0
$\begingroup$

The proton and carbon NMR spectrum datas for compound W (201 g/mol) are shown below. Based on Diagram 1 and Diagram 2, draw the structural formula of compound W according to the type of hydrogen and carbon.

Diagram 1: Carbon NMR spectrum for compound W

Diagram 2: Proton NMR spectrum for compound W

I have extracted the information from the NMR spectrum into a table.

$\endgroup$
4
  • 1
    $\begingroup$ Hi, welcome to chemistry.SE! Could you please add some more information about what you have done so far to solve the question? Maybe mention what are your guesses about the structure of the compound. $\endgroup$ – S R Maiti Apr 20 at 9:40
  • 2
    $\begingroup$ Not enough information to predict the structure. Need to have at least integration of proton NMR. Only one thing is for sure. It is a 1,4-arene. $\endgroup$ – Mathew Mahindaratne Apr 20 at 9:56
  • $\begingroup$ @ShoubhikRMaiti I'm so sorry, that's all the information I have:( I really don't know how to predict the structure based on the information given. There is no integration ;( $\endgroup$ – Lily Apr 20 at 12:23
  • 1
    $\begingroup$ What table are you referring to? And I agree with the answers that it is a horrendously bad assignment. Not only it's nearly impossible to draw a meaningful conclusion, but it also contained several typos and illiterate formatting. $\endgroup$ – andselisk Apr 22 at 9:10
2
$\begingroup$

Wow. This is a very bad question (the exam/assignment, not the OP's). There is no way that you can categorically determine the structure of this based only on the information given. This would have to be open-book providing access to chemical shift tables in the very least. Peaks are poorly labelled and what's more, I believe the molecular mass is incorrect. Anyway, let's break it down into a thought process and see what we come up with based on the information given.

What we know from first inspection: these are simulated spectra - no solvent is defined (this makes it very difficult to make predictions on labile peaks). No field is defined, so coupling constants cannot be used. Either we have to use chemical shift mapping only, or are expected to make some guestimates based on peak heights and linewidths to determine numbers of C/H. But we assume that the information provided can lead to a definitive structure.

From the 13C spectrum- we have 7 labelled sites - but is the peak labelled u actually two closely adjacent peaks?? It's a lot thicker than the other lines. I think so, and the gut feeling looking at the aromatic region of the 1H spectrum suggest a 1,4-disubstituted benzene. So I'm going to say 8 sites. The intensities (one of the joys of simulated 13C spectra is that peak intensities are quantitative) show that t and s each contain 2 equivalent carbons. So, we get 10x carbons.

From the 1H spectrum- we have a minimum of 7 sites - but again we make an assumption that the peaks at f arise from two signals - so probably 8 sites. Let's break the signals into 2 groups: those that are broad and those that are narrow. For the narrow peaks, we can estimate integrals to be 3a:1b:1e:2f:2f (if we say f is actually 2 lots of signals). For the broad peaks, we get 1c:2d:1g. So, all up: 3:1:1:2:1:2:2:1 which means 13x 1H

Now, let's work out a molecular formula based on C10H13XnYm... yuck. I'd like to think that for a problem question like this, any inclusion of a non-standard nucleus would be made abundantly clear. You can work out a range of formulae based on this mass by hand, but it's quicker using a tool like https://www.chemcalc.org/mf-finder. Whack in 201, limit it to C10H13, and you come up with junk. Add some flexibility around numbers of C and H, and you come up with mostly non-integer DBE, so alarm bells are ringing. So, let's leave the molecular mass and look closer at the NMR fragments.

From the 1H spectrum: peaks at c, d and g are labile, and in slow exchange, so probably not connected to 13C. Chemical shift information is largely useless without knowing the solvent for labile peaks. For instance, an -OH peak comes at about 1.6 in chloroform, but around 3.5 in dmso. For -NH2, the differences are even larger.

So, now let's pair up the carbons and protons to see what fragments we might have:

a/p -CH3. Could be just about anything: Ar-CH3, C(O)-CH3 are key possibilities
b/q -CH. Doublet.... q is connect to r
e/r -CH. Doublet.... r is connected to q
f/s -CH. Looking awfully like a 1,4-Ar
f/t -CH. Looking awfully like a 1,4-Ar
u and v are probably quaternary 13C-1H
c, d and g are probably labile X-H

g could be just about anything, but in the absence of any other information I'm going with something simple like a carboxylic acid. That would tie in with the chemical shift of v. the key map pairs are b/q and e/r, as they have quite distinct 13C/1H shifts with (probably) vicinal coupling. It looks small enough to be vicinal. So, this to me looks very like a 1,2-diol. But the broad peaks in the 1H spectrum suggest a ratio of 1:2, so perhaps a 1,2-olamine.

So, I think we've got the following groups:

enter image description here

How do we put them all together. Well, honestly, if you've got no other information to go on, then you could put them together any way you like and it would be close enough. I think it's too close to know whether the olamine is Ar-C(OH) or Ar-C(NH2). But we can probably guess that we have Ar-CH3 rather than Ar-COOH because the quaternary peak would shifted upfield if it were a benzoic acid (due to increased elecrton density on that quaternary). So, armed with that, I plugged in the following molecule into ChemDraw and got it to spit out a 13C prediction:

enter image description here

and I got:

enter image description here

Well, at least I know what prediction software your examiner uses. Identical to your question. The 1H prediction is not identical, giving the following in dmso:

enter image description here

You'll notice the labile peaks are indeed all over the place. But as 1H chemical shifts are not good guides for accurately determining structure alone, I am happy to ignore these. Structure determination software uses 13C chemical shifts and connectivities to determine structure, and rarely even considers 1H chemical shifts. So we have a molecular formula of C10H13NO3 with a mass of 195. I believe this is an error in the paper, although will be very interested to see another answer that can map the 13C shifts and get a mass of 201.

Hope this is helpful.

$\endgroup$
2
$\begingroup$

Interesting Problem and quite a nailbiter.

I will try to sort some of my ideas, even thought this won't be a full answer.

  1. This looks like a question from an exam. Did you have any reactions during study with hydrochloric acid and propanoyl chloride (acid chloride of propionic acid). So maybe look into different reactions of acid chlorides.
  2. As @MethewMahindaratne mentioned, it is hard without integration but for carbon we dont need integration. We have 7 Carbons, some hydrogen, pretty sure some oxygen and for sure some nitrogen. Why? Because only nitrogen is able to produce odd numbered masses. If we have CH compound, there will be always an even number of hydrogens and carbon has a mass of 12 so overall its even. When we create double bonds or add oxygen, the amount of hydrogen is decreased by 2 so it remains uneven. Only with nitrogen, we can add 1 hydrogen per nitrogen, therefore we have 1 or 3 or 5 etc. nitrogens in it.
  3. The proton g is also quite specific. This high shift in proton NMR usally comes from a carboxylic group. So we have $C_7O_2H_xN_y$.The idea of the carboxylic group is also supported by the C atom at 175 which is a really high shift.
  4. With a molecular mass of 201 g/mol this leaves quite some but not endless options. We can take of 7 x 12 (for the seven carbons) and 2 x 16 (for oxygen carboxylic group) and 1 (for hydrogen carboxylic group) leaving us with a mass of 84 split along oxygen, hydrogen and nitrogen.
  5. This is a lot. On the bottom line it is mentioned, that W consists of certain things. halogenides which usally make compounds heavy are not included there.
  6. Appart from the carboxylic group we are left with 6 carbons which could give the idea of a benzene. But I think it is not benzene because we could only attach $NH_2$ or $OH$ groups onto it and even with 5 $NH_2$ groups we get not enought mass. In addition, aromatic C atoms have high shifts usally 120-150. But we only have 3 others (next to our carbon from the carboxylic group) so not enought for a six membered ring.
  7. But also not enought for just one alkene which I would prefer. Alkene shift ist around 110-140 and fits well.
  8. Now that is all I could provide without crazy speculations.

I can throw in some suggestions but they are realy wild and you need to look in deeper.

  1. The big peek a looks a bit like some alkohols I worked with so maybe a hydroxylic group?
  2. Maybe Carbon Atom u contains 2 or even 3 peaks? (the line looks thicker than the others)= that would contribute to the idea of a substituted benzene ring and provide more carbons so we could add other groups.
  3. Could $H_f$ be two dupletts instead of a quartett so that could come from a double bond?
  4. With integrals you could look at the singuletts and determine wheather it is an amine (singulett with an integrall of 2)
  5. Maybe my whole carbon acid idea is wrong and it is an ester or something.
  6. If we are allowed to include hallogens, then it is pretty sure a benzene with a carboxylic group maybe hydroxylic and a hallogen atom.
  7. If you have more information, maybe you can use the sdbs data base https://sdbs.db.aist.go.jp I was not able to find an entry that fits.

Hope this brings the project forward.I am very happy to hear, what the compound was when you get a result.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.