Stronger ESI signal for deuterated substances

I observe a constant phenomenon in my ESI measurements, which indicates that deuterated substances give a stronger signal than the same substance in non-deuterated form. Unfortunately I can't make any sense of it, because I can't see any reason why a deuterated substance should be protonated any better. Since I have not seen or read anything about this phenomenon so far, I would be very pleased if someone could clarify whether this is something well known, or whether my samples just happen to show the same artifact over and over again. If it is a known phenomenon, I would also be happy to hear an explanation.

Further bits of information: LC-MS/MS in ESI mode, same acquisition method for both forms of the compound, observed in EPI scans, increase in intensity c.a. 3-fold.

To make this very clear, the two compounds (deuterated and non-deuterated) are measured separately in separate vials with a blank in between, so no co-elution can occur. One measurement is done with a solution of 100 ng/mL estradiol in acetonitrile and one with 100 ng/mL estradiol-d5 in acetonitrile. They are not mixed!

• Is your deuterated internal standard co-eluting in LC? How many Ds there? – M. Farooq Jun 18 '20 at 12:54
• @M.Farooq Although the deuterated substance is of course intended as IS, it does not fulfil this role in the measurement mentioned above. In fact it is only a reference measurement with one substance each dissolved in 0.1% FA and an acetonitrile gradient. So the two compounds are measured individually. The compound we're talking about is estradiol and estradiol-d5 respectively. – Sam Jun 18 '20 at 13:01
• @M.Farooq As I said, it cannot be a related to co-elution as the two compounds are measured separately in separate samples and I even ran a blank in between the two samples to make sure that none of the previous compound was left to be able to suppress the signal. This must be caused by something else. I have to thank you for your reference, though. Even if it's clearly not the case here, it's nevertheless interesting for a later step, when the deuterated estradiol is in fact used as an IS. – Sam Jun 18 '20 at 14:20
• Great q. After I wrote my answer I realized maybe you are using an ion-trap instrument. It does not do beam-type CID, but resonant CID. I think a similar but slightly different version of isotope effects during fragmentation could happen there. – Curt F. Jun 18 '20 at 17:26
• That's true, I'm using the Sciex QTrap 6500. I'll look into the ion effects on resonant CID. – Sam Jun 18 '20 at 19:20

1 Answer

There are a couple reasons this could be happening.

The most interesting possibility: isotope effects during ion fragmentation

You are performing LC-MS/MS, i.e., your mass spectrometer is chemically reacting the estradiol and estradiol-$$d_5$$ with gas molecules in the collision cell of the instrument. This process, known as CID or collisionally induced dissociation, will break chemical bonds in the parent molecule and form fragment ions.

On triple-quad and qTOF instruments, the CID is of the "beam" or non-resonant type. This means a fairly large amount of energy is imparted to parent ions. In general, this means that a wide variety of fragment ions are formed. This is true of estradiol; publicly available fragmentation spectra show many fragment ions for this compound (see several GNPS spectra for example).

Deuteration is likely to change the relative propensity of the different fragmentation pathways. For example, these reactions could represent the fragmentation of a parent ion $$\ce{P}$$ into two different fragments $$\ce{F1}$$ and $$\ce{F2}$$.

$$\ce{P -> F1}$$ $$\ce{P -> F2}$$

Isotopically normal $$\ce{P}$$ might give an intensity ratio of say 1 to 1 for $$\ce{F1}$$ and $$\ce{F2}$$. But due to deuterium isotope effects, the pathway to $$\ce{F1}$$ may be much less favored in deuterium-labeled $$\ce{P}$$. This could lead to an enhanced signal for $$\ce{F2}$$ as a much higher fraction of the $$\ce{P}$$ ion remains available to proceed down that fragmentation pathway.

However, a three-fold increase is pretty large. I was able to find good measurements of various deuterated tocopherols, and judging from Table 1 in that paper, the response ratio for $$d_0$$ (unlabeled) tocopherol vs. $$d_9$$ is 11.8/0.967/10, or 1.22, not 3. For measurement of trimethylamine N-oxide, authors note "the intensity ratios of the two product ions are different between TMAO and $$d_9$$-TMAO due to the difference in bond energy among C and H and C and D (deuterium)" but did not quantify this difference. Eyeballing Figure 1, the intensity ratio of the light and heavy fragment is about 2.5:1 for $$d_0$$ but 1.4:1 for $$d_9$$. None of those values are close to 3-fold. But maybe estradiol and its site of deuteration and its fragmentation pathway differ and that a value of 3-fold is possible.

Errors in sample preparation

Maybe the concentration of one of your samples is off. How many times have you independtly prepared the standards and repeated the measurement?

Ways to tell

If you can see a signal with a single quadrupole (i.e. do MS1 instead of MS/MS), do you see the same three-fold difference? If so, error in sample prep is likely to blame. If not, then isotope effects during ion fragmentation are to blame.

• Thank you very much for your explanations and ideas! I have, in fact, done a triplicate measurement, but I didn't prepare several independent solutions (shame on me!), so a sample prep. error is possible. As I'm pretty confident in the technique I use to dilute the solutions, I am sure though, that the concentrations will not be so massively off that the 3-fold increase could be explained by this. I could imagine, however, that they are in fact off and that this, combined with the isotope effects described by you, adds up to the 3-fold increase. I'll check this! Again, thanks a lot! – Sam Jun 18 '20 at 19:18
• Would be very curious to hear how the results turn out. If you have a chance to prepare independent samples or to try MS1 (if you're not sure how you could monitor the transition of 255 -> 255 for d0 and 258 -> 258 at a collision energy of 0, which would bascially do the same thing). In fact, you could also pretty easily measure a "survival curve" by monitoring these transitions at a range of different collision energies, and plotting the dropoff of intensity as a function of collision energy. – Curt F. Jun 19 '20 at 0:51
• I'll prepare three independant samples of estradiol, for estradiol-d5 it's a bit trickier. The sample was already dissolved in acetonitrile when I bought it from Sigma, but I suppose that the concentration is correct. I'll combine this with an EPI with a low CE (5 is the minimum on my device). Unfortunately I cannot do an MS1. Estradiol is hardly ionizable and thus the signals are pretty low. In MS1 measurements I'd need a concentration waaayyy higher than I could possibly prepare. Thus I have to use an EPI experiment. I'll report! – Sam Jun 19 '20 at 7:33
• Very cool, I'm excited to hear what you find! You can also make "survival" curves for actual MS2 ions, ramping collision energies, if that gets you better signal. If you use an actual fragment ion, the maximum won't be at 0 (or 5 CE), but at some other CE. The great thing about the survival curves is they should be different if isotopic effects are responsible for your issue, and the data is insensitive to concentration. You normalize each curve to a reference condition. So you can use samples you already made -- even if there are errors in sample prep, it won't matter. – Curt F. Jun 19 '20 at 18:10
• Just as a suggestion, how did you prepare the non-labelled estradiol stock solution? I read that your deuterated solution is already provided in acetonitrile by Sigma. Maybe you ran into a solubilization issue in the course of the stock solution. Estradiol is not a very soluble compound. Maybe a quick jump to the spectrophotometer would help and have a look at the absorbance of the stock solution from Sigma and your at a similar concentration in acetonitrile also. – PLD Jun 23 '20 at 15:29