# Could I get a hint on the structure of this molecule based on HNMR spectra?

Chemical formula is $$\ce{C7H6O3}$$. The IR spectrum has a peak at $$\pu{3238 cm-1}$$, which I believe could be due to an alcohol, phenol, or carboxylic acid. There's an additional peak at $$\pu{3015 cm-1}$$, which I believe is due to either an alkene or aromatic group. The presence of aromatic and phenol peaks makes me think there's a phenol group in this compound.

What I'm stuck on are the four protons in the 7-8 ppm range. Especially the one split into a quartet. If I'm right about the IR, I just can't think of a way to get one proton with three neighbors to split with in a phenol.

That is not a quartet, but probably two overlapping signals. You always have to pay attention to the signal ratios and not just the number of split signals. The ratio here is 2:2:2:1, that is not a quartet ratio. A quartet would be 1:3:3:1.

There is a certain distortion in strongly coupled systems, the "roof effect" you can also see in this spectrum. This distorts the ratios a bit from the ideal ones, but not so much as to explain the ratio in the pseudo-quartet here.

• Surprisingly, the three left peaks of the quartet end at the exact same position. What signal would it be then? I think the person who scaled that spectrum cut off the quartet so that the roof effect is still identifiable as well as the very broad signal at 11.5 ppm. If the quartet would be fully shown, the peak at 11.5 ppm would not be visible. Additionally, the doublet and and triplet build the left part of the roof and the quartet (or what's left of it) builds the right part of the roof. Jan 23 '15 at 12:07
• @Matthew You're right, it looks a bit cropped. I still don't think it is a real quartet, even with the roof effect it doesn't look right. The second peak from the left also looks like it has a small shoulder, I would guess this is a doublet and a triplet overlapping. Individual integrals for each signal would be nice, but we don't have that here. Jan 23 '15 at 12:27
1. Does the IR have any stretches near ~1700? This would be a great way to rule in (or out) a carbonyl. It looks like there is one from the above graph, and you have a broad peak ~ 11 ppm, which should stand out as a very specific functional group.

2. Mad Scientist is correct, that is not a quartet at 7 PPM, but is a doublet and a triplet partially overlapped.

3. Here's your hint: you have TWO oxygen-containing functional groups and an aromatic system (It looks like you already knew that from your sketches in the photo). If the aromatic system contains 4H, and is there are two functional groups, it is likely a di-substituted benzene ring. Which of the three possible types (ortho, meta, para) could give rise to two doublets and two triplets in the splitting?

P.S. I believe the spectrum is correctly referenced. The tiny peak near 7.26 ppm is likely chloroform from the solvent - it is much smaller than the other peaks and chloroform is the solvent of choice for most organic NMR.

Looking at your picture you also seem to have annotated an IR peak at around $\mathrm{1700~cm^{-1}}$ which is indicative of the carbonyl group. Together with the IR peak at $\mathrm{3238~cm^{-1}}$ and the very broad NMR peak far downfield this supports your supposition that there is a carboxylic acid present.

Looking at the molecular formula you now only have one $\ce{O}$ and one $\ce{H}$ unaccounted for so it must be a phenol group. As mentioned by Mad Scientist the pseudo-quartet on the NMR spectrum is in fact an overlapping doublet and triplet. You have two doublets and two triplets in total so you should think about which of ortho, meta and para substitution would give you this splitting pattern.

I once had a similar example in which benzene was used as solvent but the spectrum was not referenced to its peak at 7.16 ppm. If the quartet is really a quartet you might think about the meaning of that singlet at 7.26 ppm.