Propose a structure for the following compound, C₇H...

Question

Propose a structure for the following compound, C₇H₁₂O₂ with the given ¹³C NMR spectral data:

Broadband decoupled ¹³C NMR:19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ

DEPT-90:28.0, 129.8 δ

DEPT-135:19.1 δ (↑), 28.0 δ (↑) , 129.8 δ (↑) , 70.5 δ (↓) & 129.0 δ (↓)

Accepted Answer

Hey, guys, let's take a look at the following practice Question. So here it says proposal structure for the following compound C seven H 12 with the given Carbon 13 NMR spectral data. Now, before we look at all these numbers and values, let's first talk about what we must always do. We must find the HD. So we're here. We're gonna use the formula of two seats plus and minus h all in brackets plus two divided by two. So that's gonna be seven carbon. So that's 14 minus 12 hydrogen plus two divided by two, which gives us two plus two. So we have a nice HD of to now. We're gonna need room to work this out, guys. So I'm gonna take myself out of the image. Now, what is so important about carbon 13 NMR? The thing is, carbon 13 NMR is usually a useless method for determining are determining are unknown compound because most carbon 13 and a mars are done under broadband decoupled conditions. What does that mean? That means that all the carbon signals will appear as single. It's all of them will just be straight lines, and here it helps toe unq lever, the graph, the chart. But the bad thing is that everything looks like a single it It's hard for us to determine what exactly do we have, So that's where depth comes in. Handy depth stands for distortion. Lis enhancement. Polarization, transfer. Now, why is this so important? It's important because depth helps us determine if we have ah c A C h a c h two or ch three. It actually tells us, in a way, which kind of carbon we have now if we take a look at Depp 90 90 is important because Depp 90 on Lee looks at CH groups and the C H group is called your methane group. Okay, so if you have a 90 signal, that means you are automatically a CH group here. Depth 1 35 1 35 looks at ch two ch and CH three. So it takes a look at your meth in group. It takes a look at your methylene group and it takes a look at your metal group here. If you're a carbon with no protons, then you're not going to appear in either depth 90 or depth 1 now here talking a little bit more about depth. 1 35. We're going to stay 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 CH two group. Okay, so we have to pointing down here. If you're pointing up, then you're known as a positive peak. Positive peaks can be either ch or ch three. So how do we determine which one do we have? Well, here I give a step 90 signals. So here, if you're 28 here and you're still 28 here, that means that you have to be a ch 1. 29.81 29.8 means you're a C h. This one here is pointing up, but it's not part of depth 90. So that must mean it's us ch three group. So we've tagged the identity of the carbons with protons. At least now let's go up to the broadband e couples and see if there any signals in there that we don't have in the depth regions. So this one we already checked off as being a C H three. This one we checked off is being a C H 75 75. So this is our CH two 1 29. That's our CH two 1 29.8 that their CH and 65.78 never appeared in 90 or depth 1 Which means it's a carbon without any protons. Now, here's the thing. If you have a signal above 150 ppm or Delta and that means that you are a Carbondale group. Okay, so we know for right now that we have a Carbondale group, which would explain one of my eye HDs Next. We're gonna say, Here is if you have a signal between about 1 22 about run 1 50 on that indicate that you have all keen hydrogen out king carbons. Okay, so between 1 21 50 or so, that's all King carbons. And this we're gonna say we're going to stay here. If this ch two is now a double bonded carbon and all keen carbon, that means it has to be terminal, so it has to be at the end so we can draw it over here Ch two double bond ch. Which is connected to this carbon. Neil. Alright. Next. So we've taken care of what we've taken care off this signal. This signal in this signal, what else do we have left? We have the 19.1, the 28 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 CH two is 70.5. Normally single bonded carbons are between zero and 50. So why is this one much higher than 50? Remember, what increases your chemical shift? It means you're close to an electoral negative group. So that's ch two. Must be close to an electro negative group. Not just the carbon. Neil. There's two oxygen's in here. So we assume that this is an Esther and that ch two must be right next to it. Now, here, we're gonna say we have this taken care off, so that's out. So what do we have left? We have a signal around 28 1 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 1234 carbons already taken out. We have How many hydrogen is we have to and both oxygen. So what do we have left? We have C three h seven. We know we have to draw a method group. We know we have to draw. See each group. That's two carbons involved. And four hydrogen. Which means we still have left behind a metal. So what is this telling me? That must be that must tell me that this 19.1 is not coming from one metal. It's actually coming from two metals. So the rest of the structure would look like this ch connected to two metals. Remember, Carbon must make four. Bonds would make sense that ch must have been connected to two metal groups. Now, granted, this is a lot of information. Carbon 13 and Amar 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 NMR structural determination, it's always best to do it with depth 90 and depth 1 35 together that gives you the identity of ch ch two and CH three. Usually we wouldn't get on Lee 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 functional groups present as well. So using all three of those together is the best way to figure out the structure of your unknown compound.

Difficoltà con Organic Chemistry?

Propose a structure for the following compound, C₇H₁₂O₂ with the given ¹³C NMR spectral data:
Broadband decoupled ¹³C NMR:19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ
DEPT-90:28.0, 129.8 δ
DEPT-135:19.1 δ (↑), 28.0 δ (↑) , 129.8 δ (↑) , 70.5 δ (↓) & 129.0 δ (↓)

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Set di pratica su Structure Determination - Molecular Sentences

Difficoltà con Organic Chemistry?

Propose a structure for the following compound, C7H12O2 with the given 13C NMR spectral data: Broadband decoupled 13C NMR:19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ DEPT-90:28.0, 129.8 δ DEPT-135:19.1 δ (↑), 28.0 δ (↑) , 129.8 δ (↑) , 70.5 δ (↓) & 129.0 δ (↓)

Guarda il prossimo

Set di pratica su Structure Determination - Molecular Sentences

Propose a structure for the following compound, C₇H₁₂O₂ with the given ¹³C NMR spectral data:
Broadband decoupled ¹³C NMR:19.1, 28.0, 70.5, 129.0, 129.8, 165.78 δ
DEPT-90:28.0, 129.8 δ
DEPT-135:19.1 δ (↑), 28.0 δ (↑) , 129.8 δ (↑) , 70.5 δ (↓) & 129.0 δ (↓)