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
|Ch. 1 - A Review of General Chemistry||4hrs & 47mins||0% complete||WorksheetStart|
|Ch. 2 - Molecular Representations||1hr & 12mins||0% complete||WorksheetStart|
|Ch. 3 - Acids and Bases||2hrs & 45mins||0% complete||WorksheetStart|
|Ch. 4 - Alkanes and Cycloalkanes||4hrs & 18mins||0% complete||WorksheetStart|
|Ch. 5 - Chirality||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 6 - Thermodynamics and Kinetics||1hr & 19mins||0% complete||WorksheetStart|
|Ch. 7 - Substitution Reactions||1hr & 46mins||0% complete||WorksheetStart|
|Ch. 8 - Elimination Reactions||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 9 - Alkenes and Alkynes||2hrs & 10mins||0% complete||WorksheetStart|
|Ch. 10 - Addition Reactions||3hrs & 33mins||0% complete||WorksheetStart|
|Ch. 11 - Radical Reactions||1hr & 57mins||0% complete||WorksheetStart|
|Ch. 12 - Alcohols, Ethers, Epoxides and Thiols||2hrs & 34mins||0% complete||WorksheetStart|
|Ch. 13 - Alcohols and Carbonyl Compounds||2hrs & 14mins||0% complete||WorksheetStart|
|Ch. 14 - Synthetic Techniques||1hr & 28mins||0% complete||WorksheetStart|
|Ch. 15 - Analytical Techniques: IR, NMR, Mass Spect||7hrs & 18mins||0% complete||WorksheetStart|
|Ch. 16 - Conjugated Systems||5hrs & 49mins||0% complete||WorksheetStart|
|Ch. 17 - Aromaticity||2hrs & 24mins||0% complete||WorksheetStart|
|Ch. 18 - Reactions of Aromatics: EAS and Beyond||4hrs & 31mins||0% complete||WorksheetStart|
|Ch. 19 - Aldehydes and Ketones: Nucleophilic Addition||4hrs & 54mins||0% complete||WorksheetStart|
|Ch. 20 - Carboxylic Acid Derivatives: NAS||2hrs & 3mins||0% complete||WorksheetStart|
|Ch. 21 - Enolate Chemistry: Reactions at the Alpha-Carbon||1hr & 59mins||0% complete||WorksheetStart|
|Ch. 22 - Condensation Chemistry||2hrs & 13mins||0% complete||WorksheetStart|
|Ch. 23 - Amines||1hr & 43mins||0% complete||WorksheetStart|
|Ch. 24 - Carbohydrates||5hrs & 56mins||0% complete||WorksheetStart|
|Ch. 25 - Phenols||15mins||0% complete||WorksheetStart|
|Ch. 26 - Amino Acids, Peptides, and Proteins||2hrs & 54mins||0% complete||WorksheetStart|
|Purpose of Analytical Techniques||6 mins||0 completed|
|Infrared Spectroscopy||16 mins||0 completed|
|Infrared Spectroscopy Table||32 mins||0 completed|
|IR Spect: Drawing Spectra||41 mins||0 completed|
|IR Spect: Extra Practice||27 mins||0 completed|
|NMR Spectroscopy||10 mins||0 completed|
|1H NMR: Number of Signals||27 mins||0 completed|
|1H NMR: Q-Test||28 mins||0 completed|
|1H NMR: E/Z Diastereoisomerism||8 mins||0 completed|
|H NMR Table||24 mins||0 completed|
|1H NMR: Spin-Splitting (N + 1) Rule||25 mins||0 completed|
|1H NMR: Spin-Splitting Simple Tree Diagrams||11 mins||0 completed|
|1H NMR: Spin-Splitting Complex Tree Diagrams||8 mins||0 completed|
|1H NMR: Spin-Splitting Patterns||8 mins||0 completed|
|NMR Integration||17 mins||0 completed|
|NMR Practice||14 mins||0 completed|
|Carbon NMR||7 mins||0 completed|
|Structure Determination without Mass Spect||58 mins||0 completed|
|Mass Spectrometry||12 mins||0 completed|
|Mass Spect: Fragmentation||29 mins||0 completed|
|Mass Spect: Isotopes||32 mins||0 completed|
|IR Spect: Frequencies Considering Solution Effects|
|IR Spect: Structure Determination|
|1H NMR: Proton Exchange|
|1H NMR: Fast Proton Exchange (D2O)|
|13C NMR: Cumulative Practice|
|2D NMR - COSY|
|Mass Spect: McLafferty Rearrangement|
|Structure Determination with Mass Spect|
Concept #1: The (M + 1) Peak
In this video we're going to cover the role of isotopes in mass spectrometer, so before we go any further let's just all remember that an isotope would be an atom that has the same number of protons meaning it's the same element but it has a different number of neutrons so what that means is that isotopes are going to have different weights from each other depending on which isotope it is? Now why would that be important for mass spectrometry because guys we're analyzing weight so I need to know if there's atoms out there that are the same atom with different weights, I need to take that into consideration, right? So we're going to talk about a few different types of peaks that result purely because there's isotopes present. So the first one is the M+1 peak so we've talked a lot about M-1 M minus a lot of numbers but we haven't talked yet about the pluses, pluses happened because of isotopes, OK? And the M+1 peak is probably the most famous situation where this happens and it's due to the isotope of carbon 13, remember that carbon is usually 12 but it turns out that 1.1 percent of all the carbon in the universe in fact 1 percent of all the carbon that's making up your body has an extra neutron in it it's carbon 13 so what this is going to do is it's going to add a small very small because it's so just only 1 percent but distinctive M+1 peak that proportional in size to the number of carbons in your compound so remember in our intro video how we were looking at methane and methane had an M to Z of 16 but we saw this tiny little peak at 17 and I told you don't worry about it yet that is where it was coming from, it was coming from the carbon 13 isotope well guys it turns out that this isn't something that you just have to mentally know it's also something you need to make to do some calculations based on because it's such a consistent ratio that we can actually build equations to solve problems with this so the equations that we're going to cover today are one calculating the height of M+1 so sometimes you're going to be asked to estimate how tall would this M+1 be based on what the structure is and we're never....you know you're not expected to get it perfectly right because this is an approximation but we can get pretty darn close. The second equation we're going to cover is using the M+1 peak the height of it to go back and look at how many carbon did you originally have in our molecule? So let's go ahead and start with the first, OK?
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
Alright guys. So, why would we ever get an M plus 2 peak? Well, it turns out that there's two atoms that because of their isotopic ratios they happen to give very distinctive M plus 2 peaks and those two atoms are the halogens chlorine and bromine, okay? Now, why do they give M plus 2? because the way they naturally occur in the universe is in isotopes that have a difference of two in weight. So, let's solve for chlorine first. So, guys you might remember from, well, from your periodic table, whatever, that the atomic weights or the molecular weight of chlorine is actually 35.5 about 35.5, okay? But the reason that's 35.5 is because about 3/4 of it is chlorine 35 and about 1/4 of it is Corinne 37. So, about 75% 35 and about 25% 37. So, if you average those together you get 35.5, okay? Now, you don't want to do that math, it's fine, but the interesting thing is that on a mass spectrum this is going to give us a very interesting ratio where it's going to look like an approximate three to one ratio on a mass spectrum, so that means wherever your molecular ion is, if you add two to that molecular ion you're likely to see a peak, that's about one third of the size of the of the molecular ion. So, we're just going to do this molecule as an example, I'm not going to write mass-to-charge numbers because it doesn't matter, but let's say that this is your molecular ion here. So, I'm just going to put that this is my molecular ion, right? So, this is my molecular ion? Well, then I would expect to find an M plus 2 that is about a third of the size, this would be my M plus 2 and why is that? Well, once again because about 75% of my chlorine atoms were chlorine-35 and about 25% are giving it the extra plus 2 weight of 37.
Now, the reason that this is helpful for us guys is because, when you see that very distinctive three-to-one ratio, that's M plus 2 you automatically know that there's a chlorine in your molecule. So, it's like, it's a really easy read it's like, Oh, I know there's a chlorine here because I have this pattern and this pattern only happens with Chlorine awesome, so that's chlorine and it turns out bromine also has an M plus 2 but is different because the isotopic abundance is different, it turns out that bromine on the periodic table has a weight of about 80, okay? But it's not really 80, what it really is, is about 50% is 79, and about 50% is 40, about 50% is 81, I'm sorry. So, basically about half of it is 79, about half of it is 81. So, they blend that together to get the number 8. So, what this means is that for bromine we're going to get an M plus 2 about a one to one ratio. Now, I do see that the percentages are a little bit off, it's like 50.7, 49.3, but guys those are so close. So, we're just going to call it one to one, I hope you're okay with that. So again, I'm just going to write that, let's say this is my molecular ion here, right? So, this would be my molecular ion, what I would expect is if there's a bromine present then I should have an M plus 2 that is about the same exact height and it's two units over and this would be due to the bromine. So, this would be my M plus 2 and I know that there's a bromine there because that means that about half of it was bromine 79, and about half of it was bromine 81, okay? Now, really quickly, I know that I drew them differently. So, actually feel weird about that, I'm going to change it, I hope you brought your eraser because what I'm trying to show is that the 81 increased by two units, that's why I'm putting M plus 2, this is your M plus 1 that I'm not really drawing that part, but over here we should move this peak to be right here, okay? So, you can see how it's two units shifted, not just one, okay? What I'm trying to show you is that your M plus 2 should be about a third of the size of chlorine and it should be about the same exact size with bromine. So, if you see this pattern you know bromine is present, if you see that pattern you know chlorine is present, and by the way, I'm ignoring the M plus 1, I'm not even worrying about that because this, I'm not talking about M plus 1 right now, I'm just talking one M plus 2, would there be some M plus 1? sure but we are not talking about that right now, cool? Awesome. So, now we're done with M plus 2, the biggest thing is that you know how to identify i,t whether it's chlorine or bromine, let's move on to the nitrogen rule.
Concept #3: The Nitrogen Rule
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.
So, for example, the number one odd number being one, right? If there's one nitrogen present that means that nitrogen wants to have three bonds, meaning that when I file together my molecular weight I'm going to always have to add three at the end or some odd number of bonds. So, my entire molecular weight is that is likely to be odd, if there's one nitrogen present whereas if there's an even number. So, for example, two bonds, well, now you got two nitrogens, I'm sorry, two nitrogen's, those two nitrogen's like to form three bonds each. So, 3 plus 3 equals 6. So, it's an even number again. So, one of the easiest ways to look at this is just these two molecules, we've got butane, which we saw earlier has an mc of 58 but if you take the butane and you replace one of the carbons with an NH2 what's going to happen is that your molecular weight now turns odd because this nitrogen likes to form three bonds. So, what's happening is that now you're just adding one, you're adding a molecular weight of one to it, okay? Now, if you're saying, Johnny, how does that math work? Well, first of all, you can calculate yourself, you'll see that I'm right. What's happening guys is that carbon weighs 12, right? Nitrogen weighs 14. So, by turning one of the carbons into nitrogen, how much molecular weight should I be adding? 2, I should be adding 2, but then there's a difference carbon likes to have how many H's? it likes to have four, right? Nitrogen like says how many H's? three, right? So, what that means is that when you add it all together, every carbon you add should be adding something around 16, if you were to do a CH 4 but every nitrogen you add should be adding about 17 because it's a little bit heavier but it likes to have less hydrogen. So, basically, that's the gist of it but what you should know is that if you have an odd number of molecular weight then it should be an odd number of nitrogen's and then if you have an even number then you have an even number of nitrogen's, even if that means 0, 0 is also an even number so it just could mean I don't have any nitrogen's presen. Awesome guys. So, let's go ahead and flip the page and do some practice.
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
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.
Practice: Draw the expected isotope pattern that would be observed in the mass spectrum of CH2Br2. In other words, predict the relative heights of the peaks at M, (M + 2), and (M + 4) peaks.
Enter your friends' email addresses to invite them: