[luscofusco]

I'm doing this problem and I don't know how to continue.

So far I managed to get from the ¹H NMR 6 hydrogen atoms or a multiple of 6. From the ¹³C NMR 6 or 12 carbon atoms depending on symmetry.

Then with the IR spectra:

cm-1	Intensity	Group
3100-3000	weak	H-C(sp2)
3000-2900	weak	H-C(sp3)
1600	strong	C=C(aromatic)

Can the 2500 cm^-1 be a O-H from carboxylic acid and then 2000-1900 cm^-1 the C=O from that carboxylic acid?

I'm not sure how to read the mass spectra, I get the [M]⁺ at 171 and [M+2]⁺ at 173. I'm not sure if the 172 peak is relevant or not.

Overall I guess that the compound might be some benzene acid or similar. Any help appreciated.

[Babcock_Hall]

An aromatic ring often gives several signals in the general vicinity of 1450-1600 wavenumbers, but it depends upon the substitution pattern.  An aromatic ring may also produce several weak signals in the vicinity of 1600-2000, although some of these can be masked by a strong signal in the same region.  Offhand 1900-2000 wavenumbers is too high for the C=O stretch.

[luscofusco]

Thank you.
Could the 2500 signal be produced by the aromatic ring?

[Babcock_Hall]

I assume that you have access to a table of IR absorption frequencies.  In using such information one also needs to keep in mind the depth and breadth of the peaks.  That might help with deciding whether or not the 2500 wavenumber signal comes from an -OH group.

It would be good if you could provide some more of your own thinking, as per the Forum Rules.  There is quite a bit of information in the two NMR spectra and the mass spectrum.

[luscofusco]

Yes, I use the slides that the teacher gave us, but I don't understand them very well, and a book. I'll keep working on it, thank you.

[Babcock_Hall]

I suggest starting with the M+2 peak at m/z 173.

[Babcock_Hall]

The ratio of the M+2/M peak intensity is a useful piece of information.

[Babcock_Hall]

Thank you.
Could the 2500 signal be produced by the aromatic ring?

I am not sure what is producing it.  However, you have other data that can lead to a structure.

[luscofusco]

Sorry for the delay, this is what I've managed to do so far:

The M+2/M intensity is about 1/3 so Cl is present.
The M peak is odd so it contains N too (teacher said that if there's N it's only one for these exercises).
Then I divided 171 by 13 and I got C₁₃H₁₅ formula, then I needed to substract the Cl and the N so it is a CH₂ for the N and C₂H₁₁ for the Cl.
This gives C₁₀H₂ remaining, so there should be more H than the initial guess. My guess was 6 or a multiple of 6 but I don't see how to "guess" it now. Do I just try 12, if it doesn't work 18 and so on? Or just add 6H to that formula? Maybe my initial guess is wrong but adding the H in the NMR spectra is 6 no matter what you do.

I'm stuck with the formula, I think if I manage to get it the remaining part should be easier.

[Babcock_Hall]

If you can identify the functional group of which the nitrogen is a part, you should be all set.  The IR, NMR chemical shifts, and mass spectral fragmentation data can all be used.

[MOTOBALL]

For your MS data, you need to be aware of the  number of “Rings + Double Bonds” in the molecule; this can be calculated from the molecular formula.
You should google this.

Regards

[Babcock_Hall]

The peaks at 125 m/z and 127 m/z are helpful.

[Orcio_87]

@luscofusco Did you considered one of the O₂N-C₆H₄-CH₂Cl ?

IR spectra fits for the para isomer, but does not fit with the H-NMR.

Meta isomer fits with H- and C-NMR but have some different IR spectra.

All isomers fit with yours MS spectrum.

[luscofusco]

Hello @Orcio_Dojek I did consider a nitro group but my first option was a secondary amine. Yours makes a lot more sense, thank you for your feedback.

[MOTOBALL]

@luscofusco Did you considered one of the O₂N-C₆H₄-CH₂Cl ?

IR spectra fits for the para isomer, but does not fit with the H-NMR.

Meta isomer fits with H- and C-NMR but have some different IR spectra.

All isomers fit with yours MS spectrum.

Orcio,

You do them no favours by spoon-feeding them the answer!!!!
We should steer them in the right direction with hints, to allow them to work out the answer themselves.

Regards,
Motoball