Structure Determination without Mass Spect

Concept: Concept: Building Molecular Sentences

Video Transcript

So guys, now we're going to discuss what is basically the Holy Grail of our analytical technique section. 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, NMRs, 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.
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. 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. So 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 we even begin getting creative in 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 that IR, from the molecular formula. We're going to put it all together in a very ordered way. Where 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.
This isn't going to 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.
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 IHD, which is a skill that we learned in organic chemistry one and that I have included in this section so you can review it. The IHD is just basically going to tell us about double bonds, rings triple bonds, etcetera.
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 saw that 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 is there 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. 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. Well, that's one of the 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 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 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 2H. What would that tell me? If my integration was 2H for a shift of 9.1 that tells me that I actually don't just have one aldehyde, I actually have two aldehydes. The reason is because every aldehyde only has one hydrogen that results in the 9 to 10 range.
If I have two hydrogens, that must mean that I have two aldehydes. So these are the kind of clues that we gather right away. And 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 proton NMR peak or the proton NMR signal to carbon ratio. Basically, this is just a test of symmetry.
What I do is 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 ½, then that suggests it's a symmetrical compound. Whereas, if that number tends to be above ½, then it's probably going to be a pretty asymmetrical compound.
The logic behind here being that let's say that you have a molecule that's like C6H14. And you've got your proton NMR. It starts at 0 it ends at 13. And all you have is one peak. Let's just say. You have one peak. Well, what that's going to suggest to me is that. I only have one signal. So that's going to be the one in my fraction. And I have six 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 fraction that's so below ½, it's 1/6th instead of ½. Does that kind of make sense? Basically, my fraction is just a measure of how symmetrical is this molecule probably or how asymmetrical is it.
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. It's not like a tried and true method. 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. 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 fraction and it happens to be exactly ½. Let's say it happens to be 2/4. That's not very useful to me. 2/4 is ½ 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 this is very asymmetrical and I would start to think of ways that I could arrange this in a very asymmetrical way.
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 three signals, you should only draw structures that could yield three signals on 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 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 two, three or four different drawings that are possible, but not more than that.
This is where you draw out everything that fits in 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 built 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 give you data from a proton NMR. These numbers, by the way, are the shifts. So the 2.2, 9.4. You guys should know what 4H and 2H 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 two 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 two peaks in the proton NMR. Go ahead and take a stab at it. I'll be right back. 

Concept: Example: Building a molecular sentence

Video Transcript

Alright guys so I'm going to start off with the IHD, so the IHD can be calculated using the formula IHD equals (2N+2-H)/2, OK? Where N is equal to the number carbons so here it's N is equal to 4 that means IÕve got 10-H, now H is equal to the number of hydrogens or hydrogen equivalents I've got 6 hydrogens 2 oxygens the oxygens so it's literally (10-6)/2 that's going to give me 4/2 which is equal to 2 IHD, OK? 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 so the IHD didn't tell me that much yet it just provided like a framework a basis for what I'm looking for. Alright so now it's time to actually gather the clues remember I told you you look at four different things for clue building, we're going to look at the NMR shifts, the IR, the splitting and the integrations so let's see what we can gather so let's just look at NMR first do you guys see anything suspicious about these shifts and 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, OK? So I'm just going to write down my clues here and then I'll put it all together so I'm suspecting aldehyde, OK? What else I've got a shift of 2.2, 2.2 is actually in what range that's in our ZCH range right and 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 basically has a carbonyl or that has a benzene, OK? So we also said you know is possible our would look like this, OK? 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, OK? Cool so we've kind of suspected those shifts give 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 is a carbonyl and it's looking like I have a complex carbonyl, right? 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 aldehydes, OK? Now you guys might notice a discrepancy here which is that I told you guys that aldehydes have a wave number 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, OK? Just really depends on the textbook, on the way your professor wants to ask it so for my purposes 1720 and 1710 are the exact same thing, OK? So this definitely confirms that I have an Aldehyde, OK? What my peak at 28950 tell me? Nothing, it just tells me that I have SP3CH bonds which hopefully recall me saying that every single molecule has that so that doesn't really help me, OK? So 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 the splitting pattern I taught you, OK? 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 the 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 2H that means that I must have two aldehyde presents in order to give me two 9.4 shifted hydrogens, OK? 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 have now 2 aldehydes, does this conclusion correlate with my IHD? Remember that my IHD said that I have 2 IHD so do we now know where those IHDs 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 then start building it, OK? So what I'm going to say is that I'm looking for a 4 carbon Dialdehyde, is there anything else that we can say? Do you have splitting patterns? No, that's really all we can say right now, it's acyclic could we say that it's acyclic? Because if it had a ring it would have an IHD of three, right? So we could even stick in parentheses acyclic, right? So it's an acyclic dialdehyde, 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 dialdehydes that you can draw with only four carbons so we're doing great.

Now, Let's look at symmetry now 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 carbon which is 4 that gives me a ratio or fraction simplifies down to 1/2 remember that 1/2 was really actually my cut off between symmetrical and Asymmetrical meaning that this symmetry trick is pretty much worthless right now, it's not going to help me because I you guys that symmetry really only hopes at the extremes it helps me if it's very symmetrical or very unsymmetrical but in this case it's since it's right down the middle I'm just going to ignore it, it didn't work out for me this time, OK? Finally now that I did symmetry we have to restate the number of protons or proton NMR peaks we need so we would say we need a 4-carbon acyclic dialdehyde with how many proton NMR peak 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.... IÕm 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 don't even know where to begin, meanwhile you have like this beautiful little sentence here that perfectly captures what you're trying to draw, OK? So now this is the part where we get creative and we actually draw structures, OK? And I'm going to take over from here, Ok? And you're just going to see how I do it but what we find out is that this sentence is so good it's so strong that there's not that many structures we could draw so first of all let's start off with how many different 4 carbon chains can we make that are acyclic? Well 2, we could make a 4-carbon chain that is straight chain and I'm sorry guys are really running out of room because we're at the end of the page but I'll make it work and we can also have a branched 4 carbon chain like that, OK? So now 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 the chain meaning that there's only one place I can put it which is here and here, OK? 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 not poly-ketone so No the aldehyde by definition has to be on the edge so there we go, OK? Now with my other four carbon chain the branched one, would I be able to...Where can I put the aldehydes there? Yeah well, I could put that on the ends of these corners, OK? So that's another possibility so 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 are 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 used to screen first so my question to you is let's look at the first compound let's says 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 created on the other side so I would have B and A so I would only get two signals so that's a check mark. For structure B would I get 2 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 two signals with this one? Nope this is automatically crossed out which means preprocess of elimination this has to be my correct structure, OK? But we're not, I'm 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 a 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 in you know with the N+1 rule? Well is it adjacent to any nonequivalent hydrogens? Yes, it's adjacent to 2 right here so 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 B, is proton B next to any non-equivalent hydrogens? Well its next to two hydrogens here so are we going to 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 going to split here, even though there's two hydrogens why would that be so? Why am I telling you that if you go to the right there's 2 Hs there you're actually not going to count them towards N+1? Because guys remember in order to split not only do your hydrogens have to be adjacent which is right and left they also have to be nonequivalent, these hydrogens 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, OK? So that's the whole deal with adjacent and nonequivalent, 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 keeps 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, OK? Now if I go to my left do I have any adjacent nonequivalent protons? Yes right here I've got a proton that has a different letter it's A and it's adjacent it's got a carbon right next to it but it's only one of them so that means my N equals to 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 with my proton NMR? Yes it does, ok 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, OK? So 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, 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, OK? Likewise does it make sense that my doublet would have a shift of 2.2? Well notice that Proton B is right next to carbonyl and 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 so that's exactly right. Now finally do the integrations make sense? Does it make sense that we have 4 Hs...I'm sorry let me start with 2 Hs, does it make sense that we have 2 Hs for my aldehydes? Yes, does it make sense that I have 4 Hs for my doublets the 2.2s? Well 1 2 3 4 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 mess but I'm just going to draw it one more time for you guys just so you can have it clear in your notes, the answer was this now I know that a lot of you guys here you're doubting you're saying Johnny you know what I guess that makes sense but I lost you in 10 minutes ago and there's no way I could have done all of this? Guys I told you this is not going to be easy this is something that you have to practice and 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 and 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, OK? So anyway guys I hope that section made sense and best of luck practicing with the problems let's go ahead and just wrap up this topic.

Concept: Practice 1: Proposing Structures from 1H NMR data

Video Transcript

Hey guys, let's take a look at the following practice question. So, here it says, propose the structure for the following compound that fits the following proton NMR data. Now, the key to any structure determination question is first we must figure out the ihd, so the ihd formula that we're going to use is 2C plus N minus H plus 2 divided by two, we have three carbons involved. So, it's going to be 2 times 3, which is 6, there are no nitrogen's 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, plus 2 divided by 2. So, 6 minus 8 gives us negative 2 plus 2. So, we have an ihd of 0, which means there is no double bonds, no triple bonds, no ring, so this is a single bonded carbon, 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. So, here we're going to say we have 6H. Now, 6H means that we have most likely two CH3's. Now, it could also mean that we have three CH2's, 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'd be very unlikely because this 2H here most likely means we have one CH2 and that would give up 4 carbons involved, if we did three CH2's and one CH2, plus it's really hard to make three CH2 groups chemically equivalent. So, it's very unlikely that we'd have this. So, we're going to assume we have this two CH3's and one CH2.

Now, somehow both those CH3's are the same. So, there must be. So, symmetry involved and that CH2 is all by itself as its own separate signal. Now, what we next need to realize here is the multiplicity, here they're both singlets, so it must mean either their neighboring carbons don't have any hydrogen's or they're not next to any neighboring carbons. So, since these are the only two types of carbon systems within our structure what's most likely happening here is that both those groups are not next to carbon, what are they next to? they must be next to oxygens. So, we can assume that our structure most likely looks something like this, we have a CH3 next to an oxygen and plus one O, that's why it's a singlet, why is this CH2 so deshielded? why is it 4.57? it must be next to the most number of oxygens, so the CH2 must be next. Remember, their symmetry involved. So, there's another oxygen here and finally there's another CH3, both these CH3's would be the same exact signal, this would be our second signal, that's CH2 because it's in between two oxygens its most deshielded, that's why it's 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 two or three different molecules before you figure out what the best one is in this case would be this ether, this diether here.

Concept: Practice 2: Proposing Structures from 1H NMR data

Video Transcript

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 so the first thing we want to do is figure out the IHD So ((2C+N-H)+2)/2 we have 2 carbons so 2 divided by 2 is 4, no nitrogens but 4 hydrogens so that becomes 0+2 which is 2 divided by 2 so we have an IHD of 1, now 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 that we have a signal which is unique only one type of proton will give us a signal between 11 and 13 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 what it does is it actually replaces the hydrogen with deuterium, now what's the significance of this? Here this is a protons so it's going to have a signal but once we exchange with deuterium, 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, anyways since our signal is 11.5 we know we have a carboxylic acid so we know what we have them in this portion of our molecule and then the question becomes easier because here are formula is C2H4O2 the carboxylic acid that we know we have has what in it? It has a carbon in it, it has 1 hydrogen in it, it has both oxygen is 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 so here we have acidic acid or ethanoic acid so this would be the identity of our unknown compound. So remember once you figure out the IHD what we typically do is we go to the integration values to figure out what types of hydrogens groups are present but if you have a unique signal such as carboxylic acid so you focus on that and we know we have that portion and from there we can deduce what the rest of the molecule would look like.

Concept: Practice 3: Proposing Structures from 1H NMR data

Video Transcript

Hey guys, let's take a look at the following practice question. So, 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. So, here we're going to use the formula for ihds 2c plus N minus H and brackets plus 2 divided by 2, we have 10 carbons. So, we're going to plug that in. So, that's 2 times 10, which is 20, there are no nitrogens. So, ignore the N, hydrogen's we have 14, plus 2 divided 2. So, we're going out 20 minus 14, which is 6 plus 2 divided by 2, which gives us an ihd of 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, and 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 them around 7.0-7.2 or so, So, here we have an ihd of four and we have a chemical shift around seven, so that definitely indicates that benzene is evolved. So, here's benzene, we know we have that. Now here what we need to realize next is, benzene, we're saying for H here, which means that it has four hydrogens on it, which means it's a di substituted benzene ring. So, we're taking 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 two methyl groups, 3H most likely means we have one CH3 group and one H here most likely means we have one CH group.

Now, realize, when it comes to multiplicity patterns, what kind of pattern gives us a septet with an integration of one hydrogen and a doublet with an integration of six 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 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 hydrogen's, which means it has four H's on it, which means it's di substituted and the two substituents were isopropyl and just a methyl.

Concept: Practice 4: Proposing Structures from 13C NMR data

Video Transcript

Hey guys let's take a look at the following practice question so here it says propose a structure for the following compound C7H12O2 with the given carbon 13NMR spectral data. Now before we look at all these numbers and values let's first talk about we must always do, we must find the IHD so right here we're going to use the formula of ((2C+N-H) +2)/2 so that's going to be 7 carbons so that's 14-12 hydrogens+ 2 divided by 2 which gives us 2+2 so we have an IHD 2. Now we're going to need room to work this out guys so I'm going to take myself out of the image. Now what is so important about carbon 13NMR? The thing is carbon 13 NMR is usually a useless method for determining our unknown compound because most carbon 13 NMRs are done under broadband decoupled conditions, what does that mean? That means that all the carbon signals will appear as singlets, all of them will just be straight lines and here it helps to unclutter the graph the chart but the bad thing is if everything looks like a singlet it's hard for us to determine what exactly do we have so that's where DEPT comes in handy, DEPT stands for Distortionless Enhancement Polarization Transfer, now why is this so important? It's important because DEPT helps us determine if we have a C, a CH, a CH2 or a CH3, it actually tells us in a way which kind of carbon we have. Now if we take a look at DEPT 90, DEPT 90 is important because DEPT 90 only looks at CH groups and the CH group is called the methane group so if you have a DEPT 90 signal that means you are automatically a CH. Here DEPT 135, DEPT 135 looks at CH2 CH and CH3 so it takes a look at your methane group, it takes a look at your methylene group and it takes a look at your methyl group, here if you're a carbon with no protons then you're not going to appear either DEPT 90 or DEPT 135. Now here talking more about DEPT 135 we're going to say here if you have a signal an arrow that points down that's known as a negative peak and the thing about negative peaks is that negative peaks indicate that you have a CH2 group, OK? So we have 2 points down, here if you're pointing up then you're known as a positive peak, positive peaks can be either CH or CH3. So how do we determine which one do we have? Well here I gave us DEPT 90 signals so here if you're 28 here and you're still 28 here that mean that you have to be a CH 129.8, 129.8 means that you're a CH, this one here is pointing up but it's not part of DEPT 90 so that must mean it's a CH3 group, so we've tagged the identity of the carbons with protons at least.

Now let's go out to the broadband decouple and see if there are any signals in there that we don't have in the DEPT regions so this one we already checked off as being a CH3, this one we checked off as being a CH 70.5 70.5 so this is our CH2, 129 that's our CH2, 129.8 that's our CH and 165.78 never appeared in DEPT 90 or DEPT 135 which means it's a carbon without any protons, now here's a think if you have a signal above 150 ppm or delta then that means that you are a carbonyl group, OK? So we know that for right now that we have a carbonyl group which would explain one of my IHDs. Next what we're going to say here is if you have a signal between about 120 to around 150 then that indicates that you have alkyne hydrogens alkyne carbons, OK? So between 120 to 150 that's alkyne carbons and this is what we're going to say we're going to say here if this CH2 is now a double bonded carbon, an alkyne carbon that means it has terminals so that means it has to be at the end so we can draw it over here CH2 double bond, CH which is connected to this carbonyl.

Alright Next so we've taken care of? We've taken care of this signal, this signal and this signal what else do we have left? We have the 19.1 the 28 and the 70.5 we have to take care of those signals as well and figure out where exactly do we find them. Now we're going to say here that this CH2 is 70.5 normally single bonded carbons are between 0 and 50 so why is this one much higher than 50? Remember what increases your chemical shift? It means you're close to an electronegative group so that CH2 must be close to an electronegative group not just the carbonyl there's two oxygens in here so we assume that this is an ester and that CH2 must be right next to it, now here we're going to say we have this taken care of so that's out so what we have left? We have a signal around 28 and one at 19.1 now if we deduct what we've already figured out from the original formula we can see what we have left to draw. So we have 1 2 3 4 carbons already taken out, we have how many hydrogens? We have 2 3 5 and both oxygens, so what we have left? We have C3H7, we know we have to draw a methyl group we know we have to draw a CH group, that's 2 carbons involved and 4 hydrogens which means we'd still have left behind a methyl so what is this telling me? That must tell me that this 19.1 is not coming from one of methyl it's actually coming from two methyls so the rest of the structure will look like this CH connected to 2 methyls because remember carbon must make 4 bonds so it makes sense that CH must have been connected to two methyl groups. Now granted this is a lot of information Carbon 13 NMR is very difficult but here I'm showing you step by step the approach that you need to take in order to figure out this compound this is considered a very challenging type of question and remember carbon 13 NMR by itself is really not that great and if you're going to do carbon 13 structural determinations it's always best to do that with DEPT 90 and dept 135 together because that gives you the identity of CH, CH2 and CH3 usually we wouldn't only get all the information on carbon 13 NMR we usually have it connected with proton NMR and that gives us a better complete picture of what's going on and throwing in IR gives us an in-depth look at the different function groups present as well so using all 3 of those together is the best way to figure out the structure of your unknown compound.

Concept: Practice 5: Proposing Structures from 13C NMR data

Video Transcript

Hey guys, let's take a look at the following practice question. So, here it says, proposal structure for the following compound C5H10O with the given carbon 13 NMR spectral data. So, 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 in 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, if you have a positive peak, which means it can be either a CH or CH3. Now, it can't be a CH3 because that carbon can't have three hydrogen's on it because if it did it would be making five bonds, carbon at the most can only make four 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 coming CH or CH3, this negative peak here, this downward arrow means it can only be a CH2. So, we know we have a CH2 for sure and then here this is a CH or CH3. So, we know we have at least a CH2. So, we could try to do here is plug it in CH2. Now, let's make some other assumptions here.

So, so far how much of our formula have we used? our formula is C5H10O, we've used two carbons and we've used three hydrogens and the oxygen leaving us with C3H7O. Now here, if we made one of these of CH1 of them a CH3. So, we have one of each then we could be using two carbons and four hydrogens, which will leave us left with what? a CH3, which means what? which means we must have one CH and two CH 3's. So, here we just have to draw this, we're going to say here we have a CH connected to two CH3's. So, this here, would be our unknown compound, again carbon 13 NMR structural determination is much more difficult, not very much information is given, usually would it get a question like this as a standalone question, we will 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 can form connections and see how exactly could this molecule combine together from that unknown molecular formula.

Concept: Practice 6: Proposing Structures from multiple data

Video Transcript

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. So, what we need to do first is we need to figure out what is the ihd of my unknown compound. So, here we're going to say it is 2C plus N minus H, all in brackets, plus 2 divided by 2. So, here we have four carbons, 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 hydrogen's here, plus 2 divided by 2. So, it's going to be negative 2 plus 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 and 3,600 centimeters inverse, this would indicate that we have an alcohol present and remember OH groups are always singlets and the integration of this OH group would be 1, so that means it's either this one or this one. Now, can't be this top one because that's a known attack, here OH groups are always singlets. So, we know that this here, represents an OH.

Now, let's look at the other integration to figure out what else we have, we have 6H here, which most likely means we have two methyl groups, this one H can't be H because all H's are always singlets. So, this would be a CH and then here, this here, would be a CH2. So, if we look, we've got all four carbons that we need. Now, we have to put this all together in a way that will give us those multiplicities. So, we're going to say here is, we must have a CH2 next to the OH. Now, why is that's CH2 a doublet? using the N plus 1 rule this would be one, the neighboring carbon would have to have one hydrogen on it so that N would be one and that's how it become a doublet, that means the CH has to be right next to it and then how is that CH a nonatet. Nonatet means 9. So, 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 two hydrogen's on it, so that means we need another six, where those additional six come from? it would come from two methyl groups that are connected to that CH group. So, 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 group, then we looked at the integration that tell us the types of hydrogen groups that we have, multiplicity next just tells us how they piece together to give us nonatets, doublets etc. and if you need to, 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're the only ones between 6.5 and 8.0, aldehydes because they're between 9 and 10 and carboxylic acids, which are between 11 and 13, since we don't have any of those chemical shifts we don'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.

Concept: Practice 7: Proposing Structures from multiple data

Video Transcript

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, we always want to do first is figure out what exactly is my ihd. So, here we're going to say 2C plus N minus H, all in brackets, plus 2 and all that's divided by two, we have four carbons. So, we're going to plug that in. So, that's 2 times 4, which is 8, we have one 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 I tell us we have 2950, 2950 represents sp3 hybridized carbons, if there were any double bonded carbons involved then we would have a signal between 3,000 and 3150, since we don't that means there are no double bonds present therefore this must be a ring. So, remember ihd could be a double bond or a ring can't be a ring here though. Well, can't be a double bond here, because a double bond would be 3,000 to 3150.

Now, here we have a signal around 3400 centimeters inverse and because we have a nitrogen here means we have an amine, so the amines 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 will have an integration of 1h. So, here this 1h integration means that we have an NH present, we're going to say next is somehow we have 4-h, 4-h, which most likely means we have two CH2's. So, we have to form our 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 two carbons. So, here the best possible way to do this structure it would have to look like this, it'd have to be symmetrical. So, we have an NH here, it'd be connected to two CH2's and those two CH2's be connected to other CH2's, if we look these two look the same and they're both connected in the same way. So, they'll represent 1 proton signal and then these two CH2's look the same and they're connected in the same exact way. So, there will be our second signal, doing N plus 1, these would be one neighboring carbon has two hydrogens on it. So, that'd be 2 plus 1 that'd be a triplet and then the same thing would be done here, this would be 1, neighboring carbon is this carbon, we can't count this as the neighboring carbon since they have the same exact signal. So, your neighboring carbon would have 2 hydrogen's, which give us 3, so it also would be a triplet. So, in this case this would be our compound, a cyclic secondary amine.

Structure Determination without Mass Spect Additional Practice Problems

A compound with the molecular formula C7H14O2 is responsible for the fragrant smell of bananas. In either acidic or basic water, it decomposes to acetic acid and an alcohol. Given the NMR spectra below, draw the structure of this compound.

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What is the structure of the compound in the following 1H-NMR spectrum with the molecular formula C6H12 ? Relative integration is shown.

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Determine the likely structure for a compound A (C6H10O), which is found to decolorize bromine in carbon tetrachloride. Its spectral data is as follows:

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Deduce the structure of an unknown compound with molecular formula C5H12O using information given by its infrared spectrum.

Intensity (peak):  Frequency (cm–1):

m                         3300

m                         2900

m                         2800

m                         1465

m                         1450

m                         1375

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Draw the molecule with a molecular formula of C 5H8O2 described by the spectra below.

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A compound with a molecular formula C4H4N2 has the following 1H-NMR spectrum.

Show the structure that is consistent with the spectra in the labeled box. Show work for partial credit

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Given an organic molecule with the molecular formula C9H9N and the following IR and 1H NMR spectra, determine the structure of the molecule. 

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A compound with the formula C8H10O produces six signals in its 13C NMR spectrum. The 1H NMR spectrum is tabulated below. Provide an unambiguous structural formula for the compound from the data provided. 

1.19 ppm (3H triplet)

2.58 ppm (2H quartet)

4.65 ppm (1H broad singlet)

6.70 ppm (2H doublet)

7.05 ppm (2H doublet)

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A compound with the formula C7H14O has strong IR signals near 2900 and 1720 cm –1. The 1H NMR spectrum is tabulated below. Provide an unambiuous structural formula for the compound from the data provided.

1.07 ppm (6H doublet)

2.77 ppm (1H septet)

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Draw the structure of the compound that produces the 1H-NMR spectrum below. The molecular formula is in the upper left of the spectrum. Make sure to put your final answer in the box provided.

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Which of the following structures with molecular formula C8H9Br is consistent with the the data below:

1H NMR: 1.3δ (triplet, I=3H), 2.65δ (quartet, I=2H), 7.2δ (doublet, I=2H), 7.5δ (doublet, I=2H)?


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Provide a structure for the compound with molecular formula C10H12O2 and with the following spectroscopic data.

IR: 1680 cm–1, 2750 cm–1, 2850 cm–1

1H NMR: 1.3δ (triplet, I = 3H), 3.6δ (quartet, I = 2H), 4.5δ (singlet, I = 2H), 7.3 (doublet, I = 2H), 7.7δ (doublet, I = 2H), 9.9δ (singlet, I = 1H)

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Identify the following carbonyls with molecular formula C4H8O from their  1H NMR.

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Predict the structure of the compound that gives rise to the following  1H NMR spectra. 


A) Molecular Formula: C4H8O

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An unknown compound has the molecular formula of C6H11O2Cl. The NMR spectrum of the compound in chloroform is presented below. The IR spectrum shows a strong peak at 1735 cm -1. What is the structure of the molecule? Draw your final solution below.

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A compound with molecular formula C12H24 exhibits a 1H NMR spectrum with only one signal and a 13C NMR spectrum with two signals. Deduce the structure of this compound.



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