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Ch. 15 - Analytical Techniques: IR, NMR, Mass SpectWorksheetSee all chapters

# Mass Spect: Isotopes

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Sections
Purpose of Analytical Techniques
Infrared Spectroscopy
Infrared Spectroscopy Table
IR Spect: Drawing Spectra
IR Spect: Extra Practice
NMR Spectroscopy
1H NMR: Number of Signals
1H NMR: Q-Test
1H NMR: E/Z Diastereoisomerism
H NMR Table
1H NMR: Spin-Splitting (N + 1) Rule
1H NMR: Spin-Splitting Simple Tree Diagrams
1H NMR: Spin-Splitting Complex Tree Diagrams
1H NMR: Spin-Splitting Patterns
NMR Integration
NMR Practice
Carbon NMR
Structure Determination without Mass Spect
Mass Spectrometry
Mass Spect: Fragmentation
Mass Spect: Isotopes

Concept #1: The (M + 1) Peak

Transcript

The first equation says that if I want to try to calculate how tall my M+1 is going to be I'm going to need to multiply the number of carbons times the percentage 1.1, now you might be saying Johnny if the chances of having carbon 13 are 1.1 percent doesn't that mean that the M+1 is always going to be 1.1 percent of the original? And that's not true at all actually the M+1 peak gets bigger and bigger and bigger depending on the number of carbons we have in our structure and the reason guys is simple something like methane where it has only one carbon in it what are the chances of having it carbon 13 in methane? Exactly 1.1 percent so we would be multiplying 1 carbon times 1.1 and we would get an estimated height of the M+1 peak at 1.1 percent, makes sense but how about a molecule like decane, guys remember decking has 10 carbons in it, right? So doesn't it have a higher chance of having a carbon 13 in it? Sure because now any of those carbons could be a carbon 13 not just one of them it could be any of them and it's going to increase the weight for the whole molecule so that's why you have to multiply that 1.1 percentage by the total number of carbons in the molecule so in this case 10 times 1.1 guess how tall that peak is going to be? Well you do the math you could type it into your phone or your calculator that's fine it's going to be 11 percent and guys that's what seen and observed in the mass spectrum, for methane look how tiny that M+1 peak is for decane look at how much bigger it is, why is it so much bigger? I'm sorry I'm right in the way why is it so much bigger guys? Because now there's all those different chances basically what it's saying is that 11 out of 100 times one of your carbon is going to be a carbon 13 because you have so many different carbons there that the chances the probability of one of those carbons being heavier just increased by 10 versus the first one, making sense so far? Thankfully It's a super easy equation to use it's always going to be 1.1 which is the odds by number of carbons, OK? Now I do want to just say one thing for all of you guys that are reading the textbook line by line which is awesome by the way but I wanted to say that this is an approximation because there are other isotopes that will increase the weight for M+1 for example nitrogen has an nitrogen 15 isotope, Sulphur has an isotope, phosphorus has isotopes so what that means is that the bigger the molecule gets the less accurate this is if we're talking about like a 100 carbon molecule that has a bunch of oxygens and nitrogens and sulphurs this approximation probably isn't going to work very well but for a small molecule or pretty much any that you're going to get in your textbook that you're having to do this we're going to use reasonably sized molecules where this works reasonably well, awesome. So that is the end of the first equation now it's going to the second one which looks more confusing but it's actually not bad at all.

Because the M+1 peak is going to continue to grow the more carbons that we have in our structure couldn't we also go backwards and say well based on the height of the M+1 I can make a guess of how many carbons I have? And that's exactly what this equation lets us do so what the equation says is that you put your M+1 which is your small peak over your molecular ion which is the one that's scaled to 100 so you would say this is how tiny this is my tiny M+1 I'm putting that over 100 because that's my molecular ion, now your molecular ion isn't always 100 because sometimes your molecular ion will be smaller than your base peak but I'm just basically saying you're comparing them against each other you're saying M+1 over M then you multiply it by a 100 to bring up the percentage to like a 100 percent and then you divide by 1.1 which is the likelihood of finding a carbon and what this is going to do is it's going to give you the total number of carbons in your structure, again it's an approximation It's not perfect but it works really well so for this first one I filled I put some blanks in there so that we could do it together, OK? So what should be the number that I'm putting here? Well the top number should be the value of my and plus one which in this case we've just calculated is 1.1, right? Cool so it's 1.1 by the way it's also visible in my mass spectrum that it's 1.1 cool, what's the number that I should put at the bottom? Well the number I should put at the bottom is the height of the molecular ion which in this case the height of the molecular, it goes right up to 100 it's my base peak so I'm going to put 100, OK? But again your molecular ion isn't always a 100 sometimes it could be 30 or it could be 50 depending on what it where it is compared to the base peak. Great so guys just a quick glance at this shows you that all the numbers are going to cancel out, right? 1.1 cancels out with 1.1, 100 with 100 so this is just going to basically tell me that there's one carbon present which is exactly right, right? We know that methane has 1 carbon so it seems like a big exercise just to find that number but it's going to get more complicated so this is going to be a very hopeful equation when the molecules get bigger and when your peaks are different sizes so let's move on to the next one what should be the number that I put in my N+1 at the top? I should put 11 because 11 is the height of my M+1, what should I put at the bottom? Once again, my molecular ion goes all the way up to 100 so I'm going to put 100, OK? And guys this time I'm actually going to use my calculator so just to make sure everyone's following I'm going to do 11 divided by a 100 that's given me 0.11, OK? Now I'm going to multiply that by 100 so I'm going to say 0.11 times 100 is giving me 11 so now I'm at eleven because I just multiplied by a 100 and now I have to divide by 1.1, divided by 1.1 and the number I got was 10 so guys I just use this equation to determine that I have 10 carbons is that the correct number? Totally we nailed it. OK So again this isn't always going to give you like the perfect number sometimes it will give you 9.1 or 10.1 but then you would round you would round to the nearest whole number and you would say that's the number of carbons I likely have in my structure, cool? Awesome guys so that's it for the M+1 peak now let's move onto the M+2 peak.

Concept #2: The (M + 2) Peak

Transcript

Concept #3: The Nitrogen Rule

Transcript

Guys, the nitrogen rule actually has nothing to do with isotopes, I'm just throwing it in here, because it's another very helpful form of structure termination for mass spec and it's usually taught in the same area as isotopes. So, we might as well just learn it here. So, guys nitrogen has a property that's different from carbon that makes it helpful for mass spec, which is that carbon always likes to form four bonds, right? So, since the iso form four bonds, if a molecule's made just out of carbon or hydrogen, carbon hydrogen, you know that it's going to be an even number for the molecular weight because you always have those four bonds. So, it's always going to be CH4 or C2H6, there's always multiples of like two or four but nitrogen likes to form three bonds, meaning that the molecular weight could be odd or it could be even and that's going to tell us how many nitrogens are present. So, what the nitrogen rule says is that an even molecular weight of our parent ion, basically the molecular ion, right? indicates an even number of nitrogen present, okay? Whereas an odd molecular weight of a parent ion indicates an odd number of nitrogen's present.

Practice: Propose the number of carbons for a compound that exhibits the following peak in its mass spectrum:

(M)+• at m/z = 72, relative height = 38.3% of base peak

(M+1)+• at m/z = 73, relative height = 1.7% of base peak

Practice: Predict the approximate height of the (M + 1) peak for the molecule icosane, molecular formula C20H42.

Concept #4: Draw expected isotope pattern Intro

Transcript

Draw the expected isotope pattern that would be observed in a mass spectrum of CH2Br2, in other words, predict the relatives heights of the peaks at, M, which is molecular anion, so I'm just going to put plus radical, M plus 2 and the M plus 4 peaks. Now, this one could definitely catch you off guard because, where did this come from? I've never talked about M plus 4, I have only talked about M plus 1 and M plus 2 but guys it makes sense that notice we have multiple halogens in this case, we have two bromines, so where does the M plus 4 come from? well, that would be the version of the molecule where both of the bromines are bromine 81, are the bromine that's the highest isotope, if both of them are giving us M plus 2 peaks then it's going to add to a M plus 4, so what we are looking for is ratio, okay? so I'm actually going to just coach you through this so you can try one more time on your own and then I'm going to tell you the answer.

And for me to coach you what I would say is, think about probability, what are the chances that both of the bromines are the small isotope, the 79, what are the chances that one is 79, one is 81, so that'd be M plus 2 and what are the chances that both of them are 81, and just so you know, for any question involving two halogens on the same atom we can basically use a punnet square system like what you'd have use in genetics, where genetics is used, in genetics you talk about phenotypes and genotypes so you'd say like, these are the chances of having this genotype, what's the genotype going to be, well, it's the same principle but for bromines because we're talking about probability, so you would say, what are the chances of having these bromines together versus these, so I'm going to let you try to arrange the chemistry punnet square the way you think is best and then I'm going to teach you how to do it, it's very easy, just have to learn how to do it once and then you'll know it forever, hopefully, so go ahead and try to do it and I'll give you the answer.