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Fragmentation of NH4 adduct dufferent from Na adduct?

Posted: Tue Mar 03, 2015 9:32 am
by bunnahabhain
Dear all

I hav a somewhat strange mass spectrum of ivermectin. The sample is a new standard, so I trust it is really ivermectin :)
The ESI+ spectrum shows no M+H ion, weak M+NH4 ion and strong M+Na ion. I know I can't get rid of Na, as soon as I use any glassware there is No ions present. So for maximum sensitivity I should use the Na ion for MRM method development. I recorded daughter spectra of the M+NH4 ion and compared them to daughter spectra of the Na ion, and there is no mass peak that is common to both. How can that be? BTW, the ions of the NH4 adduct daughter spectra can easily be linked to fragments from the molecular structure, and described in literature, the Na adduct fragments not.

See pictures below:

daughters of M+NH4 (= 893)
Image

daughters of M+Na (= 898)
Image

Thanks for any hints
Jörg

Re: Fragmentation of NH4 adduct dufferent from Na adduct?

Posted: Tue Mar 03, 2015 12:28 pm
by lmh
It's OK, it's to be expected. Firstly the two adducts may fragment completely differently anyway (the charges may be in different places on the ion, depending on adduct, for example, and this will affect reactivity; ivermectin has some bits that look like sugars, and sodium adducts of oligosaccharides do fragment differently to ammonium adducts: I believe the sodium can bind between two sugar moieties, reducing the likelihood of a fragmentation of the glycosidic bond between them).

Secondly, the fragments from a sodium adduct will very often be sodium adduct ions themselves. So, for example, the sodium fragment 329 could be the same thing as the ammonium fragment 307 (mass difference is 22). Typically the ammonium adduct ion is lost from all the ammonium fragments, which will look like hydrogen-adducts of bits of molecule. Sodium only tends to be lost from sodium adducts as part of a neutral fragment when the fragment was actually some really acidic group, able to leave as a sodium salt, if that makes sense?