Structure Determination without Mass Spect - Online Tutor, Practice Problems & Exam Prep

Structure determination is a critical analytical skill involving the synthesis of data from molecular formulas, NMR, and IR spectra to deduce molecular structures. Begin by calculating the Index of Hydrogen Deficiency (IHD) to identify unsaturation. Analyze NMR shifts, integrations, and splitting patterns for functional group identification, such as aldehydes. Assess symmetry through the ratio of NMR signals to carbon count. Finally, draw structures that match the gathered clues and confirm the correct structure using spectral data. This systematic approach enhances efficiency and accuracy in structure determination.

Building Molecular Sentences

So guys, now we're going to discuss what is basically the holy grail of our analytical techniques section and that is the skill of structure determination. 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, NMR's, and IR spectra. 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 to freak out because this is a very complex skill. We're having to synthesize 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 going to tell you that is the first way to lose points on the exam because you're going to run out of time. You're going to draw a bunch of structures that aren't correct, and it's just not efficient. We need to be much more strategic in 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 clues as possible from that NMR, from the IR, from the molecular formula. We're going to 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, guild that takes 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. Let's go ahead and just talk about the steps to build a strong molecular sentence. The very first step is to determine the IHD which is a skill that we learned in Organic Chemistry 1 and that I've included in this section so you can review it. The IHD is just basically going to tell us about double bonds, rings, triple bonds, etc. Then we're going to use the NMR, the IR, splitting patterns, and integrations. We're going to 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 things. 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 if you learned if you saw that you had an 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 10 range and that would be an aldehyde. So immediately I would be suspecting there is an aldehyde. So now what if I look at my IR spectrum and there's also a peak there at 1710. Then would that confirm my suspicion that I have an aldehyde or would it deter me? It would confirm it because remember, aldehydes have a carbonyl peak at 1710. So then that would kind of confirm the aldehyde suspicion. Now, what if I look at my splitting patterns and I notice that in my NMR, I actually have a triplet and a quartet present. Already, I know that I'm looking for aldehydes that have some kind of ethyl group on them. That's a big deal. Now, what if I look at my integration? What kind of information do you get from integration? You get the number of hydrogens. Why is that important? Well, what if I have that shift at 9.1 for my NMR, but it actually has an integration of 2 H? What would that tell me? If my integration is 2 H for a shift of 9.1, that tells me that I actually don't just have 1 aldehyde. I actually have 2 aldehydes. The reason is that every aldehyde only has 1 hydrogen that results in the 9 to 10 range. So if I have 2 hydrogens, that must mean that I have 2 aldehydes. These are the kinds of clues that we gather right away. You have to get good at learning where to find those clues. 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. Basically, this is just a test of symmetry. 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. If that number turns out to be less than 1 half, then that suggests that it's a symmetrical compound. Whereas if that number tends to be above 1 half, then it's probably going to be a pretty asymmetrical compound. The logic behind here being let's say that you have a molecule that's like C6 H14. Right? And you've got your proton NMR. Right? It starts at 0. It ends at 13. And all you have is 1 peak. Let's just say you have 1 peak. Well, what that's going to suggest to me is that I only have one signal. So that's going to be the 1 in my fraction. And I have 6 carbons. That's going to tell me that a lot of these hydrogens are exactly the same as each other. Actually, they're all exactly the same if I'm only getting one signal. The only way that they could all be the same is if the molecule is symmetrical. This must be a pretty symmetrical molecule if it's giving me a ratio or a fraction that's so below 1 half. It's 1 6th instead of 1 half. Does that kind of make sense? Basically, my fraction is just a measure of how symmetrical this molecule probably is or how asymmetrical it is. Now because of the fact that I literally made this up, it's something that I've used for many years but it's also not for sure. You know, it's not like a tried and true method. So what that means is that you can never just rule out a 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. For example, symmetry in straight chains can be very difficult to visualize. So I don't want you guys to just go draw structures just based off of this rule. Just use it as a helping hint more than anything else. Also, just another point, it tends to be really helpful at the extremes and not very helpful in the middle. If you take your carbon, you know, if you take your fraction and it happens to be exactly 1 half, let's say that it happens to be 2 over 4, That's not very useful to me. 2 over 4 is 1 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 over 10, let's say. That would tell me it's extremely symmetrical. Or when I have a really, really high number like 8 over 10, then that would also tell me because it's very asymmetrical. Enough about that rule. The last thing is that you restate at the end, after you've gathered all these clues, after you've gathered your symmetry, you restate the number of proton NMR signals needed because you should only be drawing structures that actually have the number of signals needed. For example, if your proton NMR only has 3 signals, you should only draw structures that could yield 3 signals in a proton NMR. It's a waste to draw a structure that doesn't give that number. At this point, this is when you get creative. Unfortunately, this is a word that many of you are scared of. 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 2, 3, or 4 different drawings that are possible but not more than that. This is where you draw out 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. 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 going to actually give you guys this molecule yet. I just want you guys to learn how to build that strong molecular sentence ahead of time. Basically, what we're doing here is I've given you a molecular formula. I've given you data from an IR and I've given you data from a proton NMR. These numbers by the way are the shifts. So the 2.29.4, you guys should know what 4 H and 2 H are. Those are integrations. You should recognize what the IR is. These are all major clues. There's a ton of clues going on around here. What I want to do is I'm going to go ahead and stop the video and I'm going to have you guys go step by step and I want you guys to figure out the IHD. I want you to figure out every clue possible that you can 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 only draw structures that have 2 peaks in a proton NMR. At that point, I'll kind of take over from there. But I just want you guys to build the molecular_sentence and then only draw possible structures that fit that sentence and have 2 peaks in the proton NMR. Go ahead and take a stab at it and I'll be right back.

Building a molecular sentence

Alright, guys. So I'm going to start off with the IHD. The IHD can be calculated using the formula IHD = 2 n + 2 - h 2 , where n is equal to the number of carbons. I would say that n is equal to 4. That means I've got 10 minus H. Now H is equal to the number of hydrogens or hydrogen equivalents. I've got 6 hydrogens, 2 oxygens. The oxygens don't count, so it's literally just 106 over 2. That's going to give me 4 over 2, which is equal to 2 IHD. Now if you guys recall from IHD, that means that either I have 2 double bonds, 2 rings, a combination of that. I could even have a triple bond. The IHG didn't tell me that much yet. It just provided like a framework, a basis for what I'm looking for. Now it's time to actually gather the clues. Remember I told you that you look at 4 different things for clue building. We're going to look at the NMR shifts, the IR, the splitting, and the integrations. Let's see what we can gather. 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 aldehyde. I'm just going to write down my clues here, and then I'll put it all together. I'm suspecting that. What else? I've got a shift of 2.2. 2.2 is actually in what range? That's in our ZCA range. Remember that anything around 2 was kind of in the area of that it's either an H that's next to a carbon that has, you know, that basically has a carbonyl or that has a benzene ring. We also said allyl is possible. Allyl would look like this. Okay. So we're looking for something like that. 2.2 is kind of distinctive. It's right in that 2 range. It's going to be one of these three things. Cool. We've kind of suspected those shifts gave us a little bit of information. Does our IR tell us anything? Does it confirm any of this? Well, definitely what it's looking like I have a carbonyl,

And it's looking like I have a complex carbonyl because I've got both a peak in the 1700 range, the carbonyl range. But I've also got this peak at 2700, which is distinctive of Now you guys might notice a discrepancy here, which is that I told you guys that aldehydes have a wavenumber of 1710 and here I have 1720 and you are going to completely ignore that because it doesn't matter. Just so you guys know, all the values I told you, they could all change by a little bit. It just really depends on the textbook, on the way your professor wants to ask it. For my purposes, 17/20 and 17/10 are the exact same thing. This definitely confirms that I have an aldehyde. What is my peak at 2950, Tommy? Nothing. It just tells me that I have sp3 CH bonds which hopefully you recall me saying that every single molecule has that, so that doesn't really help me. Now I've got my NMR. I've got my IR clues. Do we see any splitting patterns that we learned 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. I taught you about triplet quartet, but that's not this. So actually, splitting patterns kinda didn't work. Finally, integration. Do we see anything interesting with these integrations that might peak us into how many functional groups we have? And actually, it looks like we just hit gold because notice that my aldehyde 9.4 actually has an integration of 2 H. That means that I must have 2 aldehydes present in order to give me 29.4 shifted hydrogens. This is actually becoming pretty great.

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 2 aldehydes. Does this conclusion correlate with my IHD? Remember that my IHD said that I have 2 IHD. Do we now know where those IHGs are coming from? Yes, we do. They're coming from both of the carbonyl groups on my dual aldehydes. So that means that what I'm looking for so this is my sentence. I'm going to start building it. So what I'm going to say is that I'm looking for a 4 carbon dialdehyde. Okay. Is there anything else that we can say? Do we have splitting patterns? No. That's really all we can say right now. It's acyclic, right? Could we say that it's acyclic? Because if it had a ring, it would have an IHD of 3. We could even stick in parenthesis acyclic. It's an acyclic dialdehyde. Now it's time to look at the symmetry thing. Now we're done. That's basically all we can say so far. That's actually a lot though because there's not that many dialdehydes that you can draw with only 4 carbons. We're doing great.

Now let's look at symmetry. The way we did the symmetry thing was you take the number of HNMR signals, which is 2, and put it over the number of carbons, which is 4. That gives me a ratio or a fraction 1 2 Remember, that 1 / 2 was really actually my cutoff between symmetrical and asymmetrical, meaning that this symmetry trick is pretty much worthless right now. It's not going to 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 unsymmetrical. But in this case, since it's right down the middle, I'm just going to ignore it. It didn't work out for me this time. 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 4 carbon acyclic dialdehyde with how many proton NMR peaks or signals? 2. With 2 HNMR 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 can build it I'm sorry. My head's in the way. 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're not even going 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, 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 that there's not that many structures we control. First of all, let's start off with how many different 4 carbon chains can we make without acyclic? Well, 2. We could make a 4 carbon chain that is straight chain. And I'm sorry guys. I'm going to 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 branched 4 carbon chain like that. Now, could I could you could you also do like a 4 carbon ring like this? Would that work? No, because we said it was acyclic. So we're going to take that out. Now on the first chain, how many places could we put 2 aldehydes? Well, this is the easy part. By definition, aldehydes 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. Let me just give you an example. If I were to choose to try to put my aldehyde in the middle, would that work? That's now called a ketone. So no. The aldehyde by definition has to be on the edge. There we go. Now with my other 4 carbon chain, the branched one, where could I put the aldehydes there? Yeah. Well, I could put that on the ends of these corners. That's another possibility. I've got 1. I've got 2. Is there anywhere, any other combination of atoms that I could put these aldehydes on? On the straight chain, no. And on the branch chain, no. So actually these are kind of our only options right now. There's nothing else we can do. Now it turns out we have to remember, we need to only draw structures that are going to give us the right amount of signals. That's what we always use to screen first. My question to you is let's look at the first compound. Let's say this is structure A and this is structure B. Would structure A yield 2 NMR signals? Would structure B yield 2 NMR signals, proton NMR? For a, the answer is yes. It would yield only 2 because these hydrogens would be, let's say, peak A and this one would be peak B. And then the same exact thing would be repeated on the other side, so I'd have b and a. So I would only get 2 signals. So that's a checkmark. For structure b, would I get two signals? Well, I do have a plane of symmetry, so I would get a b. But look, I've also got this mess up here. That's going to give me c. So, am I going to get 2 signals with this one? Nope. This is automatically crossed out which means by process of elimination, this has to be my correct structure. 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 doublet and 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 aldehyde H. How would that be split? How would proton A be split with the n plus one rule? Well, is it adjacent to any non-equivalent hydrogens? Yes. It's adjacent to 2 right here. That means that for proton A, n is equal to 2. So proton A should be a triplet. That works so far. Let's look at proton Is proton next to any non-equivalent hydrogens? Well, it's next to 2 hydrogens here. Are we going to split with those? Actually guys, this is a really, really tricky and a good example. The answer is if you go to the right, you are not going to split here even though there's 2 hydrogens. Why would that be so? Why am I telling you that if you go to the right, there's 2 H's there. You're actually not going to count them towards n plus 1. Because guys remember, in order to split, not only do your hydrogens have to be adjacent which not only do your hydrogens have to be adjacent which is right and left, they also have to be non-equivalent. These hydrogens that I have, to the right are actually equivalent. They're both called proton B. 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 2 protons here to the right, those protons are the same as the ones that are being split. So they don't split. Okay. So that's the whole deal with adjacent and non-equivalent. If it has the same letter, if it's the same type of hydrogen, it's not going to split. Let's go to the right. If you go to 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. 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 carbon right next to it, but it's only one of them. So that means n equals 1 for B, which means that 1+1 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 going to take myself out of the screen really quick, so that we don't have to deal with my head being in the way. Anyway, triplet, 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. 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. Likewise, does it make sense that my doublet would have a shift of 2.2? Well, notice that proton is right next to a carbonyl. If you're next to a carbonyl, then you would be right around 2. In fact, I told you guys, I think 2.1 when we were doing our frequencies and our shifts to learn. That's exactly right. Now finally, do the integrations make sense? Does it make sense that we have 4 H's? I'm sorry. Let me start with 2 H's. Does it make sense that we have 2 H's for my aldehydes? Yes. Does it make sense that I have 4 H's for my doublets, the 2.2's. Well, 1, 2, 3, 4. It looks like it makes sense. So guys, everything makes sense here. This is confirmed to be the structure. I know this is a huge, huge mess but I'm just going to 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, and there's no way I could've done all of this. Guys, I told you this is not gonna be easy. This is something that you have to practice. Don't worry, we're going to 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 can 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 the problems. 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:C₃H₈O₂                                                  ¹H NMR:3.36 δ (6H, singlet) 

                                                                                         4.57 δ (2H, singlet)

Hey guys. Let's take a look at the following practice question. It says, "Propose a structure for the following compound that fits the following proton NMR data." Now, the key to any structural determination question is first we must figure out the IHD. The IHD formula that we're going to use is C=2c+nh+2. We have 3 carbons involved, so it's going to be 2×3=6. There are no nitrogens in my formula, so we ignore the n. H represents hydrogens and halogens. We don't have any halogens, but we do have 8 hydrogens, +2. 6-8 gives us negative 2, plus 2. We have an IHD of 0, which means there is no double bonds, no triple bonds, no ring. This is a single bonded carbon system only. Now, what we want to do next is we're going to look at the integration because the integration tells us the types of structures that are involved within my molecule. Here we're going to say we have 6 H. Now, 6 H means that we have most likely 2 CH3s. It could also mean that we have 3 CH2s. It could mean that. But we're going to assume that it doesn't because here there's only 3 carbons within my compound. So it would be very unlikely because this 2 H here most likely means we have 1 CH2. That would give us 4 carbons involved if we did 3 CH2s and one CH2. Plus, it's really hard to make 3 CH2 groups chemically equivalent. It's very unlikely that we'd have this. We're going to assume we have these 2 CH3s and 1 CH2. Now, somehow, both those CH3s are the same. So there must be some symmetry involved. And that CH2 is all by itself as its own separate signal. What we next need to realize here is the multiplicity. Here, they're both singletons, so it must mean either their neighboring carbons don't have any hydrogens or they're not next to any neighboring carbons. Since these are the only 2 types of carbon systems within our structure, what's most likely happening here is that both those groups are not next to a carbon. What are they next to? They must be next to oxygens. We can assume that our structure most likely looks something like this. We have a CH3 next to an oxygen, n plus one rule. That's why it's a singlet. A singlet. Why is this CH2 so deshielded? Why is it 4.57? It must be next to the most number of oxygens. The CH2 must be next. Remember, there's symmetry involved, so there's another oxygen here. And then finally, there's another CH3. Both these CH3s would be the same exact signal. This would be our second signal. That CH2 because it's in between 2 oxygens, it's most deshielded. That's why its chemical shift is higher at 4.57. So these are the types of things you must run through your mind in order to figure out the best possible combination. Sometimes, you may draw 2 or 3 different molecules before you figure out what the best one is. In this case, it would be this ether, this diether here.

Propose a structure for the following compound that fits the following ¹H NMR data: 

Formula:C₂H₄O₂                                            ¹H NMR:2.1 δ (singlet, 1.2 cm) 

                                                                                   11.5 δ (0.5 cm, D2O exchange)

Hey guys. Let's take a look at the following question dealing with structural determination. So here we say, Propose a structure for the following compound based on the proton NMR data that I give to you. The first thing we want to do is figure out the IHD. So IHD= (2C+N-H+2) 2. We have 2 carbons, so 2 times 2 is 4. No nitrogens, but 4 hydrogens. That becomes 0 + 2, which is 2 divided by 2, so we have an IHD of 1. Now, what we do next is we take a look. We have these integration values. But in this question, we really don't need them because here what we need to realize is we have a signal which is unique. Only one type of proton will give us a signal between 11.0 and 13.0. That means that this is a carboxylic acid. Now here, I also include in this question D2O exchange. Now, D2O is deuterated water. Now remember, deuterium is an isotope of hydrogen. Now, when they're saying D2O exchange, that indicates that we have an OH group present because what does this do? If we have an OH group present and we use D2O on it, what it does is, it actually replaces that signal as deuterium is NMR inactive. So, that signal will disappear. So if we have an unknown and they say they use D2O on it and the signal disappears, that's an indication that we have an OH group present either from an alcohol or from a carboxylic acid. Anyway, since our signal is 11.5, we know we have a carboxylic acid, so we know we have at least this portion of our molecule. Then the question becomes easier because here our formula is C2H4O2. The carboxylic acid that we know we have has a carbon in it. It has 1 hydrogen in it. It has both oxygens in it. So, what do we have left? A CH3 that's not being used, which would mean that this compound here is connected to a CH3. Here we have acetic acid or ethanoic acid. This would be the identity of our unknown compound. Remember, once you figure out the IHD, what we typically do is we go to the integration values to figure out what types of hydrogen groups are present. But if you have a unique signal such as carboxylic acid, then you focus on that and we know we have that portion. From there, we can deduce what the rest of the molecule would look like.

Propose a structure for the following compound that fits the following ¹H NMR data: 

Formula:C₁₀H₁₄                                      ¹H NMR:1.2 ppm (6H, doublet) 

                                                                             2.3 ppm (3H, singlet) 

                                                                             2.9 ppm (1H, septet) 

                                                                             7.0 ppm (4H, doublet)

Hey guys. Let's take a look at the following practice question. Here, we're supposed to propose the structure of the unknown compound based on the NMR data that I provide. Now remember, the first thing you always want to do is figure out your IHD. Here, we're going to use the formula for IHD as C2+N-H+2/2. We have 10 carbons, so we're going to plug that in, so that's 2×10=20. There are no nitrogens, so ignore that part, resulting in 20-14+2=8/2=4. Now, usually, when we get an IHD of 4 or higher, this most likely indicates that we have benzene involved. But we can't be absolutely sure until we look at a chemical shift to see if we have benzene or not. Remember, benzene hydrogens are unique. Benzene hydrogens are the only ones that we're going to find within the region of 6.5 to 8.0 ppm. Usually, you'll see it around 7.0-7.2 or so. Here, we have an IHD of 4 and we have a chemical shift around 7. That definitely indicates that benzene is involved. Here's benzene. We know we have that. Now here, what we need to realize next is, for benzene, we're saying 4H here, which means that it has 4 hydrogens on it, which implies that it's a disubstituted benzene ring. We've taken care of the benzene part. Next, what we're going to do is we see that we have 6H, 3H, 1H. 6H here most likely means we have 2 methyl groups. 3H here most likely means we have 1 CH3 group. And 1H here most likely means we have 1 CH group. Now, realize when it comes to multiplicity patterns, what kind of pattern gives us a septet with an integration of 1 hydrogen and a doublet with an integration of 6 hydrogens? Remember that only results if we have isopropyl. Because here, we know it's isopropyl because these two methyl groups are the same. If we look at their multiplicity, they'll come off as doublets and that CH will come off as a septet. So remember those patterns. Next, here we have a CH3 that's a singlet. Why? Because it must be down here. So this must be the identity of our compound and adding up everything together gives us this formula. So again, first figure out the IHD, we saw that we had a unique chemical shift of 7 which indicates that we have benzene. Benzene has an integration of 4 hydrogens which means it has 4 Hs on it which means it's disubstituted and the 2 substituents were isopropyl and just a methyl.

Propose a structure for the following compound, C₅H₁₀O with the given ¹³C NMR spectral data: 

Fully Broadband decoupled ¹³C NMR and DEPT:206.0 δ (↑); 55.0 δ (↑); 21.0 δ (↓)& 11.0 δ (↑).

Hey guys, let's take a look at the following practice question. Here it says propose a structure for the following compound, C5H10O with the given carbon 13 NMR spectral data. Here I give us the fully, broadband decoupled carbon 13 NMR as well as depth information. Remember, fully decoupled broadband just means that all the carbons appear as singlets, which really is useless. Depth helps us to identify if we have any CH, CH2, or CH3 groups. The problem here is I don't tell us depth 90 or depth 135, so I made this question a little bit more challenging. But here, we can make some assumptions. First of all, we have a signal around 206. Remember, if your signal is above 150, then it could only represent a carbonyl. Now, here it has a positive peak which means it could be either a CH or a CH3. Now, it can't be a CH3 because that carbon can't have 3 hydrogens on it because if it did, it would be making 5 bonds. Carbon at the most can only make 4 bonds. So that must mean it's a CH group, which means we have an aldehyde, which is always terminal. It's always at the end. So we figured out a portion of our molecule. Let's look at the other things. What else do we have? We have here a positive peak, which again can mean CH or CH3. This negative peak here, this downward arrow, means it could only be a CH2. We know we have a CH2 for sure. And then here, this is a CH or CH3. We know we have at least a CH2. What we could try to do here is plug it in, CH2. Now, let's make some other assumptions here. So far, how much of our formula have we used? Our formula is C5H10O. We've used 2 carbons and we've used 3 hydrogens and the oxygen, leaving us with C3H7O. Now, here, if we made one of these a CH, one of them a CH3, so if we have one of each, then we'd be using 2 carbons and 4 hydrogens, which will leave us left with what? A CH3. Which means what? Which means we must have 1 CH and 2 CH3s. So here we just have to draw this. We're going to say here we have a CH connected to 2 CH3s. This here would be our unknown compound. Again, carbon 13 NMR structural determination is much more difficult. Not very much information is given. Usually, you wouldn't get a question like this as a standalone question. We would combine it with proton NMR or IR as well in order to give you as much information as possible that you have the best way of getting to the correct answer. This one, I just wanted to see if you guys could form connections and see how exactly could this molecule combine together from that unknown molecular formula.

Provide the structure of the unknown compound from the given information. 

Formula:C₄H₁₀O            IR:3200-3600 cm^{-1           1}H NMR:0.9 ppm (6H, doublet) 

                                                                                               1.8 ppm (1H, nonatet) 

                                                                                               2.4 ppm (1H, singlet) 

                                                                                               3.3 ppm (2H, doublet)

Hey guys. Let's take a look at the following practice question. So here it says, Propose the structure of the unknown compound from the given information. What we need to do first is we need to figure out what is the IHD of my unknown compound. Here we're going to say it is CNH+2/2. So here we have 4 carbons, so we're going to plug that in. So that's 8. Here, we don't care about nitrogen because it's not part of our formula. So we ignore it. We have 10 hydrogens here, plus 2 divided by 2. So it's going to be -2 + 2, which is 0. So we have no double bonds, no triple bonds, no rings. What we do next is we're going to say we see that we have an IR signal between 3,200-3,600 cm-1. This would indicate that we have an alcohol present. Remember, OH groups are always singlets and the integration of this OH group would be 1. That means it's either this one or this one. Now it can't be the top one because that's a nonet. Here, OH groups are always singlets. So we know that this here represents an OH. Now, let's look at the other integrations and figure out what else we have. We have 6 H here, which most likely means we have 2 methyl groups. This one H can't be OH because OHs are always singlets, so this would be a CH. And then here, this here would be CH2. If we look, we've got all 4 carbons that we need. Now, we have to put this all together in a way that'll give us those multiplicities. So here is, we must have a CH2 next to the OH. Now, why is that CH2 a doublet? Using the n + 1 rule, this would be 1. The neighboring Then, how is that CH a nonet? Nonet means 9. How could it be 9? Here, this would be 1 + 4 to get to 9, n would have to be 8. Here, this neighboring carbon has 2 hydrogens on it, so that means we need another 6. Where would those additional 6 come from? It would come from 2 methyl groups that are connected to that CH group. This would be our formula. This would be our unknown compound. So, the best maneuver, the best stepwise strategy to this is first figure out the IHD. We use the IR to tell us the functional tells us how they are pieced together to give us nonets, doublets, etc. If you need to, bear in mind, the last thing you should look at is chemical shifts. Remember, chemical shifts, a lot of them overlap. The only useful ones are for benzene because they are the only ones between 6.5-8.0, aldehydes because they are between 9-10, and carboxylic acids, which are between 11-13. Since we don't have any of those chemical shifts, we didn't need to go that far. But the other strategies that we did, we need to employ them in order to get this question correct.

Provide the structure of the unknown compound from the given information. 

Formula:C₄H₉N              IR:2950 cm^-1, 3400 cm^{-1           1}H NMR:1.0 ppm (4H, triplet) 

                                                                                                         2.1 ppm (4H, triplet) 

                                                                                                         3.2 ppm (1H, singlet)

Hey guys. Let's take a look at the following practice question. So, here it says provide the structure of the unknown compound from the given information. So, what we always want to do first is figure out what exactly is my IHD. So, here we are going to say ( 2 C + H ) + 2 2 . We have 4 carbons. We are going to plug that in. That is 2 times 4 which is 8. We have 1 nitrogen which is 1 minus 9 hydrogens, plus 2 divided by 2. So, this would become 9 minus 9 which is 0 and then 2 divided by 2 gives us 1. So, we have an IHD of 1, which means it could either be a double bond or a ring. But here's the thing. In the IR, it tells us we have 2,950. 2,950 represents sp3 hybridized carbons. If there were any double-bonded carbons involved, then we would have a signal between 3,000 and 3,150. Since we do not, that means there are no double bonds present. Therefore, this must be a ring. Remember, IHD of 1 could be a double bond or a ring. It cannot be a ring here, though, because a double bond would be 3,000 to 3,150. Here, we have a signal around 3,400 cm - 1 . Because we have a nitrogen here, it means we have an amine. The amine that would show up would be either an NH2 or an NH. How can we tell the difference between them? Well, both would be singlets, but here their integrations would be different. This one would have an integration of 2H. This one would have an integration of 1H. So, here, this 1H integration means that we have an NH present. What we're going to say next is that somehow, we have 4 CH2 which most likely means we have 2 CH2s. We have to form a ring in a way, and we have to have an NH group. An NH group would mean that we have a secondary amine. That means that nitrogen is connected to 2 carbons. Here, the best possible way to do this structure, it would have to look like this. It would have to be symmetrical, so we have an NH here. It'd be connected to 2 CH2s and those 2 CH2s will be connected to other CH2s. Now, if we look, these 2 look the same and they're both connected in the same way, so they will represent 1 proton signal. Then, these 2 CH2s look the same and they're connected in the same exact way, so they will be our second signal. Doing n + 1, these would be 1. Neighboring carbon has 2 hydrogens, the same thing would be done here. This would be 1. Neighboring carbon is this carbon. Would be 1. Neighboring carbon is this carbon. We can't count this as a neighboring carbon since they have the same exact signal. So, your neighboring carbon would have 2 hydrogens which give us 3. So, it also would be a triplet. So, in this case, this would be our compound, a cyclic secondary amine.

Here’s what students ask on this topic:

What is the Index of Hydrogen Deficiency (IHD) and how is it calculated?

The Index of Hydrogen Deficiency (IHD) is a measure used to determine the degree of unsaturation in a molecule, indicating the presence of double bonds, triple bonds, or rings. It is calculated using the molecular formula. The formula for IHD is:

$\frac{(2C+2)-H+N-X}{2}$

where C is the number of carbons, H is the number of hydrogens, N is the number of nitrogens, and X is the number of halogens. This calculation helps in identifying the presence of rings and multiple bonds in the structure.

How do you interpret NMR chemical shifts in structure determination?

NMR chemical shifts provide information about the electronic environment of hydrogen atoms in a molecule. The chemical shift is measured in parts per million (ppm) and indicates the type of hydrogen present. For example, a shift around 9-10 ppm suggests an aldehyde proton, while shifts around 0-3 ppm are typical for alkyl groups. By analyzing these shifts, you can identify functional groups and their positions within the molecule, aiding in the overall structure determination.

What role does IR spectroscopy play in structure determination?

IR spectroscopy is used to identify functional groups within a molecule by measuring the absorption of infrared light at different wavelengths. Each functional group has a characteristic absorption range. For example, a carbonyl group (C=O) typically absorbs around 1700 cm^-1. By analyzing the IR spectrum, you can confirm the presence of specific functional groups, which helps in piecing together the molecular structure.

How can symmetry be assessed using NMR data in structure determination?

Symmetry in a molecule can be assessed by calculating the ratio of the number of NMR signals to the number of carbons. If the ratio is less than 0.5, the molecule is likely symmetrical. For example, if a molecule with 6 carbons shows only 1 NMR signal, it suggests high symmetry. This information helps narrow down possible structures by indicating whether the molecule is symmetrical or asymmetrical.

What is the importance of integration in NMR spectroscopy?

Integration in NMR spectroscopy measures the area under each peak, which corresponds to the number of hydrogen atoms contributing to that signal. This information is crucial for determining the relative number of hydrogens in different environments within the molecule. For example, if a peak integrates to 2H, it indicates that two equivalent hydrogens are present. This helps in constructing the molecular structure by matching the number of hydrogens to the molecular formula.