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

Question

Accepted Answer

OK. Hi, everyone. So for this question, let's draw the estimated carbon 13 NMR spectrum for the given compound. Now recall that a carbon 13 NMR spectrum gives us information as to how many non equivalent carbons there happened to be in a given molecule, right? So for for every signal you see that will correspond to one non equivalent carbon. Now recall that factors like molecular symmetry and rotation along single bonds can make multiple carbon atoms equivalent to each other, therefore corresponding to the same signal. But if we consider the structure of the compound given, there is no such symmetry. And in addition to there being no such symmetry, I want to draw your attention to these two metal groups that are connected to the sp two carbon of the alkene or one of them, right? Because these two methyl groups are connected to the same carbon, it would be very easy to assume that they are actually chemically equivalent. But we have to recall that rotation along a double bond is restricted. So because rotation is not possible, these two metal groups are actually non equivalent, right? They are different from each other, which means that they would therefore correspond to two different signals. Now, to showcase this, I'll be labeling each metal group A and B to showcase that they are two different signals and the carbon they're connecting to I will label AC and the second sp two carbon of the all keen I will label as deep. So at this point, right, because there's no such symmetry in this compound, all four carbons of the ring are also going to be unique. So I'll be labeling them E through G or E through H. So because I've labeled here all eight carbons each with their own letter, right. That means we're going to see a total of eight signals in this carbon 13 NMR spectrum. So now let's talk about the chemical shift. Now recall that the chemical shift is expressed in units of parts per million. And the chemical shift describes to us the location of a given signal right on the spectrum itself. So here, depending on how shielded or de shielded, a signal happens to be, it's going to appear in different regions of the spectrum itself. Now recall that electrons are what surround a nucleus and protect it from NMR effects. So a signal is described as being highly shielded when electrons are surrounding it to protect it from NR effects. Now shielded signals appear more upfield or in other words, for this to the right side of the spectrum itself. And by contrast, right D shielded signals are more exposed to NMR effects because the electrons that are supposed to surround them have somehow been drawn away. Like for example, if that nucleus is connecting to something that is electron, so d shielded signals are described to be down field down for or farther to the left in the spectrum. So first, let's start off with our methyl carbons, right? Specifically A and B now carbons A and B are primary sp three hybridized carbons and al carbons in general are some of the most shielded signals that we might come across, right. So those are the ones that are going to occupy the right side of the spectrum generally between five and 45 parts per million for all cane carpets. Now A and B specifically, in addition to being all cane carbons, they're also primary. So their chemical shifts are going to be among the lowest at approximately 18 and 25 parts per million. Now, the next carbons that I notice are E and F and even though E and F are going to fall more so on the right side of the spectrum relative to everything else, right? We have to consider the effect of the oxygen right next to both E and F, right? Because the proximity to the oxygen which is highly electron is going to be shield both E and F significantly, right? But in addition to carbon E specifically being next to the oxygen, it's also next to one of the SP two hybridized carbons of the A K, right. So that is going to actually D shield E slightly more than so, carbon E here is going to appear relatively d shielded at approximately 76 parts per million. Whereas F is going to be slightly more shielded though it is still relatively de shielded overall at 74 parts Fermi. Now, in addition to having carbons E and F in the ring, we also have to discuss carbon G because carbon G is a relatively ordinary secondary sp three hybridized carbon, right, because it's not adjacent to the oxygen and it's not adjacent to the All Keen either. So that should place carbon G here at approximately 39 parts per million an H here right next to the bromine is going to be placed at around 43 ports for million. Now, again, relatively d shielded due to its proximity to the electron romi. So now the only signals that we have left to address are C and D from our All Keen. Now C and D are going to be d shielded due to the effect of the pi bond. And generally speaking, all keen carbons tend to appear between 100 to 140 part 3 million. So now we can go ahead and apply this to our actual spectrum. So here I'm going to try and clear up some space and I'm going to go ahead and draw my spectrum here on the right side now here, right, my parts are really in values for the chemical shift are going to increase, going from right to left. So I'll be putting zero on the very right side and ill be increasing in increments of 20 parts per million. So because the most de shielded value in my spectrum goes up to 140 parts per million in the case of my keen. Right. That is where I'm going to stop. So I can actually make this line slightly smaller and I can move this a little bit more and recall to always specify your units right, in parts per million. So now, right, based on the approximate values for chemical shift, you go ahead and you place your signals accordingly. So here I'll be starting off with my all Keen carbons because those are going to be the most d shielded or the farthest to the left. So the approximate chemical shift values send it to be around 135 and approximately 127. So those are the All Keen carbons. And now my next ones are going to be E and F at 76 and 74 respectively. And I also want to mention here that because each peak corresponds to one carbon, all of your peaks should be the same height up next. I'll be putting my signals for H and G at 43 and 39 respectively. And last but not least I have A and B at around 18 and 25 and there you have. So here is your approximate carbon 13 and A R spectrum. So if you stuck on 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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