2
$\begingroup$

I've been trying to solve the structure of a molecule from a 13C and proton NMR as well as a MS spectrum, but I've been stumped, I've been staring at it for a while with no improvement.
I've attached the images below. From the MS, I understand the Mr of the molecule to be 122, the 13C-NMR suggests six different environments.

The part that confuses me the most is on the NMR. There is a 3 hydrogen doublet and a 4 hydrogen singlet.
I went on to assume that the structure might be some sort of aromatic and symmetrical, and maybe 2 carbons in the same environment so that the formula might be $\ce{C7H10N2}$, this accounts for the Mr and the fact there is an exchangeable proton peak.

The spectra I've been provided are poor quality, so I can't even work out the coupling constants, or properly view the peaks.

I still can't figure out what it might be. Can you point me in the right direction please. I have a feeling I've been confusing myself over something that's simple.

Images:

$\endgroup$
  • 1
    $\begingroup$ Some points to consider to help you along a bit. Your 4H singlet is not a singlet; it is an overlapping set of peaks. Also, what you have labelled as a d at 4.8 is definitely not a d; looks a lot like a quartet to me. Your expected molecular formula is not correct. Look at your 1H spectra an summarise what groups you have; how many aromatic protons, how many alkyl protons, and which groups can be joined together. For instance, your 3H doublet indicates a particular group, and what is is next to. The chemical shift and splitting of the 1H at 4.8 likewise tells you what and where it is. $\endgroup$ – long Mar 6 '16 at 21:20
  • $\begingroup$ The spectra provided are adequate to determine the structure, and there is no need to work out coupling constants. There are only two sets of peaks that show splitting anyway, and this provides enough information for your purposes. $\endgroup$ – long Mar 6 '16 at 21:22
  • $\begingroup$ CPE Lyon student? :P $\endgroup$ – ParaH2 Mar 6 '16 at 21:45
  • $\begingroup$ @long oh it makes more sense now that you pointed out the 4.8ppm peak is a quartet. I thought the shoulders were just due to the low resolution. This clears things up thanks. $\endgroup$ – heych Mar 6 '16 at 21:56
1
$\begingroup$

A very quick solution to this problem before my train stop.

Mass Spec gives MW of 122.

1H spectrum gives 5 aromatic, a 1H quartet at 4.8ppm, a 1H exchangeable proton at 2.4ppm and 3H doublet at 1.5ppm. The 4.8/1.5 splitting indicates a substituted ethyl chain with the C1 group having two substituents (CHRR'-CH3). The chemical shift of this 4.8ppm proton indicates some substantial substitution, such as a heteroatom. We also have a labile 1H, which could come from an alcohol or an amine.

13C spectrum shows monosubstituted benzene ring (4 carbon environments with ipso carbon downfield shifted). Also shows two alkyl carbons corresponding to the ethyl chain we have worked out above.

To summarise data to date:

  • 1 monosubstituted benzene (C6H5)
  • 1 ethyl chain, doubly substituted on C1 (CH-CH3)

That gives us a molecular formula of C8H9 which adds up to 105. That leaves 17...... which includes one labile proton, leaving a mass of 16.

As this is clearly a homework question, I'll leave you to join the pieces.

$\endgroup$
  • $\begingroup$ thank you, it seems that my confusion stemmed from thinking the 4.8ppm peak was a doublet instead of a quartet - I thought the shoulders were due to poor resolution $\endgroup$ – heych Mar 6 '16 at 22:03
  • $\begingroup$ The quartet coupling is quite broad, and this comes about because there is also some coupling through to the -OH (but not visible due to exchange broadening) and also through to the aromatic ring. $\endgroup$ – long Mar 6 '16 at 22:06

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.