Mass Spect:Fragmentation - Online Tutor, Practice Problems & Exam Prep

In mass spectrometry, ionization potential determines the ease of electron removal, influencing fragmentation patterns. Lone pairs on nitrogen are most easily ionized, followed by aromatics, double bonds, and lone pairs on oxygen, with alkanes being the hardest. Fragmentation often leads to more stable ions, with common peaks like M-15 (loss of CH₃) and M-29 (loss of C₂H₅). The base peak reflects the most stable fragment, while the molecular ion may be less prominent due to fragmentation. Understanding these concepts aids in predicting mass spectra outcomes.

Ionization Potentials

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. Before we can truly understand why some fragments are going to be more favored than others, we need to grasp this concept called ionization potential. Ionization potential tells us how likely an electron is to be knocked off by an 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 are more difficult to knock off. And some of them are going to be held very loosely, so that means they are the first ones to go when they hit this electron beam. These patterns, these ionization potentials can help us to predict what is going to be the major cation that is formed or what is going to be the major radical cation that is formed after ionization has taken place.

Guys, here I've just made a very simple trend that I'd love for you guys to memorize. And all it 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 as having a very reactive lone pair. It's very loosely held. It's very easy to ionize that lone pair. So if you have a nitrogen with a lone pair on your sample, that lone pair is the most susceptible.

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. The next one would actually be aromatics. Aromatics, these are the general category of benzene and benzene-like molecules. And it turns out that any of the single bonds directly attached to an aromatic are actually relatively easy to ionize. Aromatics tend to give a radical cation that doesn't involve breaking the ring because that ring is very stable. So we want to keep that ring intact. We end up just breaking off one of the ends, one of the single bonds attached to it that is coming off of it. In terms of ionization, we wouldn't ionize the actual bonds of the ring. We would ionize maybe one of the hydrogens or something that is attached to the benzene ring.

For similar reasons, double bonds come next. This is what we call vinyl. A vinyl position means directly attached to a double 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 is on a double bond rather than something that is not on a double bond. Then we get lone pairs attached to oxygen for similar reasons as nitrogen, just less reactive, less susceptible. So this one would have a higher ionization potential.

And finally, the hardest one, guys, the ion of last resort would simply be a single bond that is attached to other single bonds. In this case, notice that I'm not attached to a double bond. I'm not attached to a ring. I mean, I'm not attached to a benzene. I'm just attached to something that is 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. An alkane, similar to the methane that we used in our intro video, would be one of the more difficult ones to ionize because there's really 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. There's no extra stabilizing factors for an alkane.

All right. Awesome guys. So that's ionization potentials. Now let's move on to simple fragmentation mechanisms.

Simple Fragmentation Mechanisms

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. 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. So it's going to basically fall apart and deconstruct itself, blow up, in a sense, so that it can become more stable. Even though there are 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. So that means that every radical cation or molecular ion can separate into a radical and a cation. 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. The same exact molecule. We know that this has an m/z equal to 16. But it turns out that there are two different ways that 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 will work is that right now there's a radical in between two atoms, but that radical would choose to go to one or the other. So it would either 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 fishhook arrow because only one 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. Now notice that not both of these are going to be detected by the mass spectrometer. We said what types of ions are detected by mass spectrometry? Only positive charges. So that means that the m/z 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's 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 fragments it. 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 that means that now my m/z because it's only detecting the positive charge is going to be equal to 1. Okay? So these are the 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/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. The m/z 15 is very common, and the other one of m/z 1 is almost not observed at all. So why is that, guys? Well, it has to do with the fact that a positive charge on a carbon is going to be more stable than a positive charge on a hydrogen. So this has to do with in terms of predicting fragments, we would always choose the fragment that's going to give the more stable carbocation.

Now, by the way, just to point out, are methyl carbocations very stable? Not at all. This is not the most stable carbocation ever, but it's better than a hydrogen. You may also recall that some of your professors or 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 that this is happening for very short periods of time. This is happening on the level of nanoseconds or even less. So what that means is that it is possible to get these carbocations because they are such a fast process that by the time it hits the detector, it's already gone. So, I know that it doesn't really jive with what we've learned about carbocations before, but keep in mind, this is all a 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 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'm 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/z -1. Well, there are other radicals that are formed that you see very commonly, and these would be nicknamed according to the 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/z -15 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 an m/z -15. m/z 15 meaning whatever your molecular weight is, subtract 15 from that and there’s likely going to be a peak there. We could say the same thing of O, for example. O, 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 m/z -17 if an O is present. By the way, the way we get 17 is that oxygen is 16 and hydrogen is 1. All right. And you're going to see that 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 the m/z 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. 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 the ones that you should just be on the lookout for. Now there is one that kind of stands out that I just need to point out really quick, which is water. This H2O, why does it stand out? Because notice that it's the only one that 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. This is 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 woul

Clarification on the Base Peak

Really quick, I want to make a clarification before you guys start typing this into the question box. A few of you 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 to that question is you wouldn't. It takes very complicated math and physics to model what happens inside one of these machines. 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 form radicals that are going to come off and then make cations of this size. 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 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.

Draw the most likely ion fragment of the molecule

• A.
• B.
• C.
• D.

Draw the most likely ion fragment of the molecule

• A.
• B.
• C.
• D.

What would be the value of the base peak (m/z)? 

• A. 48
• B. 57
• C. 68
• D. 75

Here’s what students ask on this topic:

What is ionization potential in mass spectrometry?

Ionization potential in mass spectrometry refers to the energy required to remove an electron from a molecule, forming a radical cation. This process is crucial because it influences the fragmentation patterns observed in the mass spectrum. Molecules with lower ionization potentials, such as those with lone pairs on nitrogen, are more easily ionized. In contrast, molecules like alkanes, which have higher ionization potentials, are harder to ionize. Understanding ionization potential helps predict which fragments will be most abundant in the mass spectrum.

How do fragmentation patterns help in identifying compounds in mass spectrometry?

Fragmentation patterns in mass spectrometry help identify compounds by breaking the molecular ion into smaller, more stable ions. Each fragment corresponds to a specific mass-to-charge (m/z) ratio, creating a unique pattern or 'fingerprint' for the compound. By analyzing these patterns, chemists can deduce the structure of the original molecule. Common fragments, such as M-15 (loss of CH₃) and M-29 (loss of C₂H₅), provide clues about the molecular structure, aiding in compound identification.

Why is the base peak often more prominent than the molecular ion peak in a mass spectrum?

The base peak is often more prominent than the molecular ion peak in a mass spectrum because it represents the most stable and abundant fragment ion. During ionization, the molecular ion often fragments into smaller, more stable ions. These stable fragments are detected more frequently, resulting in a higher peak. The molecular ion peak, representing the intact ion, is less prominent because the molecule tends to fragment under the high-energy conditions of the mass spectrometer.

What are common fragmentation patterns observed in mass spectrometry?

Common fragmentation patterns in mass spectrometry include the loss of small, stable groups from the molecular ion. Examples include M-15 (loss of a methyl group, CH₃), M-17 (loss of a hydroxyl group, OH), and M-29 (loss of an ethyl group, C₂H₅). These patterns are predictable based on the stability of the resulting ions. For instance, the loss of a methyl group is common because it forms a relatively stable carbocation.

How does ionization potential affect fragmentation in mass spectrometry?

Ionization potential affects fragmentation in mass spectrometry by determining how easily a molecule can lose an electron to form a radical cation. Molecules with lower ionization potentials, such as those with lone pairs on nitrogen, ionize more easily and fragment more readily. This leads to specific, predictable fragmentation patterns. Conversely, molecules with higher ionization potentials, like alkanes, are harder to ionize and may show less fragmentation or require more energy to do so.