July 3rd, 2008
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Resonance Spectroscopy
Resonance Spectroscopy
nmr spectroscopy plz plz help!!?


im given cupling constants and relative integrals next to each resonance on the spectrum. I have a proton and a carbon 13 nmr. There are 4 peaks on carbon, and 3 different peaks on the 1H nmr. The 3 peaks are a triplet, a sextet, and another triplet with 1.5H. they all have J=7.1 Hz, and aside from the third triplet, they are 1H. Please tell me how to find the structure of the compound if the molecular weight is 114.

(CH3-CH2-CH2)2C=O old guys like me call it dipropyl ketone but the politburo say it is 4-heptanone
Sextet? Haven't invented CH5 yet so must be a CHa3-CHb2-CHc2 system where JHH to central CHb2 is same for Ha and Hc. The rest is easy. The C resonance of CO will be down around 200 and much weaker than expected (no NOE).
Confirmed:

http://riodb01.ibase.aist.go.jp/sdbs/

Put in 4-heptanone and click H NMR



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Nuclear magnetic resonance spectroscopy

NMR Spectroscopy homework Help!?


the question is: A proton has a resonance 90 Hz downfield from TMS at a field strength of 1.41 T and and rf frequency of 60 MHz.
What will the shift be in Hz and ppm in a 200 Hz NMR and in a 500 MHz NMR?

frequency of 90Hz/60 MHz = 90/60,000,000 = 1.5 ppm so it is shifted downfield 1.5 ppm

1.5 ppm is always 1.5 ppm no matter what field and MHz NMR. Makes life easier that way.

200 Hz (you said 200 Hz RIGHT, not 200 MHz) x 1.5 ppm = 0,003 Hz

500 MHz x 1.5 ppm = 750 Hz

remember ppm is parts per MILLION:) so multipling by 1.5 ppm is multiplying with 1.5/1,000,000

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