chem. 230 9

1. Chem. 230 � 9/10 Lecture

3. Specialized Topic Presentations I See new handout Key points: Select partner and possible topics by 9/21 On 9/21, sign up for topic I suggest you quickly research topic soon to see if topic is too narrow or broad Since working in pairs, I�m now requesting 2 homework problems ~30 min. presentation (time may change slightly) Be prepared to ask/answer questions following presentation

4. Some Questions From Last Lecture What are advantages and disadvantages of Soxhlet extractions? Suggest a way to isolate a polar organic compound from ionic compounds in urine. It is desired to isolate CN- from CO32- by adding Ag+ and monitoring [Ag+] electrochemically. Assuming initial concentrations of [CN-] = 1.0 x 10-3 M and [CO32-] = 5.0 x 10-3 M and given Ksp values for AgCN and Ag2CO3 are 2.2 x 10-16 and 8.2 x 10-12, respectively, what would be the target [Ag+]? Will this separation be very efficient? Can acetic acid be isolated from all solutions in water?

5. More Questions A 1.00 L sample of sea water is analyzed for phenols. The 1.00 L sample is passed through a solid phase extraction cartridge to trap the phenols. Then 25.0 mL of methanol is used to remove the phenols and then reagents are added that convert the phenols to methoxyphenols. The methoxyphenols are extracted by adding 25 mL of water to the methanol and extracting with two successive 25 mL portions of hexane. The hexane portions are combined, evaporated, and redissolved in 2.0 mL of hexane. An aliquot is then determined by GC and found to contain 22.1 mmol L-1 of a particular phenol. What was the original conc. of that phenol in sea water (in nmol L-1) if it is assumed that all transfers were 100% efficient? How could the sensitivity of the method be increased? The total NH3 (NH3 + NH4+) concentration of a water sample is determined by NH3 in the headspace above a sample. A water sample at a pH of 8.1 was found to have a headspace pressure of 2.4 x 10-7 atm. If KH = 1.6 x 10-5 atm m3/mol (at that T) and Ka(NH4+) = 5.6 x 10-10.

6. Liquid-Liquid Extractions One of more common simple separations Often used to introduce partition theory Equipment is simple (separation funnel or vials + syringes) Two liquids must be immiscible (form two distinct phases) Lower phase is more dense (usually water or chlorinated hydrocarbon) Most common with water (or aqueous buffer) and less polar organic liquids

7. Liquid-Liquid Extractions Partition Coefficient Kp = [X]raffinate/[X]extractant Kp depends on thermodynamics of dissolving X in two phases Most common rule for solubility is likes dissolve likes More polar compounds exist in greater concentration in water Koctanol-water values can be found in reference tables (octanol is assumed to be the raffinate)

8. Liquid-Liquid Extractions Effect of organic solvent on Kp: Large or small Kp values mean effective phase transfer Less polar organic solvents (e.g. hexane) will give largest Kp values (assuming aqueous extractant phase) for non-polar analytes (e.g. a triglyceride), but smaller values for analytes of modest polarity (e.g. a phenol) For analytes of modest polarity, a more polar organic solvent will give a larger Kp value (e.g. ether or ethyl acetate) When separating two analytes, finding a solvent with better selectivity also is important a = separation factor = (Kp)A/(Kp)B (where (Kp)A> (Kp)B)

9. Liquid-Liquid Extractions The fraction of moles extracted depends on both Kp and volumes of phases k = partition ratio or retention factor (in chromatography) For this example we will assume an organic raffinate k = norg/naq k = Kp(Vraf/Vext) = Kp(Vorg/Vaq) Q = fraction extracted (to extractant or aq. phase) Fraction left in raffinate = 1 - Q Q = 1/(1 + k)

10. Liquid-Liquid ExtractionsDemonstrations Methylene Blue (charged organic dye) Which phase will it be in: hexane or water? Iodine (I2) Which phase will it be in: hexane or water? Will Kp (organic extractant) get larger or smaller by changing water to ACN? Or by changing hexane to ethyl acetate?

11. Liquid-Liquid Extractions To increase the fraction extracted, either Kp needs to be changed or the volume ratio Example: If a compound is extracted from octanol to water and Kow = 0.8, how much volume of water is needed if the compound is in 10 mL of octanol and the extraction should move 90% of the compound?

12. Liquid-Liquid Extractions A different example, methylethylketone (MEK), CH3CH2COCH3, has Kow = 24, if 25 mL of aqueous MEK is extracted with 10 mL of octanol, calculate Kp, Q, and 1 � Q Kp = [MEK]aq/[MEK]octanol = 1/Kow = 0.042 Q = 1/[1 + Kp(Vaq/Voctanol)] = 1/[1 + 0.042(25/10)] = 0.91 1 � Q = 0.09

13. Liquid-Liquid Extractions Let�s look at multiple extractions in more detail MEK example 91% MEK in 1st extraction to octanol 9% MEK left in water In 2nd extraction of octanol (2nd water portion), fraction of original in water = Q(1 - Q) = 0.91*0.09 = 0.085, fraction of original in octanol = Q2 = 0.82 In 2nd extraction of water portion by octanol, fraction in octanol = (1 � Q)Q = 0.085, fraction in water = (1 � Q)2 = 0.009 If water is extracted twice with octanol and octanol fractions combined, fraction in octanol = 1 - (1 � Q)2 = 0.99 (99%)

14. Liquid-Liquid Extractions Efficiency of Multiple Extractions vs. Single Extractions Back to Previous example (compound with Kow = 0.8) but using 3 successive 10 mL aliquots of water. Note: successive extractions can be done in different ways (cross-current vs. counter current) In this example (assuming only 2 components), the 10 mL aqueous aliquots normally would be combined

15. Liquid-Liquid Extractions Additional Equilibria Crown ether example Complexed metals become more organic soluble

16. Liquid-Liquid ExtractionsAdditional Notes Crown ethers and extractions Extraction depends on: Crown ether �hole� Size of metal cation Crown ether partition coefficients (rxn 3 and 4)

17. Liquid-Liquid ExtractionsOther Methods for Transferring Metals Metal � Ligand Complexes Best transfer for neutral complexes with large organic ligands Most common ligands are L- form and bidentate (2 bonds per ligand with metal) Reactions: HL (aq) ? HL (org) HL (aq) ? H+ + L- Mn+ + nL- ? MLn (aq) (note this can be broken into n steps) MLn (aq) ? MLn (org) Best at intermediate pH and for metals with large complexation formation constants Ion � Pairs (e.g. Na+--O3S(CH2)5CH3)

18. Liquid-Liquid ExtractionsAcidic/Basic Organics Ions have very small Kow values (assume = 0) Form of acids and bases in neutral species depends on pH Distribution Coefficient = KD KD = [X]total raffinate/[X]total extractant Monoprotic acid example (HA extracted from organic to water) HA = only organic phase species (a little different in text example) HA and A- possible in aqueous phase

19. Liquid-Liquid ExtractionsAcidic/Basic Organics HA example continued Kp = [HA]org/[HA]aq = constant KD = [HA]org/([HA]aq + [A-]) � not constant Ka = [H+][A-]/[HA] So KD = [HA]org/([HA]aq + Ka[HA]aq/[H+]) KD = Kp/(1+ Ka/[H+]) At pH << pKa (high [H+]), KD = Kp At pH >> pKa, KD << Kp (better transfer to aq phase)

20. Liquid-Liquid ExtractionsAcidic/Basic Organics HA example

21. Liquid-Liquid ExtractionsAcidic/Basic Organics In general, only un-ionized formed will be organic soluble (applies to multi-functional compounds also) For weak bases, only base form B, not BH+, will be organic soluble For weak bases, low pH increases partitioning into water

22. Liquid-Liquid ExtractionsSome Questions How can the following compounds be separated?

23. Liquid-Liquid ExtractionsSome Questions

24. Liquid-Liquid ExtractionsSome Questions To extract Al3+ to an organic phase with an HL type ligand, which complexation constant is most important? Amino acids can act as bidentate ligands for metal transfer. Why does the pH have to be greater than pKa2?

25. Liquid-Liquid ExtractionsOne More Question Benzylamine (C6H5CH2NH2) is a weak base with a Kow of 12. The conjugate acid of benzylamine has a pKa of 9.35. Benzylamine is being extracted from 10 mL of water buffered to a pH of 8.00 (raffinate phase) to octanol. (assume no other reactions occur in water or octanol) Determine the distribution coefficient. Calculate the fraction of benzylamine extracted into octanol if 20 mL of octanol is used? How could Q be increased?

26. Advanced Extraction TechniquesSolid Phase Extraction Solid Phase Extraction (SPE) is most commonly used in place of liquid-liquid extraction but similar methods are used to trap gases also (as shown in purge and trap GC) Can be use for contaminant removal or analyte trapping Types of solid phases are similar to stationary phases used in HPLC

27. Advanced Extraction TechniquesSPE Demonstration Procedure: Clean cartridge with removing solvent, then sample solvent Apply sample; strongly retained compounds will remain on stationary phase, weakly retained/unretained compounds pass through Rinse cartridge with sample solvent Apply eluting solvent to remove strongly retained compounds It is possible to Increase solvent strength to remove compounds in several fractions

28. Advanced Extraction TechniquesSolid Phase Extraction Solid Phase Very Similar to HPLC packing particles Smaller column Larger particles (allowing low pressure elution) Some Types Silica Based (octadecyl or C18, phenyl, aminopropyl, etc.) Ion Exchange (normally charged group on polymeric solid) Others (e.g. graphitic carbon)

29. Advanced Extraction TechniquesMore on SPE next time