Describe the ¹H NMR spectrum you would expect for e...

Question

Describe the ¹H NMR spectrum you would expect for each of the following compounds, indicating the relative positions of the signals:

d.

Accepted Answer

Right. Hi, everyone. So first question, let's describe the expected proton and Mr spectrum and indicate the relative position of the signals for the given compound. So on the right side of the screen here, what you'll notice is that I drew the same molecule again, but I decided to go ahead and highlight all of the protons that were given in this molecule, right. So now let's start talking about um proton and Mr, right, recall that every signal in a proton in Mr spectrum is split into a very number of peaks, right? So the number of peaks, if you recall in a given signal is equal to N plus one, we're in is equal to the number of equivalent protons that are nearby, right, that are adjacent. So now I'd like to point out really quickly that this molecule does in fact have an element of symmetry to it. So that therefore will actually decrease the amount of signals that you expect to see in the proton and Mr spectrum, right? Because notice how in my five carbon chain, I can draw a line down my third carbon and split the molecule in half, so to speak, And from there, I'll notice that I actually have a total of three unique signals because my two methyl substituent are going to be equivalent on both sides. Right. So in my diagram, on the right side, I'm actually going to change the color of my protons here to blue to indicate that they are all similar to each other. So my blue protons here correspond to one signal and those are all my methyl groups. Now, additionally, right, I have an additional methane carbon that has just one proton and there are two of them, but this molecule is symmetrical, right? So those two methane protons are going to correspond to the same signal which I'm going to indicate in green. And lastly, right, the methylene carbon at the center or carbon three in my parent chain that is going to correspond to its own unique signal, which I'm going to indicate in red. So now for each signal, right, for each position, we have to indicate what it's going to look like, right? Is it going to be a quartet, a triplet, et cetera. So let's start first with our methyl protons. And here all of our methyl protons are next to just one non equivalent proton, right, which is the methane proton that's directly adjacent to it. So because of that, right, my, my splitting here is N plus one, my number of peaks. So I have one equivalent adjacent proton plus one equals two, which means that this signal is going to be a double it, right? Because recall really quickly that equivalent protons do not split each other, right. So when we're trying to figure out the multiplicity, we're not interested in equivalent protons, only non equivalent ones. So now for our methane proton, the methane proton is surrounded by quite a few protons. And specifically, they're surrounded by two Meho groups, each, right. So for each individual methane proton, we have six neighbors, so to speak. So N equals six, now six plus one equals seven, which means that for the methane proton, our signal is going to be a septet. And lastly for methylene protons, our methylene protons are surrounded by the two methane protons. So that's N equals two plus one, gives me three peaks total, which corresponds to a triplet. So now that we've figured out what the spectrum is supposed to look like, we have to understand the relative positioning of the signals in the given compound. Now here recall really quickly that on the left side of the spectrum are the more d shielded protons and on the right side are the more shielded. Now, what does it mean to be de shielded versus shielded? Right, de shielded protons, for example, are bound to electron negative groups, right? Because electron negative groups draw electron density away from the proton and therefore expose the nucleus more. So therefore, d shielded protons are going to appear more on the left side of the spectrum. But here, right, notice how our only atoms here are carbon and hydrogen. Now carbon, it's not that electron negative, but it is more electron negative than hydrogen, right. So when a proton is surrounded by more carbons, it is relatively de shielded and therefore will appear on the left side of the spectrum. So our methane carbon is surrounded by three other carpets, right. So the methane protons are therefore going to be the most de shielded because it is surrounded by more carbons, which in this case is the more electron negative atom since we don't have any hetero atoms here. So if on the bottom here, I'm going from left to right in my spectrum, then my sept should be on the very left side since it is the most de shielded and going by that same logic, right, my triplet is going to be somewhere in between because the methylene protons are surrounded by two other carbons. And then lastly, right, your Meho groups are going to be the most shielding because the methyl carbons are attached to only one other carbon. So that is going to shield the nucleus the most. And therefore our doublet which corresponds to the methyl groups is going to be on the right side of the spectrum. So this is going to be your final answer, right, going from left to right in the NMR spectrum, we're going to see a septet followed by a triplet followed by a doublet. So with that being said, thank you very much for watching and I hope you found this helpful.

Describe the ¹H NMR spectrum you would expect for each of the following compounds, indicating the relative positions of the signals:
d.

Key Concepts

1H NMR Spectroscopy

Chemical Shift

Integration and Multiplicity

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