1 Chemical Separations What is a chemical separation? Examples: Filtration Precipitations Crystallizations Distillation HPLC GC Solvent Extraction Zone Melting Electrophoresis Mass Spectroscopy

2 Chemical Separations What is the object of the separation. Collection of a pure product Isolation for subsequent analysis for either quantification or identification –Analysis »How Much? »What is it?

3 Chemical Separations Major Industries –Petroleum Distillation –Distilled Spirits

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5 Chemical Separations Petroleum is a mixture of hydrocarbons. –The larger the molecular weight the less volatile. –So we must separate into various molecular weight fractions (different boiling points) –The results are still complex mixtures

6 Chemical Separations

7 Distillation As heat is added to the system the lower volatility compounds will boil away and can be collected. –In the spirits industry the low boilers are call foreshots (~75% EtOH) –The high boilers are called feints –Congeners - Chemical compounds produced during fermentation and maturation. Congeners include esters, acids, aldehydes and higher alcohols. Strictly speaking they are impurities, but they give whisk(e)y its flavour. Their presence in the final spirit must be carefully judged; too many would make it undrinkable.

8 Distillation What is whiskey? What is brandy?

9 Interesting Facts Bourbon - US whiskey made from at least 51% corn, distilled to a maximum of 80% abv (160 proof) and put into charred new oak barrels at a strength of no more than 62.5% abv. Organic whisk(e)y - That made from grain grown without chemical fertilizers, herbicides and pesticides. Tennessee whiskey - As bourbon, but filtered through a minimum of 10 feet of sugar-maple charcoal. This is not a legal requirement, but is the method by which Tennessee whiskies are currently produced.

10 Interesting Facts Malt whisky - Whisky made purely from malted barley. Angels' share - A certain amount of whisk(e)y stored in the barrel evaporates through the wood: this is known as the angels' share. Roughly two per cent of each barrel is lost this way, most of which is alcohol.

11 Extraction Liquid – Solid Liquid - Liquid

12 Liquid Solid Extraction (Soxhlet) whale.wheelock.edu/ bwcontaminants/analysis.html whale.wheelock.edu/ bwcontaminants/analysis.html

13 Liquid Solid Extractor (Bunn)

14 Solvent Extraction (Liquid-Liquid) We want to partition a solute between two immiscible phases. –Oil and vinegar –Ethyl ether and water –Hexane water –Octanol and water. The two solvent should have low miscibility Have different densities – avoid emulsion formation.

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16 Solvent Extraction

17 Replace concentration with moles over volume and let q equal the fraction in the aqueous phase. m will be the moles of solute in the entire system

18 Define a new term for the ratio of the volumes of the phases

19 We can do a little algebra and find an expression for q

20 If it does not end up in the aqueous phase it must be in the organic phase. p is the term for the fraction in the organic p + q = 1 Giving

21 Sample Problem You have mL of an aqueous solution that is mM in compound C. This solution is extracted with 50.0 mL of diethyl ether and the aqueous phase is assayed and it is found that the concentration of compound C that remains is 20.0 mM. What is the equilibrium constant for this extraction system.

22 Solution

23 We can do multiple extraction from the aqueous phase. We end up with the following expression for what is left in the aqueous phase.

24 Example How many extractions would be required to remove 99.99% of aspirin from an aqueous solution with an equal volume of n-octanol? Since 99.99% must be removed the decimal fraction equivalent of this is This leaves in the aqueous phase. Since we have equal volumes then V r is We are able to find from the Interactive Analysis Web site that K for Aspirin is We plug these values into the q equation and the power is the unknown.

25 Solution

26 What if our compound can dissociate or participate in some other equilibrium? A compound such as aspirin is a carboxylic acid. We can represent this as HA. Do we expect the ion A - to be very soluble in the organic phase???

27 Dissociation So if we have dissociation then less will go into the organic phase. K p is the ratio of concentration of aspirin (in the un-dissociated form) in each phase. This ratio will always be the same. How do we account for the ion formation?

28 Distribution Coefficient Where C is the formal concentration of the species. C a = [HA] + [A - ] D c will vary with conditions For this compound what is that condition?

29 DcDc Since the ion is not very soluble in the organic phase then we may assume that the dissociation will not happen in that phase. This gives us the expression to the right.

30 Acid Equilibria What is the equilibrium? K a

31 With a little algebra So if you know K d and K a then you can determine D c as a function of H + (pH) However if [H + ] is much larger than K a then D c will equal K d. If the [H +] is close in value to K a then D will be related to the pH Plotting this we get.

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33 So What, Why is this useful. Well we can now move a solute (analyte) from one phase to another. This can be very useful when extracting a compound that has significant chemical differences from other compounds in solution. As a matter of fact this has been used as an interview question for prospective co-ops when I worked in industry. The question would go like this. You have carried out a series of reactions and it is now time to work up the product which currently sits in an organic solution (methylene chloride). Your expected product is a primary amine. Which of the following solutions would you extract this methylene chloride solution with to isolate your amine. Your choices are: A) Toluene. B) 0.1 N NaOH (aq) C) 0.1 N HCl (aq) D) I never wanted to work here anyhow.

34 Separation So far we can tell how one compound moves from one phase to another. What if we are try to separate two compounds, A and B Well we might just suspect that if we find a solvent system that has different values of D c for each compound we could end up with most of one compound in one phase and the other compound in the opposite phase. It is not that simple.

35 Example System I D a = 32 D b = (A ratio of 1000) V r = 1 Let's recall our equations q (fraction in aqueous) = 1 / (DV r + 1) p (fraction in organic) = DV r / (DV r + 1) V r (volume ratio) = V o / V a

36 Case I p a = 32*1 / (32*1 + 1) = 0.97 p b = 0.032*1/ (0.032*1 + 1) = 0.03 If we assume that we have equal moles of A and B to start then what is the purity of A in the Organic Phase? Purity = moles A / (moles A + moles B) Purity = 0.97 / ( ) = 0.97 or 97 %

37 Case II D a = 1000 D b = 1 V R = 1 (Ratio is still 1000) p a = 1000*1 / (1000*1 + 1*1) = 1000/1001 = Aha! we got more a into the organic, as we would expect with a higher D value. Now p b = 1*1 / ( 1*1 +1) = 1/2 = 0.5 oh-oh What do we get for purity of compound a now? purity = / ( ) = Yuck!

38 How can we get around this issue? Once we have selected the solvent and pH, then there is little that we can do to change D. What else do we have in our control????? Let's look p = DV r / (DV r + 1) Not much here except V r and in fact that is the key to this problem. Is there an optimum V r value for the values of D that we have? Yes! Our equation for this is V r(opt) = (D a *D b ) -0.5

39 Revisit the two cases So let us look at our two cases and see which will give us the optimum values. Case I D a = 32 and D b = V r(opt) = (32 * 0.032) -0.5 = ( 1 ) -0.5 = 1 So we were already at the optimum.

40 Case II Revisited Case II D a = 1000 and D b = 1 V r (opt) = (1000*1) -0.5 = = Which mean that when we do our extraction we will extract _______ mL of organic for each _______ mL of aqueous.

41 Purity for Case II What is our purity for this system? p a = 1000*0.032 / (1000* ) = 32/33 = 0.97 and p b = 1*0.032 / (1* ) = 0.032/1.032 = 0.03 Purity of a then is 0.97/ ( ) Which will give us the 97% purity we had for Case I with with the V r of 1.

42 Can we improve this purity? If we were to extract again then we would just remove the same proportions. We would get more compound extracted but it would be the same purity. What if we were to take the organic phase and extract it with fresh aqueous phase. We know that one of the two compounds will end up mostly in that aqueous phase so we should enhance the purity of the other compound in the organic phase.

43 Back Extraction Called that since you are extracting back into the original phase.

44 Back Extraction Case I Example Let's look at the numbers. D a = 32 D b = V r = 1 p a = 0.97 p b = 0.03 q a = 0.03 q b = 0.97 Let’s prepare a table.

45 Initial conditions prior to starting back extraction. Before ShakingAmount AAmount B Organic Phase Fresh Aqueous Phase00

46 Now we extract – shake shake shake How much goes to the Aqueous phase q which is 0.03 for A and 0.97 for B How much goes to the Organic phase p which is 0.97 for A and 0.03 for B After ShakingAmount AAmount B Organic Phase(0.97)(0.97)(0.03)(0.03) Aqueous Phase(0.97)(0.03)(0.03)(0.97)

47 Now what is the purity for A in the organic phase??? Purity = Amount A / (Amount A + Amount B) = 0.97*0.97 / (0.97* *0.03) = 0.94/( ) = 99.9% What is the yield of A (fraction of the total amount that we started with)

48 Let’s do it again – Can we improve purity even more? After second Back ExtractionAmount AAmount B Organic Phase0.94* *0.03 Aqueous Phase0.94* *0.97 Purity A = / ( ) = % But our yield has dropped to 91.3%, there is a price to pay for the added purity.

49 Can We Expand This? Why Would We Want to? Such multiple extraction systems have been developed. Still a viable option for preparative work. For separations it has been replaced by HPLC Called Craig Counter Current Extraction. Special glassware is used.

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51 Craig CCE Equal amounts of organic (red) and aqueous (blue) solvents with the analyte(s) are added to the A arm of the tube via port O. Fresh Aqueous Solvent is added to each of the tubes down the apparatus.

52 Craig CCE Rock the system back and forth and to establish equilibrium. Allow the system to stand for the layers to separate. Rotate the apparatus counter clockwise about 90 o to 100 o.

53 Craig CCE Rotate Back to Horizontal

54 Starting Conditions Tube#01234 Organic Phase 0 Aqueous Phase After One Equilibrium Tube#01234 Organic Phasep Aqueous Phaseq0000 Transfer Step 1 Tube#01234 Organic Phase0p Aqueous Phaseq0000

55 Now here is what is in each tube/phase after equilibrium is reached. Tube#01234 Organic Phase pqpp Aqueous Phase qqqp000 Now we do Transfer 2 Tube#01234 Organic Phase0pqpp Aqueous Phaseq2q2 pq000

56 Now here is what we have in each tube after the next equilibrium. The total in each tube times either p or q as appropriate. Tube#01234 Organic Phase pq 2 p*2pqp3p3 Aqueous Phase q3q3 q*2pqqp 2 00 We transfer again. Transfer Step 3 Tube#01234 Organic Phase0pq 2 2p 2 qp3p3 Aqueous Phaseq3q3 2pq 2 p2qp2q00

57 Shake Again Equilibrium 4 Tube#01234 Organic Phase pq 3 p*3pq 2 p*3p 2 qp4p4 Aqueous Phase q4q*3pq 2 q*3p 2 qq*p 3 0 Transfer 4 Tube#01234 Organic Phase0 pq 3 3p 2 q 2 3p 3 qp4p4 Aqueous Phaseq4q*3pq 2 3p 2 q 2 p3qp3q0 See a trend????

58 Craig CCE How about a binomial expansion? (q + p) n = 1 Powers of the two terms in each tube will add up to n Coefficients will be found from Pascal Triangle

59 Craig CCE Or the formula F r,n = n!/((n-r)!r!) p r q (n-r) n is the number of transfer and r is the tube number. You start counting at zero!

60 Craig CCE Let's look at and example for a four tube system. D a = 3 p = 0.75 q = 0.25 D b = p = 0.25 q = 0.75 What would be the purity and yield of Compound A if collected from the last in our above example. Amount of A p 4 or = Amount of B p 4 or = Purity of A / ( ) = or 98.78% Yield of A We collect a fraction of or 31.64% Horrible Yield!

61 Craig CCE What if we collect the last two tubes?? Amount of A p 4 and 4p 3 q or *(0.75) 3 (0.25) = = Amount of B p 4 and 4p 3 q or *(0.25) 3 (0.75) = = Purity of A ( ) / ( ) = or 93.56% Yield of A We collect a fraction of = or % Purity still ok and yield is much better.

62 Craig system n= 200 transfers. D a of 2.0 and D b of 4.0 p a of p b of

63 Final Formulas(1) r max = np = nDV r /(DV r +1) To find the separation between two peaks we would use.  r max = (r max ) a - (r max ) b = n(p a -p b ) The Gaussian distribution approximation for our binomial expansion would be (when n>24) F r,n = (2  ) -0.5 (npq) -0.5 exp -[((np-r)^2)/2npq]

64 Final Formulas(2) The width of the distribution through the system would be: w = 4  = 4(npq) 0.5 Resolution would be R =  r max /w =  r max /4  or R = n  p/(4(npq) 0.5 ) = n 0.5  p / 4(pq) 0.5

65 Demo Site l_Craig2.htmlhttp:// l_Craig2.html