Being not a spin-trapping specialist, I would firstly assume from the spectrum, that the beta-H doesn't exist in your adduct. That would mean you have a side-reaction, where the beta-H is replaces by something else.

The a(N) isn't that low for POBN, togehter with the rather high g-value it reflects a quite hydrophobic environment. What is your used solvent?

Thank you very much for your insightful analysis and constructive opinion. I am now considering whether the beta-H on POBN adduct might be removed during the reaction by H2O2 (one of the products in this system, but in the concentration of ca. 10 mM ). May that serve a possible reason ?

This reaction is carried out in deionized water, providing a hydrophilic environment.

Interestingly, my recent experiment result shows that the EPR spectrum of this radical without a spin trap also presents a 9-line signal. May We assume that this radical could be a nitroxide radical containing a 5-membered ring hydroxylamine structure ?

On zooming the ESR spctrum,we observe that the three peaks are EQUALLY SPACED and are ALMOST EQUALLY INTENSE implying that the unpaired electron remains localized only on N (I=1)i.e. it does not undergo any hyperfine splitting with the neighbouring protons.

This does happens in cases where either the concentration of the radical is very less or temperature is somewhat on the higher side (the population of the electrons will be very less in the ground component).

I guess that you may observe hyperfine interaction at very low temperatures (liquid N2/He).

As regards to g being 2.0068, there should be no point of conflict. The HSC are also expected to different when the same nucleus undergoes hyperfine interaction or no hyperfine interaction with.

Finally, depending upon the reactions, it has been reported to show 12 or even 6 line pattern in free radical metabolism where HSC=15.38-15.48G.

Pls.treat it as an academic discussion; no excellence is proclaimed.