Processing a "Raw Spectrum" - A Beginner's Attempt

[Sippi7299]

Homework Statement

Identify the unknown compound with the help of an 1H NMR-Spectrum

Relevant Equations

further information about the compound:
Melting point: 78-80°C
Heteroatoms: X,O

The first picture shows the "raw spectrum" before processing it and the second one is my attempt to process it.
I was not sure about the solvent peak because it is expected to be a singulett at 7,26 ppm but I could not really identify a clear singulett so I dichtet to go with the peak that stood out the most (I'm a total beginner).

I am pretty sure its an aromatic compound because of the signal between 7 and 8 ppm.
the signal at 1,7 could be OH or I guess.
I am not sure about the one at 4,7 ppm it could be a CH2 group attached to an OH maybe.
A thing that would help me first is if I could get the integrals right so I have at least the right number of H-Atoms per peak and I'm also not sure about my peak picking in the aromatic region since a lot of signals are very near to each other in this region so I am not sure if its maybe actually just two signals.

I am very thankful for any kind of help :)

 

[docnet]

The multiplets in the aromatic region have an integration number of roughly 3.5. The singlet at 4.7ppm has an intgration number of 1, and the singlet at 1.7 has an integration number of roughly 1.5. multiply all those numbers by 2, and you have seven aromatic protons, two chemically equivalent Ph-OH groups and three chemically equivalent ROH groups. There is only one way to put them together that makes much sense, as far as I'm aware. Is that consistent with your melting point?

Imho, compound identification is something that computers will be able to do more efficiently than humans one day.

 

[Sippi7299]

thank you very much for your answer! :)
I am new to NMR and I was actually a bit surprised to find out that computers are not already able to solve the problems in an efficient way...
Anyways, regarding the spectrum, I do know, that the heteroatoms in this spectrum are O and X so one substituent must contain a halogen.
also I might add that I added the Integrals manually so there could be faults there as I've never processed a spectrum before.
May I also ask how you know that the signal at 4,7 ppm is the solent? That's something that I could not quite find out yet how to determine so I just set the signal closest to 7,2 ppm as a solvent peak as you could see above.
Again, thank your for your help

 

[docnet]

May I also ask how you know that the signal at 4,7 ppm is the solent? That's something that I could not quite find out yet how to determine so I just set the signal closest to 7,2 ppm as a solvent peak as you could see above.
Again, thank your help

Sorry, that was a typo.. I was going to ask if the solvent peak between 7-7.5 ppm contributes to the integration numbers, and then I figured it out.

 

[Sippi7299]

alright, no problem :). Do you have any idea what the X atom could be since it definitely has to be part of the compound?

 

[docnet]

If the integration number of the aromatic protons is 6 instead of 7 (It could be both since 1.76 is almost as close to 1.5 as it is to 2) then the halogen would just replace one of the protons. You just have to make sure the two OH protons are placed so that they are chemically equivalent, and draw the structure that is most sterically and electronically favorable.

 

[Sippi7299]

you have seven aromatic protons, two chemically equivalent Ph-OH groups and three chemically equivalent ROH groups.

so one substituent would be a Halogen and there would be two chemically equivalent -OH substitutes? And what about these ROH groups? I'm sorry that I can't quite follow yet since this this type of analysis is fairly new to me but I want to be able to comprehend your analysis. Thank you for the help :)

 

[docnet]

I was thinking something like this:

I am confident about the presence of a $C(OH)_3$ group and less confident about the rest of the molecule, because the integration numbers aren't exact. Even if the molecular formula is correct, the substituents could be arranged differently. Also, the two aromatic rings are not planar like in this drawing, but one is "rotated" 90 degrees relative to the other.

 

[Sippi7299]

Ah ok, now I see. Thank you very much for the help :) I will try to continue from there on

 

[TeethWhitener]

I haven’t looked closely at the spectrum, but a C(OH)₃ group will never ever ever in a million years be stable.

 

[Sippi7299]

I changed the integral values so currently it looks like this.
I still think that a 1,2 substituted benzene could be possible...like 2-bromobenzyl alcohol.

 

[docnet]

I haven’t looked closely at the spectrum, but a C(OH)₃ group will never ever ever in a million years be stable.

That is a great and valid point. I guess I've already forgotten organic chemistry .

 

[TeethWhitener]

I think the chloroform solvent peak is dramatically interfering with the integration over the aromatic region. I think that if the peak at 1.7 integrates to 3H, then the aromatic region likely integrates to 4H.

 

[Kingstepper]

I still think that a 1,2 substituted benzene could be possible...like 2-bromobenzyl alcohol.

You might be along the right lines for 1,2-disubstitution but integration of the 2 singlets are the wrong way round for this. CH2 of a benzyl alcohol should be ~4.7 but only integrates to one proton.

 

[Sippi7299]

yeah, the integration causes the problem here. The singlet at 4,7 says 1,94 for the integration, so I'd say it could be CH2 but the singlet at 1,7 has an integration value of 2,73 so way to high...
the thing is, I know that one heteroatom in the structure is O so the singlets are probably some kind of O group.
I was thinking, maybe the OH integrates so high because it overlaps with the water in the solvent? idk if that even makes senes... what do you think the substituent could be?

 

[Kingstepper]

Yes, suppose could be water in the sample though CDCl3 is usually pretty dry and surprised you'd be given a problem like this with a confusing peak like that.
Apart from this, does fit with 2-bromobenzyl alcohol as you suggest and the m.p. is spot on!

 

[Sippi7299]

yeah, the melting point fits perfectly so I just hoped that the peaks would fit too haha.
True, CDCl3 is pretty dry, would be different with DMSO but I know for sure that the solvent is CDCl3.
I'm not sure if I should stick with my answer and try to explain the high integration or try to find another substance... but with the information given, I can't think of much else

 

[Sippi7299]

one group that would fit the integration would be

but then the melting point does not fit