Assign each signal in the ¹³C NMR spectra to the molecule show...

Question

Assign each signal in the ¹³C NMR spectra to the molecule shown.

(a)

Accepted Answer

All right. Hi, everyone. So for this question, lets go ahead and assign each signal of the given compounds. Carbon 13 in a more spectrum to the corresponding labeled carbon atoms. And here our starting material or rather a compound is for promo two methyl but two in. So let's get started. Right. Recall that carbon 13 nr is useful because it helps us determine how many non equivalent carbons there are in a given molecule. No, the number of signals is going to correspond to the number of non colon carbons in your compound. So because we have five signals, then we're going to see a total of five non equivalent carbons as indicated by our labels here. Now, the location of each signal is referred to as the chemical shift and the chemical shift varies depending on whether or not a given at or a given nucleus specifically is shielded or de shielded because when a given nucleus is shielded, what that means is that there are more electrons surrounding it to protect it from the NMR effects. Now, as a result of that specific nucleus being more shielded, then the location of the signal or the chemical shift is described as being relatively up field or towards the right side of the NMR spectrum. Now, by contrast, a nucleus can also be d shielded because there would be something in close proximity to the nucleus itself that removes electron density away from it. Therefore exposing it more to an MR effects. So as a consequence of that right, the chemical shift of ad shielded nucleus would be or would appear more downfield or to the left side of the spectrum. Now, depending on what type of carbon we happen to have, we can more or less predict where it's going to fall in a carbon 13 nr spectrum, which is useful for identification such as for this question, right? So the first one that I want to take into consideration are the all can carbons because recall that all canine carbons are some of the most shielded nuclei in any given organic molecule, which means that they would therefore appear the most upfield or the farthest to the right in the spectrum. And what we'll notice here in the structure of four bromo, two methyl be two E is that there are three all cane or sp three hybridized carbons in our structure, the carbon labeled A the carbon labeled D and the carbon labeled E. So because we have three canne carbons, right? We've got a excuse me, D and E, all of them are all cane carbons because all of them are sp three hybridized which means that all of them should fall more to the right side of the spectrum. However, there's also a distinction to be made between these three as well because carbons D and E or primary sp three hybridized carbons considering that they both appear like methyl groups in their structure. So because of this carbons D and E are expected to fall in the range of approximately 5 to 45 parts per million. So the two signals here that are farthest to the right would be the signals corresponding two carbons D and E. Now it's worth mentioning that both carbons D and E are connecting to an all keen carbon, right? An SP two hybridized carbon. So that is actually going to restrict the rotation of carbon's DNE relative to the double bond. And the reason that matters is because that make carbons D and E actually non equivalent. So they would give different signals. Now, if carbon C by contrast was a single bond, right? It was sp three hybridized, then carbons D and E would be equivalent because then rotation would be possible, right? But in this case, it's not now of the three alkane carbons that we have to address, right. Carbon A is an sp three hybridized carbon next to oops next to an electron group such as Bromy. So due to its proximity to the electron bromine, the signal for carbon A is going to be more D shielded compared to D and E, which means that it would fall approximately within the range of 30 to 80 parts per million. So the signal here to the left of D and E just before 40 parts per million would be due to carbon eight. And it's at this point here that we only have left to address the two ALEN carbons, which would be signals be in C now, like I said previously, right? Single excuse me signals B and C correspond to the all clean carpets and O carbons are going to be the most D shielded in this molecule due to the presence of that Pyon. Now recall it, generally speaking, all keen carbons tend to appear between 100 to 140 parts vermillion. So the two signals here to the very right side or the left side of the spectrum that fought in between 121 140 parts per million. These two peaks would correspond to C and B and here we've accounted for all five carbons, right. So going from left to right in the spectrum, we've got CB A, sorry about that ad and finally eat and there you have it. So if you stuck around to the end, thank you so very much for watching. And I hope we found this helpful.

Assign each signal in the ¹³C NMR spectra to the molecule shown.
(a) <IMAGE>

Key Concepts

¹³C NMR Spectroscopy

Chemical Shift

Signal Assignment

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