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:

b. CH₃OCH₂CH₂CH₂Br

Accepted Answer

Right. Hi, everyone. So first question, let's draw the expected proton in a more spectrum and indicate the relative position of the signals for the given compound. So on the right side of the screen, you'll notice that I drew the same structure, but I decided to go ahead and highlight the protons that are associated to each carbon. And the reason I did that right was to understand the splitting pattern of each signal in our proton and Mr spectrum, right? Because recall that in proton and Mr An in an individual signal is split into a different amount of peaks. So the number of peaks in an individual signal is equal to N plus one and N corresponds to the number of equivalent protons on adjacent carbons. Yeah, because recall it the signal of equivalent protons do not split each other, right. So we're only interested in non equivalent protons on adjacent carbons. So here we have four carbons, right? And each of them has protons. And because there is no symmetry in this molecule, then all four carbons are going to give a distinct signal in the proton NMR spectrum, right. So when I go from left to, right, I'm going to number my carbons one through four to make referencing them a little bit easier. So now let's talk about the splitting pattern for each of them. So first I'm going to start off with the carbon that I've labeled number one on the very left side of the molecule, right? So for number one, notice how carbon one is only bound to carbon two, which has two equivalent protons. So for my number of peaks, for my number of peaks, I'm going to have N equals two plus one equals three, right, which indicates that signal one here is going to be a triplet and signal two. On the other hand is surrounded by carbon one and carbon three. Now carbon one has three equivalent protons and carbon three has two equivalent protons. So for my signal two, my number of peaks in the individual signal is going to be equal to N equals five plus one equals six, right, which indicates that signal two is going to be a multiple. Number three. On the other hand, number three is bound to carbon two and an oxygen which has no protons, right. So for number three, my only neighbors, so to speak are the two protons on carbon two. So N in this case, is going to be equal to two plus one equals three, which indicates that we have another triplet. And lastly, right, for carbon number four, notice how carbon number four is bound to oxygen and bromine, neither of which have any protons associated to them, right? So carbon four here doesn't have any neighbors. So N equals zero plus one equals one, which indicates that we have a single for signal four. So now that we understand what the spectrum looks like a little bit better. Let's go ahead and talk about the relative position of these signals in the spectrum itself, right? Because recall that the proton environment makes a huge difference in the positioning of that signal in the NMR spectrum. Because when a proton is in an electron rich environment, the nucleus is going to be better concealed, right. So therefore, it is going to be more shielded and it is going to appear on the right side of the spectrum. On the other hand, electron poor environments result in the D shielding of the nucleus which results in a signal appearing on the left hand side of the spectrum. Now, here we'll notice that one of our signals is bound to two highly electron negative atoms. In this case, carbon four is bound to oxygen and bromine on either side of it that creates an incredibly electron poor environment because both oxygen and bromine are removing electron density away from carbon four, which results in increased D shielding. So therefore, right, if I go ahead and I'm making a list of the signals that appear from left to right in my NMR spectrum, then my single it corresponding to carbon four or rather the protons on carbon four are going to be on the very left side. So from here, we'll notice that as we get farther away from the electron negative atoms, then shielding is going to increase, right. So now from here up next, so to speak, is carbon three, which is bound to only oxygen. So just one highly electron negative atom. So the protons on carbon three are going to be relatively de shielded, but not as much as the protons on carbon four, right. So up next is my triplet and then comes my multiple coming from my protons on carbon two, right? Because now I'm just moving towards the left side of the molecule farther away from the highly electron negative atoms. And lastly, right, the farthest away is going to be my protons from carbon one because it is the furthest away from those highly electron negative atoms. So therefore my triplet corresponding to my protons on carbon one is going to be the most shielded. And therefore, it's going to appear on the right side of the spectrum. So from here, I'd like to recap a little bit and kind of rewrite the answer to make it look a little nicer. So when we're going from left to right in our proton and Mr spectrum, on the left side, we're going to see a single, followed by a triplet, followed by a multiple, followed by another triplet. And there is your final answer. So if you made it to the end, thank you very much for watching. I appreciate it. 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:
b. CH₃OCH₂CH₂CH₂Br

Key Concepts

1H NMR Spectroscopy

Chemical Shift

Integration and Multiplicity

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