[Sophia7X]

Hopefully those pictures are clear because I scanned them with my iPhone.

So I think I see 2 O-H's (intramolecular) at around 3000. Or maybe that at ~2900 is C-H (alkane). What is the peak at 3200? At 1700, it looks like C=O. There are supposed to be 4 structural features?

[orgopete]

It is difficult to identify a compound only from its IR spectrum. If there is additional spectra or other information, that would help more than an IR table of absorptions.

[Sophia7X]

OK, here's some additional info. Also, I believe the molecular formula is C8H8O3 based on a calculation in a previous part of the problem, if that's any help.

[orgopete]

C8H8O3 looks correct to me. Let's gather what you know.

How many carbons are giving the hydrogen absorptions between 6.5-8.0 ppm? How many carbons are left over and how many of them have hydrogens attached? How many hydrogens? What is the peak at 3.9 ppm?

What is the peak at 1690 cm-1 in the IR? Do any of the oxygen atoms have a hydrogen attached?

At this point, if you try to connect these pieces together, you should have a limited number of possibilities. It may be helpful to write out all of the possibilities. I don't think you should be able to come up with more than 8-10 depending. If you put it together as I have, then it will be two.

[yesway]

In case of simpler compounds (containing only C,O,N,H...) I've found it helpful to determine the double bond equivalents (degree of unsaturation) from the molecular formula. Combine this with the information of the 13C-NMR to find some probable carbon skeletons.

[orgopete]

In case of simpler compounds (containing only C,O,N,H...) I've found it helpful to determine the double bond equivalents (degree of unsaturation) from the molecular formula. Combine this with the information of the 13C-NMR to find some probable carbon skeletons.

Agreed, I normally do this as well. In this case, it seemed too simple to require this step. I don't want to say more at this point else I spoil the problem for the original poster.

[Sophia7X]

C8H8O3 looks correct to me. Let's gather what you know.

Hmm, I don't know much. :p
I've been reading about spectroscopy online, and it's still really confusing to me.

How many carbons are giving the hydrogen absorptions between 6.5-8.0 ppm? How many carbons are left over and how many of them have hydrogens attached? How many hydrogens? What is the peak at 3.9 ppm?

It looks like 3 carbons but apparently there is a doublet at 6.9 ppm and a triplet at 6.8 ppm. So 4 carbons (I thought around 6.8 was a single peak?)

I am also confused why 10.8 ppm is supposed to be OH when it could be OH, COOH, or CHO according to the chart.

3.9 is O-CHx, I'm assuming it's O-CH3.

What is the peak at 1690 cm-1 in the IR??

Looks like C=O
But my answer key says "the following is not thought to be part of the solution
Broad peak at 3200 cm-1: C-H
Broad peak at 1700 cm-1: C=O
Sharp peak at 2900 cm-1: O-H
Sharp peaks around 1600 cm-1: benzene"
What does this mean?

[orgopete]

I thought it easy to recognize an aromatic ring with absorptions in the 7 ppm range. If so, that is six carbons. There are only two more carbons to place and one is the C=O.

A quick recap, C6H?, a C=O, two more oxygen atoms, and another CH?. There are a total of eight hydrogen atoms. The additional carbon can be a CH2 or CH3. If you look at the chemical shift, it is attached to an oxygen, actually an oxygen attached to a C=O. If there is an OH, it must be attached to the benzene ring as there aren't any other possibilities.

There are three ways these can be put together, but one of them is symmetrical and would not show eight peaks in the 13C-NMR. So there are two possibilities, ortho and meta. What are they?

[RaInBowDaSh1488]

I found the MS very helpful for this problem. IMO the fragmentation might be very useful in helping you ID what functional groups are present.

[orgopete]

I found the MS very helpful for this problem. IMO the fragmentation might be very useful in helping you ID what functional groups are present.

Although this may be true, I don't use that information very often. This reminds me of a problem I had put on a test and which many students did not answer correctly. In class, I brought in the pieces that were clearly identifiable from the NMR. Everyone was still puzzled by what the answer was. In exasperation, I handed a student the molecular model fragments for the answer. Because two of them were terminal fragments (I don't remember what they were, but something like a methyl group and a phenyl group) and a bifunctional piece in between. It was only with pieces in hand did he realize that they could be connected to give the structure.

In this instance, if you have a phenyl ring (NMR), an OH (IR), a C=O (IR), and a OCH3 (NMR), I ask, how many ways can you put these together?

[Sophia7X]

so it would have to be ortho because of the intramolecular hydrogen bonding with O-H and carbonyl oxygen

Well, looks like I need a lot more practice with this.

[orgopete]

I'd be leery about assuming it was ortho because of hydrogen bonding unless you have been able to rule out intermolecular hydrogen bonding. Virtually all alcohols and carboxylic acids intermolecularly hydrogen bond.

You could rule out para as it would only have six signals in the 13C spectrum due to symmetry, there are eight. That would leave ortho and meta. You could compare the IR with authentic spectra and compare the fingerprint region, "peak for peak" to find a match. The other method is to look at the coupling of the aromatic protons. In the meta isomer, H2 will have a small coupling constant, J = 1-2 Hz. The ortho isomer will have larger couplings consistent with a larger coupling, J = 5-7 Hz.

[Vidya]

IR and NMR clearly indicate that there is an aromatic ring , -COOH and  methoxy group
Now splitting pattern of NMR from 7-8 indicates that there are 4types of H which is possible if it is ortho or meta.If it is meta then we should get one singlet in the center due to shielding and deshielding  effects of -OCH3 and -COOH.I can make out that it is missing.Moreover if the splitting pattern is carefully observed from 7-8ppm in HNMR that is in favor of otho ring.