An unknown compound (C₇H₆O) gives the IR spectrum shown here. ...

Question

An unknown compound (C₇H₆O) gives the IR spectrum shown here. At what chemical shifts would you expect to see signals in the ¹H NMR spectrum?

Accepted Answer

All right. Hi, everyone. So this question says that a student synthesized a compound whose molecular formula was found to be C six H 60 and its IR spectrum is given below what chemical shift would be observed for its signals in proton NMR. So here like the question says, right, we have an infrared or IR spectrum of this compound C six H 60 on the screen. Now, Ir if you recall is useful for helping us identify what functional groups may be present in a molecule. However, before we start analyzing the infrared spectrum in more detail, your first step in structure determination problems should be to find the IHD. So here I'm going to write down my molecular formula on the bottom here thats C six H 60 and I'm going to start by writing the formula for the IHD. Now to find the IHD, you're going to multiply the number of carbons by two and then add two to that. In addition, you're going to add any nitrogens you may happen to have while subtracting both the number of hydrogens and the number of halogens if you have any. And then you take the entire first part that I mentioned previously and divide that by two. Now it just so happens that this compound doesn't have any nitrogen atoms or halogens. So I'm simply going to ignore those terms. And now I can plug in my values, right? I have six carbons. So it's going to be two multiplied by C which is six. I'm going to add two to that and I'm going to subtract the six hydrogens I have and then divide everything by two. So two multiplied by six is 12, subtracted by six is six. And then when I add two to that, I get eight divided by two, which ultimately gives me an IHD of four. Now recall that an IHDF four is indicative that there is a benzene ring in this molecule, right? Because recall it ring structures as well as double bonds add to the id of a molecule. So an IHD of four indicates benzene because of one ring structure as well as three double bonds. So we can confirm the presence of a benzene ring using the data provided in the IR spectrum because recall that we can search for two different peaks in the IR spectrum. We can look for the SP two carbon hydrogen stretch from the bending ring itself and we can also find the carbon carbon stretch of the ring itself. Now, the SB two carbon hydrogen stretch is generally found in close to 3100 inver centimeters or aromatic compounds. And it looks like we do have such a peak if we consider the left side of the spectrum. So we do have a peak that confirms this and we also have a peak at approximately 1600 or 1600 inver centimeters confirming the presence of the carbon carbon stretch. So approximately this one more to the right side of the spectrum because we do see a peak at approximately 1600 to confirm the carbon carbon stretch of the aromatic ring. So with this in mind, we can get a better idea on what the compound looks like. But now I want to draw your attention to this relatively large peak farthest to the left side of the spectrum itself because it's a very characteristically large peak that spans between approximately 3200 to 3600 parts per million. Now, the shape and size of this peak is characteristic if you recall of hydroxy stretches. So what this peak proves is that this compound has a hydroxy group in it. And that makes sense considering there is oxygen in the molecular formula of this compound. So the combination of a hydroxy group and the presence of a benzene ring indicates that this compound is most likely going to be phenol. So when it comes to the proton NMR spectrum of phenol, we're going to see a few distinct peaks, right? One is going to come from the hydroxy proton of phenol and the others are going to come from the benzene ring itself. Now, the question is even though there are five aromatic protons in this molecule, right, Neno, how many signals are actually going to appear? Because recall that when molecules have an element of symmetry to them, it actually reduces the amount of signals that you would expect to see. So because FOL does have an element of symmetry to it, we're going to see less than five signals coming from the aromatic ring. So if the, if the hydroxy group here is on carbon one of the aromatic ring, then I have protons on carbons two through six. However, due to their proximity to the hydroxy group itself, the protons on carbons two and six are equivalent and therefore correspond to one signal. Now, the same can be said for the protons on carbons three and five because those two protons correspond to the same signal because they are considered equivalent due to the symmetry present in venal. And then the proton on carbon four corresponds to its own signal because that one has no equivalent. So in total, we should expect to see four distinct signals, right, one signal comes from the hydroxy proton and then three are going to come from the aromatic ring. So even though there are five protons on the aromatic ring, there are only going to be three signals due to the symmetry present. Now recall that the hydroxy proton is going to appear at approximately nine parts per million due to its proximity to the benzene rate. And generally speaking, signals from aromatic rings or aromatic protons tend to appear between six and seven parts per million in the spectrum itself. So the three signals from the aromatic ring should be found somewhere in between six and seven parts per million and there you have it. So let me scroll down one more time so I can show you the final answer. Our final answer is that in the proton NMR spectrum, there will be one broad peak at nine parts per million for the hydroxy proton and three peaks between six and seven parts per million for the five aromatic protons. So with that being said, if you stuck around to the end, I thank you so very much for watching and I hope you found this helpful.

An unknown compound (C₇H₆O) gives the IR spectrum shown here. At what chemical shifts would you expect to see signals in the ¹H NMR spectrum?
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Key Concepts

Infrared (IR) Spectroscopy

¹H Nuclear Magnetic Resonance (NMR) Spectroscopy

Aromatic Compounds

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