. You shouldn’t expect to account for all the peaks, however.
Ethers are not easily differentiated by their infrared spectra, but they tend to form predictable fragments in the mass spectrum. The following compounds give similar but distinctive mass spectra.
Both compounds give prominent peaks at m z 116, 73, 57, and 43. But one compound gives a distinctive strong peak at 87, and the other compound gives a strong peak at 101. Determine which compound gives the peak at 87 and which one gives the peak at 101. Propose fragmentations to account for the ions at m>z 116, 101, 87, and 73.
Show the fragmentations that give rise to the peaks at m z43, 57, and 85 in the mass spectrum
of 2,4-dimethylpentane (Figure 12-16).
Show the fragmentations that give rise to the peaks at m z43, 57, and 85 in the mass spectrum
of 2,4-dimethylpentane (Figure 12-16).
Show the fragmentation that accounts for the cation at m z 57 in the mass spectrum of 2-methylpentane. Explain why this ion is less abundant than those at m>z 71 and 43.
If a compound has a molecular ion with an odd-numbered mass, then the compound contains an odd number of nitrogen atoms. This is known as the nitrogen rule.
a. Calculate the m/z value for the molecular ion of each of the following compounds:
b. Explain why the nitrogen rule holds.
c. State the rule in terms of a molecular ion with an even-numbered mass.
What is the likeliest m/z value for the base peak in the mass spectrum of 3-methylpentane?
What would distinguish the mass spectrum of 2,2-dimethylpropane from the mass spectra of pentane and 2-methylbutane?
The mass spectrum of ethyl methanoate is shown below. Identify the ions responsible for the following peaks:
Which is the case when a molecular ion in the mass spectrum is also the base peak?
a) The original compound likely contained halogen
b) Fragmentation of M+ is favored.
c) Fragmentation of M+ is not favored.
d) McLafferty rearrangement has occurred.
e) none of the above.
Which compound would show major fragment ions at m/z 73 and 59?
The electron impact ionization (EI) mass spectrum for 2-propanol is shown below. Write Lewis structures for the species that give rise to peaks at m/z 45 and m/z 43. Make sure to show all C, H, O atoms, charges, lone pairs, and unpaired electrons.