SIDA issue

Discussions about sample preparation: extraction, cleanup, derivatization, etc.

7 posts Page 1 of 1
I'm planning to measure a volatile compound found in liquor in tiny quantities using stable isotope dilution analysis - by adding the d10 deuterated analog of the analyte.

The blank I'm using is spiked with same quantities of both analyte and deuterated analog, then derivatization is performed followed by SPE.

However, when I run the extract on a GC-QQQ, the integral of the deuterated derivative peak is approximately 50% lower than the one belonging to analyte derivative.

What might be the reason for that and how to avoid it?

In my opinion it's unlikely H/D exchange takes place, since all the deuteriums in the analog are alkylic, and the only remaining protons (H) sit at two -OH groups.
Do you have the same situation in the calibration curve (without the extraction procedure)? My guess is that the detection settings might be less optimal for the internal standard compared to the analyte (parent/daughter mass, collision energy,..). I don't think it's an issue provided you have similar behavior in the calibration curve.

Edit: typo
It seems that the source of the issue is the secondary isotope effect, which influences the kinetics of the derivatization reaction. That's probably why not all of the deuterated standard is converted.

The question is how to fight against that, what could be done in order to increase the conversion of the deuterated analog during derivatization step...?
Krusvar wrote:
It seems that the source of the issue is the secondary isotope effect, which influences the kinetics of the derivatization reaction. That's probably why not all of the deuterated standard is converted.

The question is how to fight against that, what could be done in order to increase the conversion of the deuterated analog during derivatization step...?


The other thing to consider is if the signal generated by the same mass of deuterated analyte is the same as that of the native analyte. I know in full scan mode that toluene and toluene-d8 do not have the same peak area counts when both are in the standard at the same concentration. Sometimes it is up to 10% difference. Does the deuterated analyte ionize as easily as the native? There are many things to consider before determining that the difference is caused by just the reaction in the preparation.
The past is there to guide us into the future, not to dwell in.
It can be either a lower derivatization yield of the deuterated analogue (giving a lower signal that reflects the lower concentration compared to the analyte) or a lower sensitivity of the derivatized, deuterated analogue (giving a lower signal that does not reflect the similar concentration). In your first post, you talk about the peak areas. What happens when you perform quantification (working with area ratios) ?
James_Ball wrote:
The other thing to consider is if the signal generated by the same mass of deuterated analyte is the same as that of the native analyte. I know in full scan mode that toluene and toluene-d8 do not have the same peak area counts when both are in the standard at the same concentration.


What is the reason for that behavior, how do deuteriums influence the change in peak area, since ionization should occur in quite a similar way? How to deal with this kind of situation when it comes to quantification of analyte in real samples, can we just consider the ratio of areas of non-deuterated analyte vs deuterated analog during quantification? I had no previous experience with that, that's why I'm asking...

Or is it maybe better to deal with MS/MS conditions here, instead of trying to modify the reaction procedure...?
We calculate a calibration factor ( also called response factor ) which is the ratio of the area of internal standard and concentration of internal standard versus the area of the target and concentration of the target.

https://msuweb.montclair.edu/~olsenk/intstd.htm This gives a good explanation and example of how to use internal standard to calculate unknown concentration, the F in the equation would be the response factor.

If you do multiple points on the curve you can calculate a response factor for each point then take an average to use to calculate the unknown concentration as long as the % relative standard deviation between the individual response factors and the average is low ( normally 10-15%RSD for most EPA methods).

There are also options if the calibration curve is a better fit to a linear regression or quadratic regression over the range of the calibration curve, but most instruments should have all the equations built into the quantification software.
The past is there to guide us into the future, not to dwell in.
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