One of the greatest differentiating powers of supercritical fluid extraction (SFE) with carbon dioxide (CO2) is the tunable selectivity of the extraction workflow. I would direct the readers to papers and books that have been written. Today I will do it with glittering generalities in which I have (again) taken (enormous) liberties with scientific theory to fit my word pictures and analogies. (Also I have taken significant liberties with the English language.)
This article illustrates how to predict the best conditions for extraction if you know the solubility of the components of botanical of interest. To illustrate this I made a fictional plant, AlphaBet, with the compounds of interest being A, B, and C.
Overview of Generic Supercritical Extraction System:
I choose a generic extraction system that is simple in design but is automated through computer control to be able vary the pressure, temperature and flow rate necessary for step gradient workflow. I will ignore the length and width of the extraction vessel and just give volume of one liter in the assumptions.
The system is best described by following the flow of the CO2; it begins with a source of liquid CO2, through a cooling heat exchanger, then through a flow meter (measures density of CO2), and next a liquid CO2 pump for (metering the flow based on grams per minute of CO2). On a separate flow path there is a pump for delivering co-solvents (metering liquids solvent such as food grade ethanol (provides faster extractions and automated cleaning of system fluidics). After the mixing T, an in line heat exchanger is used to preheat the CO2/co-solvent before entering the 10,000 psi extraction vessel (5 ml to 10,000 ml size – today we are assuming 1,000 ml) before a computer controlled back pressure regulator (this provides the ability to have automated step gradient pressure methods). While you can have a series of three collection vessels (provides ability to collect extractions at different conditions) this system shows just a single fraction collection vessel before the CO2/co-solvent exit the system.
Model for the Extraction of Components A and B from Plant Cell
The mathematical model of extraction in a bulk fluid phase and assuming that the vessel is packed with AlphaBet material is expressed by:
“The material balance for the solute in the supercritical ﬂuid, around a differential element along the axial direction of the extractor where Cf is the concentration of the solute in the SCF, u is the interstitial velocity of the solvent, ε is the void fraction of the bed, Dsf is the axial dispersion coefﬁcient of the solute in the SCF, asf is the effective solid–ﬂuid contact area for mass transfer, Ksf is the mass transfer coefﬁcient for transport of the solute through the external ﬂuid ﬁlm around the solid particles, and Cf∗ is the concentration of solute in the SCF ﬁlm that is in equilibrium with the solid surface”.
But let’s look at this, without formulas, for extracting two compounds from a single plant cell. I will call these compounds A and B. Initially they are inside an individual plant “cell” before CO2 is in the vessel (Figure 2). With addition of solvent in a specific state, supercritical CO2 at a specific pressure and temperature in this case, compound(s) will have specific solubility of the CO2 entering the plant cell. In other words, the compound (A or B) has to have an “ability” to be extracted and transported through the sample wall. It also has a miscibility (solubility) in solvent itself. The solubility will most likely be different for A and B (Figure 3). Once the materials are outside of the cell, and remember NOT ALL of the material is outside of the cells, as the solubility is a ratio or percentage (and not 100%) of being in cell and out of cell. In the final step the compounds need to be moved away from the cells and transported down the rest of the vessel and out of the system. This is determined by the flow rate of the solvent.
The compound A, for example, at a certain pressure and temperature, it has a certain concentration outside the plant, and a certain concentration inside the plant. In our plant, AlphaBet, at 200 bar, 40 degree C, this is 0.4 (40%) at equilibrium. For compound B it is 0.2 (20%).
NOTE: The extraction and the movement away from the cells are governed by at least major four separate steps and a lot of minor steps in each one. Then there are all the steps that must occur to be able to collect the material. I have just chosen some major ones, knowing that the others are also occurring as well as all the other complexities.
Extraction of A, B, and C from AlphaBet
We will assume the solubility of A, B and C at four different conditions shown in Table 1, we will do an isocratic run and then a step pressure gradient extraction, to see how this would be affect the results. I am also assuming a one liter extraction vessel and 100 mg of potential material of A, B and C.
|Compound||100 bar||200 bar||300 bar||400 bar|
For example, then we can assume that at 200 bar, when the completely extracted, theoretically would collect 40 grams of A and 20 grams of B and nothing of C. Of course, theoretical and actual are not always the same. Substitute “rarely” for “not always” and the previous sentence is more accurate.
Extraction Vessel Volume
If we assume that the vessel volume of CO2 will be one half of the volume of the vessel because the other half of the vessel is occupied by botanical material, then the extraction vessel volume of CO2 will be half the volume.. Therefore, for a one liter vessel, 500 ml will be botanical material and the other 500 ml will be CO2.
Knowing this assumption, ff we again are to look at the 200 bar conditions and our one liter vessel, and again, make some glittering generalities, and assume that we will stop after each extraction vessel volume of CO2 then we can look at what is happening along each stage.
If we therefore assume after one vessel volume that 0.4 (40%) of A will be extracted and 0.2 (20%) of B will be extracted:
100 g A – (100 g A x .40) = 100 g A – 40 g A = 60 g of A after one vessel volume of CO2
100 g B – (100 g B x .20) = 100 g B – 20 g B = 80 g of B after one vessel volume of CO2
At the end of the second 500 ml of CO2, then
60 g A– (60 g A x .40) = 60 g A – 24 g A = 36 g of A left after second vessel volume of CO2
80 g B – (80 g B x .20) = 80 g B – 16 g G = 64 g of B left after second vessel volume of CO2
This will continue with each vessel volume. You can therefore predict when you will “finish”. This will be toward diminishing returns of the amount of material. See Figure 6 for the comparison of A versus B. Note that A is extracted below 1 gram after the 9th vessel volume, while B is 21 vessel volumes.
|Compound||100 bar||200 bar||300 bar||400 bar|
Figure 6 Comparing the number of vessel volumes of A (40%) versus B (20%) at 200 bar
What would the results be if we were to take the same conditions at 400 bar? This takes into effect having compound C as part of the isocratic extraction. There are a couple of changes to note in this graph in Figure 7. The first is that the amount of material that is extracted in the early part of the extraction is higher. For example in Figure 6, taking what is left in vessel and assuming rest is collected, then 40 g A + 20 g B = 60 g material versus 80 g A + 60 g B + 40 g C = 180 g material in first vessel volume. It is important to have the full system be able to meet the expectations of the physics and chemistry of the system to match the conditions.
Another consideration is the postprocessing and final formulation of the product. With initial conditions there could also have unintended consequences from the co-extracted components and also the interaction of the extracted compounds themselves that contribute to conditions. If for example, compound C is more polar a compound than A or B, then when it is extracted and moves into the solvent phase, it will impact the very non-polar CO2 and increase the polarity of the total solvent. As the concentration of C increases as it moves throught the vessel, the solvent becomes more polar. So by the end the solvent is very different, and could extract compounds that are very polar, such as chlorophyll, that we we re not wanting.
Figure 7 Comparing the number of vessel volumes of A (80%) versus B (60%) versus C (40%) at 400 bar
Step Pressure Gradient
One of the key to eliminate these real and potenial issues is to be able to do a step pressure gradient extraction. There are several advantages this can have on the extraction process. One is that is reduces the adverse impact on the more volatile and more soluble compounds that the less soluble compounds can have on them. These compounds are usually the high value added components and can be lost by using conditions that are initially too aggressive. The second, is that with the large amount of material being extracted in the beginning it reduces the power of SFE and fractional collection tools available to chemists. The third, is that the time savings in the beginning is in minutes but it can add days to the post processing steps.
Let’s take a look at an elementary method. In this case, we could start at 100 bar, collect all of A, then move to 200 bar to collect B and then 400 bar to collect C.
Note that all of the A is collected and the 100 grams of compounds B and C are still in the extraction vessel. At this time, the users could collect A. But note that if C is truly a polar compound, it cannot effect the extraction. Only compound A and CO2 could affect the extraction.
Next, bringing up the pressure to 200 bar for 21 vessel volumes will extract compound B. A is already been removed, so we could also bring up the flow rate as it continues because as the extraction continues the amount is less and less. If we are trying to keep the rate the same, that would save time on each step.
Finally we left with compound C. Because it is the only one left, we can go directly to 400 bar to reduce the number of vessel volumes.
Adding up the number of vessel volumes would be 21 for A, 21 for B and 9 for C for a total of 51. Knowing that have increased the time of extraction has to be weighed against the time for the overall workflow. If we want to have the compound in greater purity, then it saves significant time for the overall processes. With the flow at 100 grams per minute, then it would be 51 x 5 or 255 minutes. The time for insolating the compounds would be days, so the addition of 3 hours at even a low flow rate is still days ahead of other modes. But in each step we could increase the flow rate as the amount of material being extracted leading to a significant savings in time.
This has been done successfully to selectively isolate CBN from THC . The preferential extraction of CBD from THC has been done using the same workflow due to their difference in solubility.
Time to Complete the Extraction
If we were to run the CO2 at a flow rate of 100 grams per minute and made the assumption of a density of 0.9 (close enough to 1), then that would be every 500 g / 100 g is 5 minutes. Based on that assumption, how long would it take to go from 100 grams of A to less than 1 gram? It would be 5 minutes times 9 vessel volumes or 45 minutes. If it takes 21 vessel volumes to complete B, then it would be 5 x 21 or 105 minutes to theoretically complete.
But the reality is that it is more complex than this. There are other factors to consider, but that is for next article.
Conclusions for Today:
- Based on the knowledge of solubility you can predict to the first order how to do a step gradient of an extraction that will save significant amount of post processing time by preferentially extracting the components out.
- You can be able to use the power of CO2 compression and selectivity to predict when the extraction will be completed based on the number of vessel volumes of CO2 that the matrix is exposed to and also with the knowledge that it will always be more complicated, however it will give you a good starting point for method development.
- As I always encourage, use the peer reviewed resources available to further your understanding and apply them to advance your productivity today. Stand on the shoulders of giants. There are so many research teams around the world that are using mathematics, advanced scientific instrumentation, and good laboratory practices and reporting their results in peer reviewed journals that are available in University libraries today.
Hoping you extract the best from your day!
John A. MacKay, Ph.D., is the Senior Director, Strategic Technology at Waters Corporation. He earned a B.A. in Chemistry from St. Lawrence University, and Ph.D. from the University of Vermont in chemistry focused on the synthesis for cancer fighting molecules using a cyclic -phosphazene backbone. After positions teaching organic, biochemistry and analytical courses, at Davidson College, Lyndon State College and UVM as well as supervising grant based funded student research projects, John joined Waters Technologies in July, 1983. His career has included many roles in product development and now heads up the newly formed Strategic Technology team for the Americas Business Operations. He is widely recognized as scientific expert in application of supercritical fluid for extraction and chromatography across a wide application space including botanical space.
John can be reached at [email protected], or [T] +1 508 482 2104.
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