Draw the ¹³C NMR spectrum of each molecule in Assessment 15.48.

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Alright. Hi, everyone. So for this question, let's go ahead and draw the expected carbon 13 nr spectrum for the given molecule, which happens to be Hept four I 20. So recall that the carbon 13 NMR spectrum of a given compound can give us information as to what types of carbons might be present in a molecule. And when I say types of carbons, I'm referring to really how many non equivalent carbons there are in a given molecule. No factors like symmetry and rotation around single bonds can make multiple carbons equivalent and therefore contribute to the same signal. But if we consider the structure of HEPT for I to own, there is no such symmetry, right? So if we consider all of the carbons that are on this molecule, they're all going to be distinct from each other, which means that in total, all seven carbons are going to be distinct from each other, which means that we're going to see seven different signals in our carbon 13 spectrum. So for the sake of identification, I'm going to go ahead and label all of my carbons A through G in accordance with the IUPAC system, right. So carbon A is carbon one in the parent chain and so on. So now that we have an idea on how many signals, right, we have to consider what's called their chemical shift. Now recall that the chemical shift is expressed in units of parts per million. And those are the numbers that encompass the x axis of your NMR spectrum, so to speak. Now, the chemical shift in parts per million represents the location of any given signal on your spectrum. And the location, right, the chemical shift is highly dependent on what type of carbon that particular signal represents. Right now, we can compare the location of two different signals, two or more signals by describing them either as shielded or D shield it. So to understand the concept of shielding and de shielding, right, recall that electrons are which surround a given nucleus and shield it physically from NMR effects. So a shielded signal, right, a shielded nucleus is one that is less exposed to the NMR effects. And because of this right shielded signals shield, the nuclei are said to be relatively up field. In other words, they tend to occupy the right side of a carbon 13 in a more spectrum and their chemical shift values tend to be lower or smaller. Now, by contrast, right, a signal can be d shielded if that nucleus happens to be in close proximity with anything that can withdraw electron density away from it. Therefore, exposing that nucleus more to NMR effects. So the D shielded signal is going to therefore appear relatively downfield in the spectrum. Or in other words, occupying the left side of the spectrum with larger values for the chemical shift. Now recall also that Alcaine carbons tend to be among the most shielded signals and will therefore appear farthest to the right in the spectrum. So let's start with those. And if I look to the very right side of the structure of HEPT four I two own, I can start with carbons G and F because carbons G and F are all cane carbons, they are all cane carbons and generally speaking, they would fall within the range of 5 to 45 parts per million. Now, what you'll notice is that I haven't highlighted two other Alcaine carbons, right? Because both A and C are sp three hybridized as well. But the reason why I'm referring to them separately is because the proximity of ANC to the carbon, carbon B is going to affect their chemical shift relative to G and F. Now because A and C are also all caned carbons, they're going to appear approximately between 5 to 45 parts per million as well. But we have to consider the effect of the car bal right next to both A and C, right. So because the car bono itself is highly electron withdrawing the signals for carbons A and C are going to be more downfield than those of G and F. So within the range of 5 to 45 parts per million, essentially, you have to make sure that A and C are more to the left than G or F. So from here, we can go ahead and focus our attention on carbons, D and E which are from the all kind. Now, all kind carbons also tend to be relatively de shielded because of their sp hybridization. And generally, you can find them between the range of 65 to 100 parts per million and last but not least, right, we've got carbon B, which is the car bono carbon. Now, the car bono carbon in this structure at least is going to be the most de shielded due to the fact that it's directly connected to our oxygen, right? Because our carbon carbon has a partial positive charge due to the dipole moment between itself and oxygen, which means that it's going to be highly d shielded and carbon carbons tend to be found within the range of 160 to 210 parts per million. So now that we understand where each signal is supposed to be approximately. Right. Let's go ahead and include the approximate values of the chemical shifts on the actual structure of HEPT four I 20. And I'll start off with carbon A that should have a chemical shift of approximately 29.7. Up. Next is B with A chemical shift of 206.0 followed by sea at 30.8 all of which are in units of parts per million by the way. But carbon D has a chemical shift of approximately 80.3 parts per million and E at around 82.3 F and G for that matter again, are all cane carbons. So they're going to appear farthest to the right at 12.3 points a million for F and 13.9 parts per million four G. So it's at this point, right, that we can go ahead and begin to draw our spectrum. So the first thing I'm going to do is draw my scale. So here's my skill and always specify that you're talking about units of port chari. So that's why I'm writing it on the bottom. Now, the thing about carbon 13 NMR is that there's a very wide range, right when it comes to the values of the chemical shifts. So on the very left side or right side, excuse me is my zero. And I'm going to range from zero to approximately 220 and I'll be going in 20 parts per million increments. So here I'm just adding tick marks with my values in 20 parts a million increments and I'm actually going to have to extend this. So thats why I added some more space over here. Oops, here we go. That's better. So now that I have my skill, right. I can proceed with actually drawing my peaks and I'll start off with my car bal on the very left side at approximately 206. So I'll be drawing that one just above 200. And then up next were my all kind carbons which if you recall were add 80.3 and 82.3 parts per million respectively. So they should be relatively close together from there. I've got the all cane carbons that were next to the car bal which was A and C at 29.7 and 30.8. So again, also relatively close together. And then lastly were my two other alkane carbons F and G at 12.3 and 13.9. So those are going to be the ones farthest to the right and there you have it here is your completed carbon 13 nr spectrum. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Draw the ¹³C NMR spectrum of each molecule in Assessment 15.48.

Key Concepts

Nuclear Magnetic Resonance (NMR) Spectroscopy

Chemical Shift

Integration and Multiplicity

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