Principle of Extraction Organic solvents dissolve neutral organic compounds Water dissolves ionic compounds +HOHHOHHOHHOHHOHHOHHOHHOHK =[Xorg][Xaq]partitioning coefficientK > 1 for most organic cmpdsK < 1 for most saltsXaqXorgKnote: be sure to check your ________ to determine which layer is which densitiesExtraction 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 = volumeThe 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:0.995 mol / Vorg0.005 mol / 100 mL10 =0.995 mol0.005 molVorg=xx10100 mL=1990 mLExtraction 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. Now, a total of 0.89 moles (89%) have been removed from the aqueous solution with 2 extractions totaling 40 mL. 10 =norg0.33 - norgx10020norg=0.22 molesMultiple 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 >4.2pKa 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 >10pKa 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 <4.6PhenolsCircles – aqueous phasesRectangles - organic phasesThin 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) Glass or plastic Adsorbent (stationary phase) • 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 glassSiOSiOSiOSiOSiOO OH OHHHOPhHOPhHOPhR OR'OMore Polar Less PolarAdsorbents:• Silica (acidic)• Alumina Al2O3 (basic)ROR'OLess polar substances move