So, guys, Now we're going to discuss what is basically the holy grail of our analytical technique section, and that is the skill of structure determination. So at some point this semester, you may be asked to produce a structure from scratch. That means draw out a structure from scratch using nothing but a combination of molecular formula and a Mars. And I are spectra. Okay, that means literally. All you have is a bunch of peaks and a bunch of spikes and stalactites, and you're supposed to actually turn that into a carbon structure. That is exactly the right structure. Okay, so when students see this type of question, they tend toe freak out because this is a very complex skill. We're having a synthesized tons of information. We're having to get creative, and students just start drawing every structure they can think of now for you guys. You guys are smart. You're already watching my videos. That means you're trying to get ahead, and I'm gonna tell you that is the first way to lose points on the exam because you're gonna run out of time. You're gonna draw a bunch of structures that aren't correct. And it's just not efficient. So we need to be much more strategic and how we draw these structures. That's why I always teach my students to build a strong molecular sentence before you even begin getting creative and drawing structures. The way we build a molecular sentence is by gathering clues, we're going to gather as many clothes as possible from the NMR from the I r from the molecular formula, we're gonna put it all together in a very ordered way. We're almost going to basically create, like, a mini essay on this molecule. And then from that sentence, we can then go ahead and propose structures that are actually relevant. Okay, this isn't gonna be easy, guys. This is actually a skill that takes a lot of work. And it's one of the harder things you may have to do this semester. But I promise you that by using this strategy, it's going to cut down on the learning curve big time. Okay, so let's go ahead and just talk about the steps to build a strong molecular sentence. The very first step is to determine the HD, which is a skill that we learned in organic chemistry. One and that I've included in this in this section, so you can review it. Okay, so the HD is just basically going to tell us about double bonds, rains, triple bonds, etcetera. Then we're gonna use the NMR, the IR splitting patterns and integrations. We're gonna look at all of that for major clues now. Specifically, I put all four of these things for a reason. The reason is because we tend to find extremely helpful clues with these four. Um, you know, with these four things. So, for example, NMR. What if I have a chemical shift in my NMR? That's like 9.1. I'm just giving you examples here, but there's a lot of different shifts that we learned. What have you learned? If you saw that you had a NMR shift of 9.1, what would you suspect about that molecule? Well, there's really only one functional group that results in the 9 to range, and that would be an alga hide. So immediately I would be suspecting. Is there an alga hide? Right. So now what if I look at my ir spectrum and there's also a peak there at 17. 10 then would that confirm my suspicion That I haven't Aldo Hide? Or would it deter me? It would confirm it because remember, Aldo hides have a carbon. Oh, peak at 17. 10. So then that would kind of confirm the alga hide suspicion. Right now, what if I look at my splitting patterns on? I noticed that if in my anymore, I actually have a triplet and a quartet present. Well, that's one of splitting patterns I told you guys to memorize, and that's very indicative of an ethyl group. So already I know that I'm looking for alga hides that have some kind of ethel group on them. That's big deal. Okay, now, what if I look at my integration? What kind of information do you get from integration? You get number of hydrogen. Why is that important? Well, what if I have that shift at 9.1 for my NMR, but it actually has an integration of two h. What would that tell me if my integration is to age for a shift of 9.1 That tells me that I actually don't just have one Aldo hide. I actually have to. Aldo hides. The reason is because every Alba hide on Lee has one hydrogen that results in the 9 to 10 range. So if I have to hydrogen, that must mean that I have to. Alba hides. So these the kinds of clues that we gather right away, okay. And you have to get good at learning where to find those clues. Okay, so now we've done that. Now we do something. That's kind of like a clutch prep special. This is something that you're not going to see in your textbook, but sometimes it's helpful. And that's to do something that I call calculating the NMR, the proton NMR peak or the proton NMR signal to carbon ratio. Okay. And basically, this is just a test of symmetry. So what I do is I say that you look at the number of signals that you have, and you put that over the number of carbons that you have. Okay, if that number turns out to be less than one half okay, then that suggests that it's a symmetrical compound. Whereas if that number tends to be above one half, then it's probably gonna be a pretty asymmetrical compound. Okay, um, the logic behind here being, Let's say that you have a molecule that's like C six h 14. Right? And you've got your proton NMR, right starts at zero. It ends at 13 and all you have is one peak. Let's just say you have one peak. Okay, Well, what that's going to suggest to me is that I only have one signal, so that's gonna be the one in my fraction. And I have six carbons. So that's going to tell me that Ah, lot of these hydrogen are exactly the same as each other. Actually, they're all exactly the same if I'm only getting one signal, Okay? The only way that they could all be the same as if the molecules symmetrical. So this must be a pretty symmetrical molecule. If it's giving me ratio, that's are a fraction that's so below one half. It's 16 instead of one half. Does that kind of makes sense? So basically, my fraction is just a measure of how symmetrical is this molecule? Probably, Or how asymmetrical is it? Okay, now, because of the fact that I literally made this up, it says something that I've used for many years. But it's also not you know it's not for sure. It's not like a you know, it's not like a tried and true methods. So what that means is that you can never just rule out of structure because of symmetry. I've had students that say, Oh, but that molecule doesn't look symmetrical, so that can't be it. Don't do that. This is more of a hint than anything else. It should serve as guidance, but it shouldn't serve as your final cut. Okay, for example, symmetry in straight chains can be very difficult to visualize. So I don't want you guys to just go based draw structures just based off of this rule. Just use it as a helping hint more than anything else. Okay, um, also just another point. It tends to be really helpful at the extremes and not very helpful in the middle. So if you take your carbon, you know, if you take your fraction and it happens to be exactly one half, let's say that it happens to be to over four. That's not very useful. To me. Two or four is one half that could really be anything. When this rule becomes really helpful is when I have something either, Like a very, very low number, like 1/10. Let's say that would tell me it's extremely symmetrical or when I have a really, really high number like 8/10. Then that would also tell me it is very asymmetrical. Um, and I would start to think of ways that I could arrange this in a very asymmetrical way. Okay, enough about that rule. The last thing is that you re state at the end after you've gathered all these clues. After you've gathered your symmetry, you re state the number of proton NMR signals needed because you should only be drawing structures that actually have the number of struggle signals needed. So, for example, if you're Proton NMR on Lee has three signals. You should Onley draw structures that could yield three signals in a proton NMR. It's a waste to draw structure that doesn't give that number. Okay, at this point, this is when you get creative. Okay, Unfortunately, this is a word that many of you were scared of. Um, but what I'm trying to do here is take the most creativity out of it as possible because I know that's the hard part is trying to, like, really think about what could it be? Could it be this or that? I'm going to try to give you a system so that when you do get creative, there's not that much to think about. It's literally Maybe you have 23 or four different drawings that are possible, but not more than that. Okay, this is where you draw everything that fits the above criteria. And then finally, once you've drawn all the possible structures that could fit all of those clues that molecular sentence that we built, you finally use a combination of shifts, integrations and splitting to confirm which structure is the right one. Okay, so I know that was a mouthful. What we want to do for this next example is we're finally going to get into structure termination. But I'm not gonna actually give you guys this molecule yet. I just want you guys to learn how to build that strong molecular sentence ahead of time. So basically, we're doing here is I've given you a molecular formula. I have given you data from an I r. And I've given you data from a proton NMR. Okay, these guys. Thes numbers, by the way, are the shifts. So the 2.29 point four you guys should know what? Four h and two HR those integrations, right? You should recognize what the IRS, these air, all major clues. There's a ton of clues going on around here. What I want to do is I'm gonna go ahead and stop the video on. I'm gonna have you guys go step by step and I want you guys to figure out the HD. I want you to figure out every clue possible that you could gather here. I've already given you a lot of hints. I want you to think about symmetry, if that's important. And then finally, I want you to Onley draw structures that have to peaks in a in a proton NMR. And you know, at that point, I'll kind of take over from there. But I just want you guys to build the molecular sentence and then Onley draw possible structures that fit that sentence and have two peaks in the proton NMR. So go ahead and take a stab at it and I'll be right back
Building a molecular sentence
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Alright, guys. So I'm going to start off with the HD so the HD can be calculated using a formula I H d equals two n plus two minus h over two. Okay, where n is equal to the number of carbons. So I would say the end is equal to four. That means I've got 10 minus h now H is equal to the number of hydrogen or hydrogen equivalents. I've got six hydrogen to oxygen is the oxygen's don't count. So it's literally just 10 minus six over two. That's going to give me four over to which is equal to two I HD. Okay, Now, if you guys recall from I h d, that means that either I have two double bonds, two rings, a combination of that. I could even have a triple bond, so the HD didn't tell me that much yet. It just provided, like, a framework of basis for what I'm looking for are there? So now it's time to actually gather the clues. Remember I told you that you look four different things for clue building. You were gonna look at the NMR shifts, the ir splitting and the integrations. So let's see what we can gather. So let's just look at the NMR first. Do you guys see anything suspicious about these shifts in the NMR? Absolutely. We've got a 9.4, which happens to be in the range of 9 to 10. So immediately, I'm suspecting alga hide. Okay, so I'm just going to write down my clues here, and then we'll put it all together, So I'm suspecting Ally. Okay, Um what else? I've got a shift of to to 2.2 is actually in what range? That's in our Z. See, a trained right. And remember that anything around too was kind of in the area of that. It's either on age that's next to a carbon that has, um, you know, that basically has a carbon Neil or that has a benzene ring. Okay, so we also said alos Aziz possible alot would look like this. Okay, so we're looking for something like that to point to is kind of distinctive is right in that two range. Um, it's gonna be one of these three things. Okay, Cool. So we've kind of suspected those shifts gave us a little bit of information. Does does our I r. Tell us anything. Does it confirm any of this? We'll definitely What it's looking like I have is a carbon eel, and it's looking like I have a complex carbon eel. Right, Because I've got both a peak in the 1700 range. The carbon you'll range. But I've also got this peak at 2700, which is distinctive of Aldo hides. OK, now you guys might notice a discrepancy here, which is that I told you guys that Aldo hides have a wave number of 17 10. And here I have 17. 20. And you are going to completely ignore that because it doesn't matter. Just you guys know all the values I told you because they could all change by a little bit. Okay? It just really depends on the textbook on the way. Your professor wants to ask it. So for my purposes, 17 2017 10 are the exact same thing. Okay, so this definitely confirms that I haven't Alva hide. Okay. What is my peak at 29. 50 Tommy? Nothing. It just tells me that I have s p three c h bonds, which hopefully you call me saying that every single molecule has that. So that doesn't really help me. Okay, so now I've got my NMR. I've got my eye are clues. Do we see any splitting patterns that we learn to memorize that could give the structure away? Well, I see a doublet and a triplet, and actually, that's not a splitting pattern that I told you. Okay, I taught you about triplet quartet, but that's not this. So actually, splitting patterns kind of didn't work, you know? Finally, integration. Do we see anything interesting with these integrations that might pique us into how Maney functional groups we have? And actually, it looks like we just hit gold because noticed that my Aldo hide 9.4 actually has an integration of to age. That means that I must have to Aldo hides present in order to give me 294 shifted hydrogen is okay, so this is actually becoming pretty great. So now if I were to start off my sentence, what I would say is that well, actually, really quick before I build the sentence, I want to talk about this. We just gathered all these clues. It looks like I now have to Aldo Hides. Does this conclusion correlate with my i HD Remember that my i h d said that I have to i h d So do we now know where those PhDs air coming from? Yes, we do. They're coming from both of the carbon You'll groups on my dual Aldo hides. So that means that what? I'm looking for us. This is my sentence. I'm going to start building it. Okay, So what I'm going to say is that I'm looking for a four carbon di alga hide. Okay? Is there anything else that we can say? Or do you have splitting patterns? No. Um, that's really all we can say right now. It's a cyclic, right? Could we say that it's a cyclic because it had a ring? It would have an idea of three. Right? So we could even stick in parentheses. A cyclic. Right? So it's an a cyclic die Aldo hide. Um, now it's time to look at the symmetry thing. So now we're done. That's basically all we can say so far. That's actually a lot, though, because there's not a whole lot of die. Aldo hides that you can draw with only four Carbon. So we're doing great now it took it symmetry. Now the way we did, The symmetry thing was you take the number of H and M R signals, which is to and put it over the number of carbons, which is four. That gives me a ratio that are a fraction that simplifies down toe one half. Remember that one half was really actually might cut off between symmetrical and asymmetrical, meaning that this symmetry trick is pretty much worthless right now. It's not gonna help me because I told you guys that symmetry really only helps at the extremes. It helps me if it's very symmetrical or very un symmetrical. But in this case, since it's right down the middle, I'm just gonna ignore it. It didn't work out for me this time. Okay, Um, finally, now that I did symmetry, we have to restate the number of protons proton NMR peaks we need. So we would say we need a four carbon a cyclic die Aldo hide with how maney proton NMR peaks or signals to with to each NMR signals and guys, This is exactly the kind of sentence that you need to start getting good at building. This is the kind of sentence that, if you could build him Sorry, my head's in the way. Um, if you can build this sentence, you are so far ahead of your classmates because your classmates are going to be struggling with the basics. They don't even get to know where to begin. Meanwhile, you have, like, this beautiful little sentence here that perfectly captures what you're trying to draw. Okay, so now this is the part where we get creative and we actually draw structures, okay? And I'm going to take over from here, okay? And you're just going to see how I do it. But what we find out is that this sentence is so good, is so strong, but there's not that many structures we control. So first of all, let's start off with how many different four carbon chains can we make without are a cyclic well, to we could make a four carbon chain that is straight chain. And I'm sorry, guys. I'm gonna be running out of room because we're at the end of the page. But I'll make it work. And we could also have a branch for carbon chain like that. Okay, So now could I could you? Could you also do, like, a four carbon ring like this? Would that work? No. Because we said it was basically So we're gonna take that out now on the first chain. How many places could be put to Aldo hides? Well, this is the easy part. By definition, Aldo heights always have to be terminal. They always have to be at the end of a chain, meaning that there's only one place I can put it, which is here and here. Okay. Let me just give you an example. If I were to choose to try to put my Aldo hide in the middle, would that work that's now called the key tone. So no, the alga hide by definition has to be on the edge. So there we go. Okay. Now, with my other four carbon chain, the Branch one. Um, would I be able Thio? Where could I put the algae hides there? Yeah, well, I could put that on the ends of these corners. Okay, so that's another possibility. So I've got one. I've got to Is there anywhere any other combination of atoms that I could put these Alba Heights on on the straight chain? No. And on the branch chain? No. So, actually, these are kind of a on the options right now. There's nothing else we can do now. It turns out we have to remember. We need toe Onley, draw structures that are going to give us the right amount of signals. That's what we always use. Two screen first. So my question to you is, let's look at the first compound. Let's say this is structure A and this is structure. Be would structure a yield to NMR signals? Would structure be yield to NMR signals? Proton NMR For a. The answer is yes. It would yield on Lee, too, because these hydrogen would be, Let's say, peak A and that's gonna repeat be and then the same exact thing would be repeated on the other side. So I have being a so I would only get two signals. That's a check mark for structure be Would I get to signals? Well, I do have a plane of symmetry, so I would get a be, But look, I've also got this mess up here that's gonna give me see, so I'm gonna get to signals with this one? Nope. This is automatically crossed out. Which means, by process of elimination, this has to be my correct structure. Okay, but we're not. I'm not letting you off that easy, though, because sometimes you're not going to get the answer that quick. So what I want to do is I want to. Even though we know that it has to be that structure, I want to use the rest of the information to confirm that it actually is That structure for example, would this actually have a double in a triplet? We have to analyze that. So we can really be sure. So a And I'm talking about proton A notice. It's the alga hide h How would that be? Split? How would proton a be split in? Uh, you know, with the end plus one wall. Well, is it adjacent toe? Any non equivalent hydrogen? Yes, it's adjacent to to right here. So that means that for proton A and it's equal to two. So proton a should be a triplet that works so far. Let's look at proton be. Is Proton be next to any non equivalent hydrogen? Well, it's next to two hydrogen is here. So are we gonna split with those? Actually, guys, this is a really, really tricky and good example. The answer is, if you go to the right of B, you are not gonna split here. Even though there's two hydrogen. Why would that be so? Why am I telling you that if you go to the right, there's two ages there? You're actually not gonna count them towards N plus one? Because guys remember in where to split. Not only do your hydrogen is have to be adjacent, which is right and left, they also have to be non equivalent. These hydrogen is that I have to the right are actually equivalent. They're both called Proton Beat. You can't have equivalent protons splitting the same type of proton. So even though I'm going to the right and even though I'm counting up these two protons here to the right, those protons are the same as the ones that are being split so they don't split. Okay, that's the whole deal with adjacent and non equivalent. If it has the same letter, it's the same type of hydrogen. It's not going to split. Let's go to the right if you go toe. I'm sorry. I keep mixing up my right and left my apologies. I don't know if it's the same for you, but that was to my right. Okay, Now, if I go to my left, do I have any adjacent non equivalent protons? Yes, right here. I've got a proton that has a different letter. It's a and it's adjacent. It's on. It's on a carbon right next to it. But it's only one of them. So that means n equals one for B, which means that one plus one would be a doublet. Does that make sense with the peaks or the splits that I actually saw in my proton NMR? Yes, it does. I'm actually gonna take myself out of the screen really quick so that we don't have to deal with my head being in the way. Okay, so anyway, Tripplett Doublet, does that make sense? Yeah, it does now notice that Proton A is the triplet and Proton a should be the one with the shift of 9.4. So is my triplet the one with the 9.4 shift? Yes. So that even makes sense that my triplet and my 9.4 are happening on the same proton. Okay. Likewise. Does it make sense that my doublet would have a shift of 2.2? We'll notice that proton be is right next to a carbon eel. And if you're next to a carbon eel, then you would be right around to. In fact, I told you guys, I think 2.1 when we were doing our, um you know, our frequencies in our shifts toe learn. So that's exactly right. Now, finally, do the integrations make sense. Does it make sense that we have four? H is? I'm sorry. Let me start with two ages. Doesn't make sense that we have to. H is for my Alba Heights. Yes. Does it make sense that I have four? H is for my double. It's the 2.2. Well, 1234 It looks like it makes sense. So, guys, everything makes sense here. This is confirmed to be the structure. So I know this is a huge, huge mass, but I'm just gonna draw it one more time for you guys so you guys can have it clear in your notes. The answer was this Now I know that a lot of you guys you're doubting. You're saying Johnny, You know what? I guess that makes sense. But I lost you, like, 10 minutes ago when there's no way it could have done all of this. Guys, I told you this is not gonna be easy. This is something that you have to practice. And don't worry. We're gonna give you practice. You can get better at it. The biggest deal that I'm trying to make here is always build your sentence first. If you could build a strong molecular sentence, you've almost done all of the work. Confirming is the easy part. The hard part is really just making sure that you gather all your clues ahead of time. Okay, so anyway, guys, I hope that section made sense. And best of luck practicing with the problems. Um, let's go ahead and just wrap up this topic
Propose a structure for the following compound that fits the following 1H NMR data:
Formula:C3H8O21H NMR:3.36 δ (6H, singlet)
4.57 δ (2H, singlet)
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Propose a structure for the following compound that fits the following 1H NMR data:
Formula:C2H4O21H NMR:2.1 δ (singlet, 1.2 cm)
11.5 δ (0.5 cm, D2O exchange)
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Propose a structure for the following compound that fits the following 1H NMR data:
Formula:C10H141H NMR:1.2 ppm (6H, doublet)
2.3 ppm (3H, singlet)
2.9 ppm (1H, septet)
7.0 ppm (4H, doublet)
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Propose a structure for the following compound, C5H10O with the given 13C NMR spectral data: