Can anyone identify the peaks?
(1 votes)
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You assume there is an aromatic system, as we have peaks at 6.5 to 6.7. This cannot fit for several reasons.
Only a proton, namely the left next to the lowermost OH group, has an adjacent one which may be is coupled to (and vice versa) – i.e., two dublets with a singulet in the arom. System (without identical environment). The other H atoms in the aromatic system have no adjacent H atomsThere’s something wrong. The other H atoms are likely to appear singulet at best. Somehow it doesn’t fit
Furthermore, the protons of the two OH groups in the arom. System appears at about 4.45 or so, but there is only one proton to see and this is also coupled with others in the environment (J_ab and J_ac). – The roof effect also says that the actual signal should be at 4.45.
Look at the peaks at about 6.5. The two double peaks are wonderfully symmetrical and equally apart, which means that here too a quite simple environment must be present, with the same coupling constants.
I have the feeling that we have a nitro group somewhere (-C-CH2-NO2), for this also speaks the N–O asymmetric stretch of 1550-1475 cm-1
Thank you for your answer!
I have the spectrum with the side https://sdbs.db.aist.go.jp/HNmrSpectralView.aspx?imgdir=hsp&fname=HSP46136&sdbsno=5166 compare with the molecule adrenaline. The spectrum looks the same, don’t you think? I also wondered where the peaks of the OH groups and the NH group can be seen, because the peaks are not shown there? As a singulet, this would not necessarily have to appear as an H atom on the Armenian carbon, since coupling can be coupled via 4J (and 5J). Accordingly, all peaks in the aromatic system are dublets.
and this is your fault…so pages do not always have to vote. Look at my answer, go through the fragments, look at your IR peaks.
A shift of 4.5 often indicates a nitro group in the H-NMR.
Why the OH groups are to be seen is simple because they quickly exchange protons with the residual water of your compounds, and that amines are always difficult to locate (see comment)
In your 400 HZ spectrum, you are guaranteed to have no coupling over more than 2 bonds, unless your aromatic system has a proximity to other groups that have protons.
An aromatic system usually appears at 130 in the C-NMR. The OH groups would rather shift it to 140 Hz.
It’s just my opinion, maybe I’m wrong.