Assign the aromatic signals to the product

Question

I have an HNMR of 4-hydroxy-3-iodoacetophenone with a close up of the aromatic region. I want them assigned.

I have a doubled at around 745Hz. a doublet of doublets at 690-710Hz a singlet at around 667Hz and finally a doublet (looks like two distinct peaks but unsure) at 620-645Hz.

I know there is only 3 signals possible since there is only 3 aromatic hydrogen regions so i want clarification while you assign them please.

Answer 1

I'm sure the need for an answer to this question has long passed, but there is enough information contained in your question to provide a correct assignment, though there is a little bit of ambiguity.

First of all, most chemists refer to the chemical shift in "parts per million" or "ppm" rather than Hertz (coupling constants use Hz, but chemical shifts use ppm). While you have not included the field strength of the magnet, which is necessary to convert shifts in Hz to ppm, it has to be 90 MHz. To convert shifts in Hz to ppm, you divide the shift in Hz by the field strength in MHz (hence "parts per million"). Using 90 MHz as the field strength of the magnet would mean your signals are at:

1. Doublet around 8.3 -- Ha
2. A Doublet of doublets from 7.7-7.9 -- Hb
3. A singlet at 7.4 -- Hc
4. A doublet around 6.9-7.2 -- Hd

The shift of 8.3 seems a little high, since aromatic protons most commonly come between 7-8 ppm. But using any other likely field strength (100 MHz or 60 MHz) results in worse departures from expected shifts. 60 MHz is a common magnet strength, but it gives shifts from 10-12.5 ppm, which is impossible. 100 MHz is not as bad, but results in the most shielded signal being around 6.2-6.4 ppm, which is a greater departure from expected aromatic shifts. As such, the explanation must be that the electron withdrawing acetyl substituent shifts one signal above 8 ppm, which is not impossible.

You are correct that there are only three aromatic signals, and yet your spectrum has four signals in the aromatic region! This can easily happen for phenols, when the phenolic -OH signal coincidentally comes at a similar chemical shift to the aromatic CH's. Phenols can come anywhere from 5-10 ppm, depending on substitution, solvent, concentration of the sample, and other factors like temperature or impurities present. So they are not always in the same region as the aromatics (usually 7-8 ppm), but that is not uncommon.

You have given some information about the splitting patterns for the signals, although it is a little vague. If you had given the precise shifts of each sub-peak in each coupled signal, that would enable the calculation of the coupling constants, which would give some additional certainty to the assignment. But we can do a little inference from the information you did include.

Specifically, only one signal was given as uncoupled -- the peak at 7.4 ppm (667 Hz). The phenol -OH does not have any neighbors (and, in fact, -OH protons often do not couple neighboring protons anyway), so that is most likely the singlet at 7.4, which I labeled "Hc".

The remaining three signals are the expected aromatic signals. And they have the expected coupling patterns for this molecule. For aromatic molecules, coupling constants are fairly predictable. Protons that are ortho have larger couplings -- often 7-8 Hz. Protons that are meta to each other have much smaller couplings -- usually smaller than 3 Hz. And para H's usually have no visible coupling at all (except at VERY high field strength, which this spectrum is not).

Thinking about the structure of the molecule (it would be great to be able to insert a picture, but oh well), there are H's on C2, C5, and C6 (since the acetyl group is on C1, the iodine is on C3, and the -OH is on C4). The protons on C2 and C6 are meta to each other, and should have a coupling constant of ~3 Hz, give or take. The protons on C5 and C6 are ortho, so they should have a larger coupling constant, of ~8 Hz.

That means the proton on C6 couples with BOTH C5 and C2 protons, so it must the "doublet of doublets" from 7.7-7.9 ppm. That is, what I called "Hb" is the proton on C6 -- adjacent to the acetyl group, on the side opposite of the iodine.

The assignment of the last two is SLIGHTLY ambiguous, but I think we can reason it out. We have two pieces of information -- coupling and chemical shift. The coupling information you provided was a little vague, and maybe contradictory.... You said the signal at 8.3 ppm (745) was a doublet, with no ambiguity, but said the signal from 6.9-7.2 (620-645 Hz) MIGHT be a doublet, but it was not certain. That's a little bit odd. Of the two remaining protons, the proton on C5 should be an obvious doublet, since it is directly adjacent to Hb, and has a large (ortho) coupling constant. In contrast, the proton on C2 only has a long-range meta coupling with Hb, so its coupling constant would be smaller, which would make it a less obvious doublet. Since you were confident of the "doublet-ness" for Ha, and less so for Hd, that would suggest that Ha is on C2, and Hd is on C5.

But I don't think that is correct, because of chemical shift. The acetyl group is moderately electron withdrawing, which will shift the ortho protons (on C2 and C6) downfield (to higher chemical shift). In contrast, the -OH group is very electron donating, which would shift the adjacent proton (on C5) to a lower chemical shift. (This may seem odd, since the -O is fairly electronegative. But the substituent effect of the -OH group is much more strongly determined by the presence of a lone pair which donates by resonance to the ring, and -OH is fairly strongly donating as a result, despite being electronegative.) So based on chemical shift, that would argue strongly (and convincingly) that Ha, being at 8.3 ppm, is much more likely to be adjacent to the electron withdrawing acetyl group (on C2). (Notice, the proton on the other side of the acetyl is the second-highest in chemical shift, from 7.7-7.9 ppm.) In contrast, Hd (from 6.9-7.2 ppm) is much more likely to be adjacent to the electron donating -OH group (that is, on C5).

This also gels with the width of the chemical shift ranges you provided for each signal. While you seemed unsure whether Hd was a doublet or not, you did provide a much wider chemical shift range (6.9-7.2 ppm, or 620-645 Hz), which suggests a larger coupling constant. In contrast, for Ha, you listed a single chemical shift, which suggests the two peaks are a lot closer together (that is, the coupling constant is smaller). I am not certain why you would have been less certain that Hd is a doublet -- I would have assumed that Hd would be more obviously a doublet, and that Ha might be less obvious, since the coupling constant is smaller.

Anyway, I'm pretty certain that the correct assignments are:

1. Ha (d, 8.3 ppm, small coupling constant) is on C2
2. Hb (dd, 7.7-7.9 ppm) is on C6
3. Hc (s, 7.4) is the phenol -OH
4. Hd (d, 6.9-7.2, large coupling constant) is on C5

One final note. This information was not provided, but in you are actually obtaining a NMR for an experimental sample, you can always do a "D2O wash". You first obtain the spectrum in CDCl3, or some other non-polar solvent, which will show all protons. Then, you add 5-10 drops of D2O and shake the bejeezus out of the sample in the NMR tube for a few minutes. The D2O does not mix with the nonpolar CDCl3, but it will cause any and all OH hydrogens (and NH hydrogens) to undergo chemical exchange with the D2O. This "swaps" all OH (and NH) protons for OD (or ND). Then, after the D2O separates back out to a separate layer (on top of the CDCl3, since CDCl3 is more dense than water), you re-take the NMR spectrum. Then, AFTER the D2O shake, all of the OH or NH signals will be "erased" from the spectrum -- because they are now OD or ND signals, and D does not show up on H-NMR spectra. In this particular case, the singlet at 7.4 ppm would have disappeared after the D2O wash. This is a super helpful way to unambiguously determine which signals in a given spectrum are OH or NH signals. And if/when they overlap with other CH signals in a way that obscures details you may want to see, you can use this to clarify the spectrum and allow more precise interpretation of the CH signals that were previously obscured by the OH/NH signals.

Long answer, but hopefully it makes sense! Answers without complete explanations are less helpful.

Answer 2

This is one of the Organic Chemistry problems that require visual reference to enact the needed process due to the endless possible correct answers without vital context.

Including the original spectrogram will make it possible to determine where the acetophenone, iodine, and hydroxide ion groups sit along the aromatic ring relative to one another. You need the j values provided by the spectrogram chart to calculate the hydrogen/proton coupling arrangement of these signals.

Please review the sample video links from YouTube below to get an idea of the vital information required to demonstrate the ability to solve this type of problem. They don't demonstrate the exact solution of this particular molecular structure, but they are a needed demo of the conceptual framework you need with your reference to solve this problem.

HNMR Practice Problems with Step-by-Step Solutions by @VictortheOrganicChemistryTutor will show you the graph you need to include, the overall preferred method for this kind of Organic chemistry problem, and why to do so.

Interpreting Aromatic NMR Signals by @kristopherkolonko365 is an excellent applied presentation that shows the peak patterns you can see as side-by-side comparisons to understand why the proton coupling category matters.