Question

9-fluorenol NMR: I've provided two sets of NMR's. The theoretical vs. the experimental. In the theoretical I understand all the shifts except for the 2.06, representative of the H alpha to the OH group. Methine protons usually have a shift around 1.7ppm and protons alpha to an (OH) are deshieled adding +2.5ppm, according to my textbook, which would make me think that the actual shift be 4.2ppm. Why is this not the case? Could I be mistaking this nuclei for another? I've attached my experimental NMR and I'm wondering what's causing alot of these seeemingly incorrect shifts. Is this due to impurities, and how would impurtities effect the NMR results. The huge integration 3.55ppm is 492. We used chlorofor as a solvent for NMR tubes so I was thinking maybe this caused the huge peak. Thanks. IMAGES: http://imgur.com/a/gWTXM

Answer 1

For methine protons usually shift around 1.7 ppm, this is the case when the substitutions are alkyl groups. Here the substituents are benzyl groups. So the chemical shift will be around 3.0- 3.2 ppm and when hydroxy attached which will add +2.5 more to give 5.5-5.7 ppm. However, chemical shifts vary with the deuterated solvents you used for recording NMR spectra.

Different NMR solvents give different chemical shifts and the presence of metal impurities like (Pd catalysts, Rh, Ru and Ir catalysts) in the sample can produce variation in chemical shifts.

Further, your experimental NMR spectrum is very impure. Even it has more impurity than your compound. In NMR, the aromatic region has integration value of 120 for 8 protons where as hydroxy itself has the value of 283.

Chloroform gives NMR peak at 7.27 ppm, not at 3.55 ppm. So it's not chloroform. It can be some other solvent which you used for the reaction. If you can provide the reaction what you have done (if you don't mind) then it would have been easy to analyse spectra more clearly.