I know that shouldnt be the case but this is what i observe : the [M-H+]- elutes around 6.00 and the [2M-H+]- around 11 minutes in the reversed phase. The spectra look similar and could perfectly explain the presence of the dimer in the 11 minutes. But why a non-covalently linked dimer should have such higher RT? Im working with phenolic compounds. I checked and none of the compounds is a fragment of something else.
This is just a guess, but what is the mobile phase? Are you using any modifiers?
I'm wondering if the pH of your mobile phase allows for the presence of both a charged and uncharged form of the molecule with a slow interconversion between the two. The slow interconversion leads to what appears to be 2 peaks. This can happen close to the pKa of your compound, which could be between 4 and 10 depending on the substitution of your phenol. Using reverse phase, the conjugate base of the phenolic compound should elute first, and the uncharged from would elute later. Multimers usually from in MS when the sample is concentrated, and if the second peak is larger than the first, one may see dimers simply due to concentration, or it may simply be that the compound in the first peak is already charged and there are few neutral forms to dimerize.
There are a couple of ways to test my theory-
- Change the pH of your mobile phase and see if the 2 peaks collapse to a single peak
- Reduce the injection volume or concentration and see if the dimer peak is lower in favor of the [M-H]- ion
The attached PowerPoint slide is part of a training module. Although the compounds are basic (split peak is nicotine) rather than acidic in your case, it demonstrates how one can see two peaks from one compound. Running the mixture in a solvent with 0.1% formic acid collapsed the peak nicely.
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