Hello! I was running a method for octyl methoxycinnamate in a sun cream product. The mobile phase is (40:40:5:15)ACN:MeOH:THF:H2O. The diluent is methanol. In the standard and sample a negative peak occured immediately before the analyte causing interference. I suspected the BHT preservative in the THF may have caused the negative peak so I injected 5mg/mL BHT in methanol and found a large positive peak where the negative peak eluted. Does anyone know why there was a negative peak in my runs?

What wavelength? Maybe you have detected a product that doesn't absorb at that wavelength and then you detected a negative peak (always in reference to the blank absortion)

What happens when you run: (1) a blank: MeOH? When
(2) you run mobile phase as a sample? When (3) you
collect some eluted mobile phase and run it as a
sample?

Hey guys, octyl methoxycinnamate (OMC) in a sunscreen product is one of the easiest assays one can do; sunscreen actives are present in relatively high levels, and a selective wavelength (313 nm) is used so sample matrix effects are absent. My company does sunscreen actives routinely, has validated the test methodologies. We use 2.1 mm i.d. x 10.0 cm 5u RP-18 column coupled with 2.1 mm i.d. x 3 cm 5u RP-18 guard column. If no other sunscreen actives besides OMC are present, we use Solvent A = HPLC H2O Solvent B = ACN or methanol
Flow = 0.25 ml/min.
%Solvent B = 85
Wavelength = 313 nm
Run time = 9 min. (using ACN) or 12 min. (using methanol) With these conditions, octyl methoxycinnamate will elute at about 6 minutes using 85% ACN and at about 9 minutes using 85% methanol.

For samples that contain both oxybenzone and octyl methoxycinnamate (or oxybenzone only):
Solvent A = 0.4% acetic acid in HPLC H2O
Solvent B = THF
Flow = 0.25 ml/min.
% Solvent B = 55 Wavelength = 313 nm
Run time = 8 min.

Oxybenzone and octyl methoxycinnamate will elute at about 2.5 and 5.5 minutes, respectively, using these conditions. We make up our samples and standards in DMF.

To the person who asked the question: you proofed yourself that the negative peak was due to the BHT in the mobile phase phase, since it eluted at the same time as the positive BHT peak that you got when you injected BHT. You will get a perfectly flat baseline only, when the BHT in the mobile phase is at exactly the same concentration as the BHT in the sample, which may be very difficult to do.

The wavelength I am using is 280nm. When I run a MeOH blank I see the negative peak as well. What I don't understand is why I would see such a strong negative peak at the retention time of BHT in a sample with no BHT. Since there is no BHT in any of my samples, wouldn't the difference in BHT concentration between the mobile phase and the sample show up only at the void volume?

If the BHT is retained under the chromatographic conditions that you are using and you have BHT in the mobile phase, you will get a negative peak when injecting a sample with less BHT than in the mobile phase and a positive peak, if you inject a sample with more BHT than in the mobile phase. The negative peak has been called vacancy chromatography.

It certainly appears to be a vacancy peak.
Suggestions to minimize interference:

1) Use THF w/o BHT in your mobile phase
2) Dissolve your sample in mobile phase
3) Dissolve your sample in MeOH and make
subsequent dilutions w mobile phase
4) Inject a minimum volume of sample. The area of
the vacancy peak is normally proportional to the
solvent volume.

Can you please expand on the concept of vacancy chromatography? Is the negative peak due to a small amount of BHT from the mobile phase dissolving in the sample in a lower concentration than in the mobile phase? It is difficult for me to see how there can be a response in a sample absent of BHT.

I would change to wavelength of 313 nm to increase signal-to-noise and get better specificity. I would also not use THF containing BHT for HPLC assays.

Responding to Anon of 7/15-08:19. What is
happening here (as U. Neue has explained) is that
the mobile phase has a component (BHT) which (1)
absorbs light at the wavelength selected and (2)
has a given retention time which is greater than
MeOH i.e. MeOH cannot displace BHT from the
stationary phase. The injection of MeOH therefore
dilutes the mobile phase wrt BHT: creates a mobile
phase segment that has a lower absorbtivity than
the rest of the mobile phase. The diluted mobile
phase segment moves thru the column with the same
retention time as BHT but the absorbance is less
than that for which the baseline was balanced
therefore producing a negative vacancy peak. Intro
to Moderen LC, Snyder and Kirkland is a good
reference to start with.