Principle of Extraction Water dissolves ionic compounds H H H O H O H H O H O H H H O O H H O H H O H H Organic solvents dissolve neutral organic compounds K Xaq K Xorg Xaq Xorg partitioning coefficient K 1 for most organic cmpds K 1 for most salts note be sure to check your densities to determine which layer is which Extraction Efficiency Example 1 The partition coefficient K for a compound between diethyl ether and water is 10 If you dissolve the 1 mole of the compound in 100 mL of water how much ether is needed to extract 99 5 of the compound from the aqueous solution where n moles and V volume The moles in the organic layer norg should be 0 995 99 5 of 1 mole and the moles in the aqueous layer naq will be 0 005 The aqueous volume is 100 mL so if we plug in all of the data and solve for the organic volume Vorg we find 10 0 995 mol V org 0 005 mol 100 mL V org 0 995 mol x 100 mL 0 005 mol x 10 1990 mL Extraction Efficiency Example 2 For the previous problem how much material would you extract with 20 mL of diethyl ether This time you are looking for the moles of compound that will be in the organic layer norg Since we know that the total number of moles is 1 we can substitute for naq and solve for norg Multiple Extraction Efficiency Performing multiple extractions with smaller volumes is more efficient than just one extraction with a larger volume of solvent In example 1 to remove 99 5 of the compound with one extraction it required 1990 mL of ether In example 2 0 67 moles 67 was removed with one extraction of 20 mL But what if we were to repeat the extraction on the aqueous solution again Remember there are only 0 33 moles remaining in the aqueous solution 10 norg 0 33 norg 100 x 20 norg 0 22 moles Now a total of 0 89 moles 89 have been removed from the aqueous solution with 2 extractions totaling 40 mL Multiple Extraction Efficiency So now we know that multiple extractions are more efficient but how many do we need to perform to remove 99 5 of the compound If we look again at the calculations for the two extractions this time solving for naq First extraction Second extraction Combined calculations Multiple Extraction Efficiency Now that we have a general equation for multiple extractions we can determine how many extractions it would take to remove 99 5 of the compound which would be a value of 0 005 for naq Therefore it would take 5 extractions of 20 mL each to remove the same amount that it would take 1990 mL to remove in one extraction pKa Dependence in Extractions So what type of aqueous solution do I need to extract with For organic acids In a neutral aqueous solution pH 7 the acid will exist mainly in the HA form which is insoluble in water But what if we shift the equilibrium by adding a strong base If B is stronger than A the compound will exist mainly as its ionic form and will partition into the aqueous layer pKa Dependence in Extractions So what type of aqueous solution do I need to extract with Example benzoic acid Below pH 4 2 benzoic acid is mainly protonated and uncharged Above pH 4 2 benzoic acid is mainly deprotonated and charged Therefore to extract benzoic acid into the aqueous layer I need a solution with pH 4 2 pKa Dependence in Extractions So what type of aqueous solution do I need to extract with Example phenol Below pH 10 phenol is mainly protonated and uncharged Above pH 10 phenol is mainly deprotonated and charged Therefore to extract phenol into the aqueous layer I need a solution with pH 10 pKa Dependence in Extractions So what type of aqueous solution do I need to extract with For organic bases In a neutral aqueous solution pH 7 the base will exist mainly in the B form which is insoluble in water But what if we shift the equilibrium by adding a strong acid If HA is stronger than HB the compound will exist mainly as its ionic form and will partition into the aqueous layer pKa Dependence in Extractions So what type of aqueous solution do I need to extract with Example aniline Below pH 4 6 aniline is mainly protonated and charged Above pH 4 6 aniline is mainly deprotonated and uncharged Therefore to extract aniline into the aqueous layer I need a solution with pH 4 6 Phenols Circles aqueous phases Rectangles organic phases Thin Layer and Column Chromatography What s the Point Purify compounds by taking advantage of differences in polarity Why Should we Care This it the organic chemist s favorite method for purifying organic compounds as the recovery tends to be the highest Thin Layer Chromatography Separation of a mixture of compounds based on polarity of the components Mobile phase solvent Adsorbent stationary phase Glass or plastic Concept A mobile phase solvent carries a mixture of compounds over a stationary phase Different compounds have different affinities for the adsorbent of the stationary phase Compounds with higher affinity for the adsorbent move more slowly are more retained Solute Adsorbent Interactions Forces that bind solute to adsorbent Dipole dipole Hydrogen bonding More Polar Ph O Less polar Less Polar O R R O H substances move faster O R Ph O O O Ph H H H O OH O H Si O Si O Si O Si R Adsorbents O Si Silica acidic O Alumina Al2O3 basic glass