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

Question

Accepted Answer

All right. Hi everyone. So for this question, let's go ahead and draw the possible carbon 13 and a more spectrum for the following compound. So recall first and foremost, right, that a carbon 13 and a more spectrum gives us information as to how many non equivalent carbons exist in a given molecule. So every signal you see in a carbon 13 NMR spectrum is going to correspond to one type of non equivalent carbon that's present. And recall that factors such as molecular symmetry and rotations along a single bond can make multiple carbon atoms equivalent thus corresponding to the same signal. And if we look at the structure given here, there is actually an element of symmetry because I can draw an imaginary line right, going through the benzene ring itself. So what that means is that I'm going to have less signals than there are carbons in this molecule, right? Because I'm going to see multiple sets of equivalent carbons. So in this case, right, the methyl carbon of the meth boxy group is going to correspond to the first signal which I will label eight. And so the carbon connecting the meth oxy group to the benzene ring is its own signal as well, which I will label B no, the carpet of the benzene ring that are Ortho to the one that contains a myox group are equivalent due to the symmetry that we discussed previously as are the ones that contain the isopropyl groups. So I'll be labeling only one of them, right? And I will label one set C and the other set D and the carbon that bears the bromine in the benzene ring is its own signal as well, which I will eat. Now, both isopropyl groups are going to be equivalent and both of them or rather the isopropyl group will provide two distinct signals or two new ones. One from the tertiary carbon connecting to the benzene ring and the other from the metal group. Now, it's worth mentioning that both methyl groups attaching to the carbon that I labeled f both of them are equivalent because rotation around the single bond is possible. So thats why I will label only one of them G. So now that we have an idea on how many signals we expect to see, right. In this case, seven, let's go ahead and talk about the chemical shift. Now recall that the chemical shift is expressed in units of parts per million and it's a description of the location of a given signal on the spectrum itself. No, the chemical shift is dictated by what type of carbon you happen to be examining because there are multiple factors that can shield or de shield any given signal. So the reason why I'm redrawing our starting material on the bottom here is to showcase this principle of shielding versus de shielding. Because resonance with the benzene ring is going to significantly affect the chemical shift of our carbons particularly on the ring. So for this example, we have to consider how the metox group interacts with the benzene ring and recall that it does so by donating a pair of electrons towards the ring. So as it donates electrons towards the ring, right, what it does is create a carbon ion on the carbon labeled seat. So here's a pro here are my isotopic groups and now there's going to be a double bond in between Carmen and my meth oxy oxygen, therefore giving it a positive charge, whereas there's going to be a negative charge on the carbon labeled seat. So now the anion is going to alternate throughout the ring. And in the next resonance structure, that negative charge is going to be on the carbon labeled eat. So here's a negative charge and the positive charge on oxygen still my isopropyl groups and my bromine attached. Now, the reason why I'm bringing up the resonance structures even though I could draw more, right? But for the sake of our discussion, let's just go ahead and notice the change in electron density of the carbon labeled sea and eat because recall that electrons surrounding a nucleus are which shielded from NMR effects, therefore making it relatively shielded. So because resonance structures demonstrate a high electron density at carbon C and E, they are going to appear relatively shielded in their respective range. So let's keep that in mind because we call that carbon's in a benzene ring are going to fall between approximately 120 to 150 parts per million. But carbon C and E specifically are so shielded due to the resonance structures that they're going to fall actually beneath this range. So carbon C and E would fall actually beneath the aromatic range or just under the aromatic range, specifically at approximately 109 parts per million for sea and approximately 113 parts per million 48. So even though they're both carbon on the aromatic ring, the resonance structures dictate that they're going to be incredibly shielded relative to the others. No, if carbon C and E are relatively shielded because of the resonance structures, then we also have to consider carbon B because even though carbon B is part of the aromatic ring, it's also attached to the oxygen in the meth boxy group. So therefore, right B is going to be obtained more downfield even beyond the range of aromatic carpets. So, B specifically would appear at around 159 parts per million. So just beyond this limit here for the aromatic carbons and I mentioned that earlier, you could draw more resonance structures and technically, you could, but due to the element of symmetry present in this molecule, they would be equivalent to other structures. So I just wanted to correct that, but now we can go ahead and address the remaining aromatic carbon specifically D. So here's D, now D is going to fall within the range of 122 150 parts per million, namely, it was approximately found at 144 parts per million. So now let's talk about our all canne carpets. Now, the al carbons tend to fall within the range of between approximately 5 to 45 parts per million. But carbon A right, even though it isn't all can carbon, it's an sp three hybridized carbon, it's also attached to an electron atom, name me the oxygen. So therefore, it's going to be de shielded and it's going to fall slightly beyond the general range of all cane carbons because when an alkane carbon is connected to an electron atom, they tend to fall within the range of 30 to 80 parts per million instead. So carbon A would show up at around 56 parts per million but compounds F and G would fall within the typical alkene range. So carbon F appears at approximately 30 parts per million and G at around 23 parts are really. So now we can use this right to finally draw our spectrum. So I'll be scrolling all the way down to where I have some space. And I'm going to start by drawing my X axis now because my maximum range was approximately 161 170. That's where I'm going to stop. Right. So I'll be drawing or marking 160 parts per million on the very left side, the showcase where it ends. And at this point, I'm adding new tick marks in increments of 20 parts per million ending off at zero. So now I'll be adding my peaks as discussed earlier. But there's one more thing to keep into consideration, right? Because even though carbon 13 NMR doesn't tell you how many peaks there are exactly or rather how many carbons correspond to one peak. Exactly. We can distinguish between relative amounts based on peak height. So for example, right, because four equivalent carbons would contribute to the signal G, the peak illustrating signal G should be among the tallest, right? It should have the highest ratio so that one was at approximately 23. And so that will be followed by carbon F at a round 30. Now, carbon F corresponds to two equivalents, right? Two equivalent carbons contributing to the same signal. So I have to ensure that it's going to be shorter than the peak corresponding to carbon G. And then for carbon A, carbon A falls at around 60 parts per million. So carbon a should be among the shortest, right? Because it corresponds only to one non equivalent carbon. So the idea is to just make sure that your ratios are approximate based on the peak height. So up next are my shielded aromatic carbons, right? C at around 109 which corresponds to two carbons and then e at around 113 corresponding to only one carbon. And then I've got my remaining aromatic carpets, right. I've got d falling within the range at around 143 144 corresponding to two equivalent carbons. So that's right here. And last, but not least is the one for signal B carbon B at 159 and there you have it, we have made it to the complete carbon 13 NMR spectrum. So if you stuck around to the end, I appreciate you for sticking around. 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 Shifts

Integration and Peak Multiplicity

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