We are conducting an experiment in biology class, where we let Cannabidiol (CBD) react with Liver Enzymes (CYP2C19) (And NADH, which is needed as an energy source for the reaction). First we prepare different samples (with different concentrations of CBD and one without any CBD) and put them in a water bath incubator. We then measure the amount of NADH that has reacted via UV Spectroscopy at 340nm. Like that we determine how fast different concentrations of CBD have reacted (a small extinction = a fast reaction).
Further should be noted, that the CBD was added in a solution of Ethanol and the Liver Enzymes were obtained by centrifuging cow liver 8so there is also Alcohol dehydrogenases (ADH) in the solution. The Alcohol gets broken down by the ADH and forms NADH from NAD+. So, the reaction can has a lot more NADH, than initially added.
So here is my question: CBD reacts twice with CYP2C19 (as shown in the attached picture or here: http://profofpot.com/wp-content/uploads/2016/08/Cannabinoid-metabolism-568x1024.png), so it has an intermediate product. What would be the best way to determine the concentration of the intermediate molecule (7-OH-CBD) just by using lab equipment, that you could find in a high-school (so no mass spectrometry or anything like that).
If you want to get any further detail on the experiment (we conducted it already once and are going to do it again), if you have other suggestions for the experiment or if I got anything wrong, please contact me.
(CBD is legal to obtain and consume where I am from, so this experiment was completely legal)
Thank you for your help