Phenalenyl with natural-abundance of 13C isotope

[trukhan]

I noticed that the EPR spectra of phenalenyl radical anion with the natural abundance of 13C in the examples \isotropic\phenalenyl_manual.m and \isotropic\phenalenyl.m differ markedly.
In my opinion, it would be more correct to use this code

Code: Select all

Sys.n = [6 3 6 3 3];

in the phenalenyl.m instead of

Code: Select all

Sys.n = [6 3 1 1 1];

but, unfortunately EasySpin "Cannot compute isotope patterns for equivalent nuclei."

I think this problem can be solved as follows:

Code: Select all

Exp.mwFreq = 9.5;
Exp.CenterSweep = [339.4 8];
Exp.nPoints = 10000;
A_H = [-0.629 +0.181];
A_C = [+0.966 +0.966 +0.966 +0.966 +0.966 +0.966 -0.784 -0.784 -0.784 -0.784 -0.784 -0.784];
Sys.g = 2;
Sys.n = [6 3 1 1 1 1 1 1 1 1 1 1 1 1];
Sys.A = mt2mhz([A_H A_C]);
Sys.Nucs = '1H,1H,C,C,C,C,C,C,C,C,C,C,C,C';
Sys.lwpp = [0, 0.01];
[x,y2] = garlic(Sys,Exp);
plot(x,y2);

(warning: calculation time is about 15 sec on a 3 GHz processor)

The spectrum calculated in such a way is almost identical to the spectrum calculated by phenalenyl_manual.m example.
Moreover, in this case, one can see additional small peaks, which, in my opinion, are due to the species with more than one 13C.
Is it right? Any comments?

[Stefan Stoll]

This looks right. We will update the phenalenyl example. Thanks!