# How do I find out what benzene containing compound this is with IR and NMR?

I have the following IR and NMR of an unknown substance. Can someone tell me what is a possible structure of the compound or point me in the correct direction. So far I am thinking of a benzene ring based compound with an OH group. Something like 2-phenylethanol but not sure.

Some properties:

• melting point: $$\pu{116 ^\circ C}$$.
• Composition: white powdery solid.
• Soluble in $$\ce{NaOH}$$ and $$\ce{NaHCO3}$$.

Based on the IR, the broad peak at $$\pu{3424 cm-1}$$ represents an alcohol while the 4 humps in the region $$\pu{2100-2600 cm-1}$$ is indicative of an aromatic ring (most likely a benzene ring). The medium peak around $$\pu{1660 cm-1}$$ is most likely a carbon-carbon double bond. NMR is where I get stuck; I know there should be 6 distinct types of protons and since most of the peaks are around $$\pu{7-8 ppm}$$, it is most likely the benzene ring protons, but I am not sure how to calculate J values and splitting as it was done in

• Subtract the signals of the individual peaks of your (doub|trip|...|multi)plets and multiply them by the MHz of your spectrometer. You will then have the coupling constants, in other words $J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$. – Martin - マーチン Jul 2 '14 at 16:30
• @Martin Unfortunately I am not sure of the frequency used at my university-my instructor got me this graph for my unknown sample. Is it possible to use this NMR without the JValues ? Thanks for any help you can offer ! – oneCoderToRuleThemAll Jul 3 '14 at 3:43
• Since it says bruker500, I think it is fair to assume that it operates at 500 MHz. The coupling constants are necessary to get which hydrogens are next to each other. – Martin - マーチン Jul 3 '14 at 3:48

This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant.

The first and second IR spectra are so different that it is hard to be confident about which one is correct. Are they collected in different media (one a solution and one a mull perhaps)? The first looks very low concentration

• ~3420 is clearly an -OH stretch. Probably too high to consider a N-H group of any sort. So, it could be an alcohol or an acid.
• 1682 a C=O group. Probably in the acid region, likely aromatic acid, conjugated or intramolecular H-bonded. Possibly amide, but not convincing based on other peaks.
• There are lots of other peaks, most of which could mean many things, but I do see 749 and 708 which support the NMR notion of a monosubstituted benzene. I usually think that IR is only useful to corroborate peaks of a known structure, but then I'm an NMR advocate.

NMR

Again, the first and second spectra are very different; the obvious concern is that they are of different batches of sample - were the first and second analyses taken from the same sample, or from a different batch. In any case, let us look at the second, which is only a partial spectrum, and omits some possibly very significant information, especially in the region above 10ppm, where I would like to see a broad acid peak.

The aromatic protons shown exhibit a classic mono-substituted benzene ring pattern, labelled above. There is a strong downfield shift of the ortho (H2) protons, consistent with an electron withdrawing group substitution (such as -(C=O)R). Ortho coupling for the H2 proton is difficult to determine accurately as there are 3 peaks picked. This may be a digitizer point average, but considering the average of each peak, and the coupling from the first spectrum, we get a ortho coupling of 7.2-7.9. Not overly useful, but less than 8, so we can exclude ethers and halides. Couplings are calculated by:

(8.1651-8.1507) x 500 = 7.2

Using additive analysis for monosubstituted benzene rings, we can determine the following shifts from benzene:

• H2 (8.162) 7.34 + 0.82
• H3 (7.514) 7.34 + 0.17
• H4 (7.651) 7.34 + 0.31

Looking at the published tabulated data for substituent effects on monosubstituted benzene rings (I'm using Pretsch, Buhlmann and Badertscher, but lots of other sources are available), we find one possible candidate that matches closely:

• Benzoic Acid H2(+0.79), H3(+0.14), H4(0.28)

The caveat that I would add to this is that the data presented is not the best quality, and the solution presented is consistent with the data only (not definitive). The key concern is the missing -OH peak in the NMR spectrum. Unless this sample was run in D2O, or washed with D2O, I would expect to see an acid peak around about 12 or so. Depending on the amount of water in the sample, this may be shifted somewhat and broadened, but should still be visible, even if a broadened hump in the baseline. Additional NMR experiments could be useful to confirm the structure.

• I usually do not clutter the comments section with unnecessary affirmation, but when I do, I say things like: "This is an awesome post, you're awesome, dude!" – Martin - マーチン Jul 11 '14 at 7:10
• @Martin Ha! Thanks, but don't take anything away from your own efforts, which provide a very strong stepwise approach to solving these types of problems. Unfortunately, the data sets provided are not great, and in real life we would use a range of other useful techniques to determine structure including 2D NMR (which I think is far more useful than IR for determining unknowns) – long Jul 11 '14 at 7:26
• Agreed on every term. When I was doing my organic chemistry studies we were provided several spectra, including at least a 13C, sometimes COSY, NOESY and HSQC or HMQC. They provide much more insight into the chemical bonding and I believe they are absolutely necessary for modern chemistry. Also CHN and mass spectroscopy should be done for most compound. Doubt free characterisation of a compound is an art! Thumbs up for everyone who can do it right. – Martin - マーチン Jul 11 '14 at 7:51
• @long Thanks a lot ! I tried to obtain a NMR with the more down field view but unfortunately I couldn't access the NMR machine. However, I measured the melting point to be in the range 118.6-122.4 Celsius which is close enough to the Benzoic acid mp which is 122. So this is most like it. Thanks a lot ! – oneCoderToRuleThemAll Jul 11 '14 at 14:22

I tried to solve it, but I horribly failed. But as I put some effort in it, I still share this with you. Maybe you can work it out eventually.

I have to admit that this spectrum is bad to work with, it does not contain an internal standard and the solvent signal is not marked. It is possible, that it is integrated. If this is an educational exercise, your supervisor clearly did a very bad, a horrible job.

The signals are way too small to see a splitting pattern, it is very hard to even put a number to the signals. The aromatic area should at least have been enlarged. If you have another possibility, you do never ever scale the integrals to an aromatic proton, especially not the one that mixes with other signals.

The following you did already: Have a look at typical IR bands (other source) and identify them.

• 3424 phenolic $\ce{OH}$ stretch
• 3054 aromatic $\ce{C-H}$ stretch
• around 1600 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
• 1265 aromatic $\ce{C-O}$ stretch could be ether alkoxy stretch
• 737 and 704 aromatic $\ce{C-H}$ bending, hints towards meta substituted phenyl

Now you could go ahead and identify the residual solvent signal.

• 7.14 should be chloroform $\ce{CDCl3}$
• 7.28 is also a good candidate

It can also be none of them.

If you have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table you can identify some of the protons.

• 7-8.5ppm These are aromatic protons
• 2.20ppm allylic/benzylic proton, could also still be $\ce{sp^3-C}$, it is very low field. As it is a singlet it has no neighbour carbons with a hydrogen, but it integrates to only one, or maybe two - it is hard to tell.

From the IR you identified a phenolic $\ce{OH}$ and = there should be a signal around 5ppm, which is not there. So I am very puzzled about this.

Now the splitting pattern is completely random to guess.

You can calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, as it says bruker500, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 8.15 ppm: $$J=(8.1560-8.1402)\cdot500~\mathrm{MHz}=7.9~\mathrm{Hz}$$

• 8.15 doublet, $2\ce{H}$, $J=7.9~\mathrm{Hz}$
• 7.65 multiplet, $1\ce{H}$, $J=1.4~\mathrm{Hz}$, $J=5.8~\mathrm{Hz}$, $J=6.5~\mathrm{Hz}$
• 7.52 multiplet, $3\ce{H}$, $J=8.3~\mathrm{Hz}$, $J=7.0~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$, $J=7.8~\mathrm{Hz}$
• 7.39 doublet of doublet or triplet, $1\ce{H}$, $J=7.5~\mathrm{Hz}$, $J=7.4~\mathrm{Hz}$
• 7.28 singlet, $1\ce{H}$ - or solvent
• 7.14 singlet, $1\ce{H}$ - or solvent
• 2.20 singlet, $1\ce{H}$

I am very sorry I could not be of any more help. A $\ce{{}^{13}C-NMR}$ would certainly have been helpful at least.