Partition Coefficient

Partition Coefficient

Partition

If two immiscible phases are shook in a separatory funnel they will form two separated layers. The phases can be in the form of gases, liquids and even solids, but for simplicity only the liquid phase will be used in this explanation. Suppose liquid one is water (H2O) and the second one is diethyl ether (CH3-CH2-O-CH2-CH3), the ether will float on top of the water since it has a lower density. A substance called X, soluble in both liquid was put in before the shaking took place. Suppose the particles of X are more soluble in the ether than in the water such that the scenario will resemble the picture below. A dynamic equilibrium will be established which consists of a constant rate of particlescrossing the boundary between the two liquids.  

 

  This information could be interpreted in the following equation:

          X(in solution in water)    X(in solution in ether)

This is a simple equilibrium, hence an equilibrium constant can be found.

Kc = [X in ether] / [X in water]  

This particular equilibrium constant is given the name Kcp, which stands for the partition coefficient.

The partition coefficient is dependent as all equilibrium constants on temperature but they do have specific limitations as well. For example the solutions have to be quite dilute for the constant to be accurate, the solute is required to be in the same molecular state in both solvents. This means that the solute cannot ionise, react or associate.   

Partition Coefficient

The calculations of the partition coefficient is a simple ratio of two separate concentrations. This proposes that the units that are used are irrelevant as long as they are the same in the numerator as in the denominator. The concentration can be expressed as mol dm-3 - moles per cubic decimetre, but more commonly g cm-3 - grams per cubic centimetre is used. The important distinguishing factor between the units is that the square brackets are strictly used for mol dm-3. This is to say that [X] is only to be used for standard concentration units, and in other cases “concentration of X” is a viable way.  

Lipophilicity and log P

To summarise; the partition coefficient is the ratio between the concentration of un-ionized solute in two immiscible solvents. To make this possible the pH of the water is adjusted to make the largest amount of compound un-ionized. The logarithm of the Kcp value is also known as the     log P, which is the measure of Lipophilicity. Lipophilicity is the tendency for molecules to dissolve into fats, lipids, oils and other hydrophobic solvents such as ether.   

log Poctanol/water = log ([solute]octanol / [solute]water)

When calculating the partition coefficient, the compound dissolved must be strictly be un-ionized. The same principle can be applied to a compound that ionizes completely or partially, and in this case it is called the distribution coefficient.  Like the partition coefficient the distribution coefficient is a ratio between the concentration of solutes in two immiscible solvents, but in contrast it takes into account ionized and un-ionized compounds. In practice the aqueous state is buffered to a specific pH, which will be stable when a compound is introduced. The distribution coefficient is the logarithm of the ratio between all the sum of all forms of a solute in two solvents.

log Doctanol/water = log ([solute]octanol / [solute] (ionized) water + [solute] (un-ionized) water)

The distribution coefficient can be seen as an extended form of the partition coefficient, taking more factors into account, which implies that log P = log D when ionization does not occur. As log D is dependent on pH it must be specified at what value of pH the calculations were performed.             

Practical Applications

The application of this method its applications can be seen in the chemical and pharmaceutical sciences. Normally water is chosen as one solvent while the other one is a hydrophobic, such as octanol. This makes it possible to determine how hydrophobic or hydrophilic a substance is by determining the partition coefficient. This in terms of medical studies allows to determine what distribution a drug will have inside the human body. Based on this an appropriate compartments of the human body are chosen to distribute them within. If a drug has a high octanol/water partition coefficient it is preferably distributed in an area that is hydrophobic such as the lipid bilayers. On the contrary, drugs that have a low octanol/water partition coefficient are preferably distributed in a hydrophilic source such as the blood serum.  

Lipophilicity is crucial in pharmacological practice, not only by how easily a drug can dissolve into fats as mentioned in the previous paragraph. It helps determining how strongly it will affect its target and how long it will stay in the body in its active form.   

A reliable way of determining the partition coefficient is by the “shake flask method”. The solute is dissolved by shaking the flask containing the solvents octanol and water. The concentration is then measured in each phase by UV/VIS spectroscopy

Problem Involving Partition Coefficient

Problem I

A solution of 1.56 g of X in 125 cm3 of water was shaken with 12 cm3 of ether, 0.89 g of X was moved to the ether layer. Now calculate the partition coefficient of X between ether and water.

Solution

The question asks to calculate the coefficient of X between ether and water. As the ether is stated first in the question it will come on top in the expression as follows.  

Kcp = concentration of X in ether / concentration of X in water

Concentrations are given in g cm-3 and hence will be calculated with these “non-standard” units.

Concentration of X in ether: 0.89g/12cm3

Concentration of X in water: 0.67g/125cm3

The concentration of water is not directly given, but can easily be calculated from the information that 0.89g was transferred to the ether. Hence (1.56g - 0.89g = 0.67g) is left in the water.

So:

Kcp = concentration of X in etherconcentration of X in water

     = (0.89g/12.0cm3) / (30.67g/125cm3)

     = 13.8  (3 sig fig)

Note that all units cancel in the expression and a unitless number is the result.   

Problem I a)

Now calculate the Lipophilicity (log P) of the previous question.

Solution

The log P value of the previous question is simply the logarithm of the Kcp.

log(13.8) = 1.14 (3 sig fig)

Problem II

In this problem use the same solvents, X and partition coefficient as given or calculated in previous problem.

How much of X would have been extracted to the ether if the original solution (1.56 g of X in 125 cm3 of water) was shaken with half the amount of ether.  

Solution II

Like in the previous equation we represent an expression for the concentration of both solvents.

Concentration of X in ether: m/6 g cm-3

The unknown mass will be represented as m.

Concentration of X in water: (1.56 - m)/ 125 g cm-3

Now, the information can be plotted into the expression for Kcp and solved for m.

Kcp = concentration of X in etherconcentration of X in water

13.8  = m/6 g cm-3(1.56 - m)/ 125 g cm-3        Note that all units cancel

13.8  = m/6 (1.56 - m)/ 125    

13.8(1.56-m) / 125  = m6   

21.528-13.8m / 125  = m6   

         21.528  = 125m6 + 13.8m

             21.528  = 34.633m

          m  = 21.528 / 34.633

          m  = 0.622   (3 sig fig)

Hence the mass of X extracted to the ether is 0.622 grams

Extraction of paracetamol (alvedon) from 4-aminophenol

I recently made a visit to Stockholm University to extract paracetamol, the chemical name for alvedon. Followed is a summary of the lab.

Aim

The aim of the lab is to extract one of the following products from an unknown substance and then investigate which chemical has been produced.

Method

• Take 1.1g of substance A and mix it with 3 ml of distilled water in a 50 ml boiling flask equipped with a magnetic stirrer.
• Add acetic anhydride and agitate the bottle.
• Then put a condenser on the flask and then but it in an oil bath with a temperature of 70-80°C until the solution becomes clear (this might take up to 30min).
• Let it cool gradually, first in air, then water and finally in ice.
• As the substance crystallizes filter it out using a Büchner funnel and water suction.
• Rinse the flask used with cold distilled water and pour it into the Büchner funnel to get all the crystals out to ensure the optimal yield.
• Put the crystals to an Erlenmeyer flask to prepare for recrystallization while heating distilled water in another.
• Add the water to the crystals just until the entire product is dissolved.
• Let it cool until crystals appear and then filter them off and wash with small amounts of cold distilled water and weight the product.
• Dissolve a small amount of the product produced, Ibuprofen, Acetylic acid, Paracetamol in approximately 1ml of methanol.
• Using a small capillary take small amounts of each sample and release it on a THC-plate. Between each sample be sure to clean the capillary with the use of acetone.
• Pour in a 1:1 mixture of pentane ethylacetate (about 3ml) into a TLC-chamber.
• Then place the THC-plate inside the TLC-chamber and be sure to check that the samples are not in direct contact with the liquid.
• After the THC-plate has dried, look at the samples under an UV-light and make observations. 

Discussion

It was concluded that the product created was paracetamol when a thin layer chromatography test was performed and observed under ultraviolet light (which is discussed further down the page). An unknown substance was treated with acetic anhydrate and resulted with paracetamol. A compound that behaves in this matter is 4-aminophenol and is widespread in the industrial production of this drug. Upon these observations was the build up of “Chemical A” decided.  

When 4-aminophenol is t treated with acetic anhydrate it produces paracetamol. This procedure is the last stage of an industrial production of the drug called “Alvedon”. 

Followed is a balanced equation of the reaction taken place. 

C₆H₇NO + C₄H₆O₃ --> C₈H₉NO₂ + C₂H₄O₂ (acetic acid)

4-aminophenol + acetic anhydrate à paracetamol + acetic acid

Mr 4-aminophenol - 109.13 g/mol

       acetic anhydrate - 102.09 g/mol

       paracetamol - 151.17 g/mol

       acetic acid - 60.05 g/mol       

As the experiment started, approximately 1.09 grams of 4-aminophenol was measured out to use in the experiment. This is (1.09/109.13) 9.99 x 10^-3 moles of acetic anhydrate which should result in the same amount of moles of paracetamol, as these are created in a 1:1 ratio. The theoretical yield is (9.99 x 10^-3 x 151.17) about 1.51g of paracetamol which is the amount that should be received. The actual yield at the end was 1.05g which is ≈ 70 % (69.54 %) (1.05/1.51 x 100) of the theoretical amount.

Recrystallization was performed to purify the compound received. The crystals of compound A were filtered out using filter paper. Doing a recrystalization purifies the substance but also decreases the yield as material is lost due to the finite solubility of paracetamol. This together with the product lost on the filter paper makes up for the primary loss of the substance, decreasing the yield.     

In practice it is impossible to receive the total theoretical yield (100%) as some molecules or atoms will be lost in the process. 4-aminophenol or acetic anhydrate could have been used in excess in the reaction. Having substances in beakers, boiling flasks, test tubes etc. will always result in loss of particles as they don’t get rinsed out of the container. In the reaction performed between 4-aminophenol and acetic anhydrate all molecules did not necessarily had to react with each other, leading to a lower percentage yield received. The main cause to loss of the product was in the filtration as visible amounts to the eye were left on the filter paper, drastically lowering the yield. The total amount of paracetamol could yet not have been crystallized when filtration was performed, once again leading to a loss of particles. The recrystallization performed also lowered the yield quite significantly.    

The product was analyzed using Thin-layer chromatography (TLC). The sample molecules of each drug were dissolved in a small amount of methanol and placed on the bottom of a TLC-plate. This was then placed in a TLC-chamber with a solvent at the bottom. The solvent rose up the plate by capillary action and because of different analytes they rose at different rates ending up at different positions. 

The THC-plate was exposed to ultraviolet light to see how high each sample had traveled and what color it emitted.  The effect of UV-light on fluorescence objects is more colesly discussed in the following post; 

http://www.blogger.com/blogger.g?blogID=997544598088907329#editor/target=post;postID=6376829843095828029

The conclusion was drawn that the product received was paracetamol as it emits an identical color, which in terms mean that they must have the same electronic configuration and hence be the same molecule.

The process of paracetamol acting as a pain reducer is not completely understood in modern times. It is believed that it’s vital function is to reduce the production of cyclooxygenase (COX), which causes easing of pain and reduce inflammatory symptoms. Cyclooxygenases are enzymes that produce prostanoids, which in terms cause pro-inflammatory symptoms. Currently three isoenzymes have been discovered; COX-1, COX-2 and COX-3. A research group in Lund, Sweden discovered in November 2011 that paracetamol effect the TRPA1-receptors, which are located in the spinal cord. They cause an inhibition of neural signals on the surface layer of the dorsal horn. The spinal cord is a part of the Central nervous system (CNS) and is the pathway of signals from the Peripheral nervous system (PNS) to the brain.  

Paracetamol metabolized into AM404, a molecule that mainly stops the uptake of anandime by the neurons. The effect of the substance being taken up is that it activates TRPV1, the major pain receptor of the body. AM404 also affects the sodium channels by restricting the movement of sodium ions in and out of the cell. Both these factors have tested to have a pain reducing effect, which leads to the conclusion that these factors are involved in the mechanism of paracetamol.      

Sodium channels are integral membrane proteins found in the plasma membrane which control the movement of sodium ions (Na⁺) in and out of the cell. The pain reducing factor comes from the blockage of the movement of sodium ions in neurons. Sodium ions are necessary in order to carry an electrical impulse along neurons. When a stimuli, (such as touching something on your fingertip) exceeded a certain threshold a brain wave is generated. This electrical impulse consist of the sodium-potassium pump, which are sodium ions (K⁺) and potassium ions (Na⁺) moving in and out on a membrane. As the sodium channels are not letting through sodium ions the signal is solaced, reducing the sensation of pain.