in this video we're going to focus on mass spectroscopy so here we have the mass spectrum of pentane and just to give you a little background information on how this technology works so basically you would take a sample like this one pentane and put it in a mass spectrometer and what's going to happen is that sample is going to be vaporized and then it's going to be ionized using an electron beam now when it ionizes it can fragment into radicals and cations but only the positively charged ions will make it through the analyzer and reach the detector in this graph on the y-axis we have the relative abundance of the fragments so this represents the quantity of the positively charged fragments that reach the detector and on the x axis we have the mass of charge ratio now let's say if we get a methyl cation typically the charge will be plus one so if z is one then m over z becomes the mass of the fragment so what we want to do in this example is we want to determine what fragments will form and which of these numbers these fragments correspond to if you understand this then you could use this to identify which molecule corresponds to which graph but for now let's identify the fragments that form in pentane if we put it in a mass spectrometer so let's call this carbon one two three four five now if we break the c1 c2 bond it can split up into a ch3 group and a butyl group now typically one is going to be a radical and the other one is going to have a positive charge now carbon has an atomic weight of 12 and hydrogen is about one so ch3 is 15 ch2 has an atomic weight of 14 and if we add up 14 plus 14 plus 15 plus another 14 this will give us 57 so 57 could correspond to this particular fragment now we don't always have to get a methyl radical we could get a methyl cation if we do get a methyl cation then this molecule will be a radical so this is another possibility but it's important to understand that only the positively charged fragments will reach the detector and so the methyl cation could easily correspond to the peak that we see at 15. now what about the peak at 43 that peak is known as the base peak it has a relative abundance of 100 so everything is compared to the base peak the peak at 72 is known as the parent peak also called the m plus peak and it turns out that it's simply the mass of the compound if you add up 15 14 14 14 and 15 this will give you the total molecular weight of pentane which is about 72 grams per mole and so that corresponds to the m plus peak now if we break the c2c3 bond we can get these two fragments we can get an ethyl radical or an ethyl cation but i'm only going to focus on the cations at this point or we can get a propyl radical or even a propyl cation now anytime you see a ch2 group it has a weight of 15 and the ch2 group has a weight of 14. so the ethyl cation has a weight of 29 so that corresponds to the peak that we see here as for the propyl cation if you add 14 14 and 15 that will give you a total of 43 or you could do 72 minus 29 and that'll give you 43. so the propyl cation corresponds to the base peak now here's a question for you why is the propyl cation the base peak and not the butyl cation let's compare both of them side by side if we break the c1 c2 bond we could get a methyl cation and we could get a primary butyl cation now which situation leads to the formation of more stable fragments in the first scenario where we break the c2 c3 bond notice that we have a primary ethyl cation and a primary propyl cation in the second situation where we break the c1c2 bond we have a primary butyl cation and a methyl carbocation so a methylcarbocation is less stable than a primary carbocation which means it's easier to break the c2 c3 bond as opposed to the c1 c2 bond because this will give us more stable fragments and that's why this is more abundant because it's easier to break the c2 c3 bond now it's important to understand that both of these cations can rearrange to a more stable secondary carbocation but the reason why this is the base peak is because it's easier to break the c2c3 bond due to the formation of more stable carbocation fragments this one is less stable so it's going to be harder to form and so that's why this corresponds to the base peak now let's clear away some of the stuff that we have on the board so at this point we've identified all the peaks so the peak at 15 corresponds to the methyl cation the pcat-29 corresponds to the ethyl cation the pcat-43 corresponds to the propyl cation and the pcat57 corresponds to the butyl cation at 72 it's the original compound that lost only one electron now what about the peak at 41 how can we get that so starting with the propyl cation this molecule can lose two hydrogen radicals and if it does so what's going to happen is we're going to get an allylic carbocation and this carbocation is stabilized by resonance and so that's why it's very abundant as well relative to the other fragments so it's relatively easy to form this particular carbocation it's primary and it's allylic so if this has a mass to charge ratio of 43 and if we subtract 2 from that then the mass of this particular fragment will be 41. you can add 14 ch is 13 plus another 14 that'll give you an m to z ratio of 41. now it's also important to understand how the primary carbocation could rearrange into a secondary one so let's start with the propyl cation a hydride shift can occur converting the primary carbocation into a secondary one so this is going to be ch and then ch3 with the plus charge on the secondary carbon and so the m2z ratio for this as well will be 43. so this carbocation can easily be arranged into this one as well so