Solvent Selection for Separation

1. CHEMISTRY OF SEPARATION Solvent Selection for Separation

2. Solvent Selection for Separation • Processes Requiring Solvents • Extraction • Partition • Fractional Crystallization • Extractive Distillation • Liquid Chromatography • Gas-Liquid Chromatography

3. Solvent Selection for Separation • “Peripheral” Properties of the Solvent • Factors that usually don’t affect efficiency of the separation but are of interest • Synder defines a solvent “as either a pure compound or a mixture of pure solvents.” Binary and ternary solvent mixtures afford a wider range of solvents to choose than pure solvents

4. Peripheral Properties • Boiling pt. (bp) • Normally select a solvent with bp • above that of the operation. • Which distillation technique • would you not want the above property? Boiling Point • Properties • Easily evaporated or removed • bp 10-50 C higher than the temperature of separation • Minimize accidental evaporation • Diethyl ether • Volatile samples – fractional distillation to remove solvent or sample

5. Peripheral Properties • Viscosity • Low viscosity solvents preferable (General rule) • Liquid chromatography = poorer separation • Low viscosity coincides with low bp • Exceptions: very polar solvents (alcohols) & compact molecules (cyclalkanes, aromatics, CCl4) • Low viscosity enhances diffusion which speed separation • Viscosity of a solvent mixture is usually intermediate between those of the pure solvents, ie binary mixture A & B

6. Take home – it is possible to use a viscous solvent when in a mixture Peripheral Properties Viscosity of Mixturesh =(ha)xa (hb)xb

7. Peripheral Properties Viscosity of Water-Organic Solvent Mixtures

8. Peripheral Properties Solvent Properties Affecting Detection UV Cutoff -Solvent may interfere with detection • Appendix A shows minimum UV cut off for solvents What solvents might be poor choices for use with UV detection?

9. Peripheral Properties Solvent Properties Affecting Detection Refractive Index • Maximize differences in refractive index • between sample and solvent (Appendix A) Note the relatively small differences What does this mean in relation to detection?

10. Peripheral Properties Solvent Properties Affecting Detection Specific Element Content Common Gas Chromatography Detectors: Method Element Solvent Electron Capture Cl Chloroform Flame Thermionic N Acetonitrile Flame Photometric S Dimethyl Sulfoxide

11. Peripheral Properties SOLVENT MISCIBILITY CHART Appendix C in your notes

12. Peripheral Properties • Toxicity • Flammability • Reactivity • Cost • Disposal

13. Factors Affecting Solubility and Separation If the two solvents are immiscible, they can be shaken together and they will separate. If an analyte represented as ‘x’ is placed in one of the solvents before mixing, where will the analyte be after mixing?

14. x x A A x x x x x x x B B x x x x The concentration of x in the two solvents will be given as: Cx,a, Cx,b, concentration of solute x in solvents A & B; R, gas constant (1.99 cal/oK); T, temperature (oK); DG, free energy for transfer of 1 mole solute x from solvent B to A. x

15. Solvent Selection for Separation DH (enthalpy) change for transfer of 1 mole solute x from solvent B to A. If DH positive, interaction with solvent B is stronger, the quantity on the right will be <1, and solute x will prefer phase B (Cx,b > Cx,a) In most solutions entropy effects are negligible: replace DG with DH

16. Solvent Selection for Separation Cx,b Cx,a = ?

17. Solvent Selection for Separation • Separation • If solute x has a high solubility for the extracting solvent while solute y has a low solubility, solute x will separate from solute y The same principle applies to chromatography – solute x has high solubility for mobile phase while solute y has a high solubility for the stationary phase, therefore they will separate on the column – which one will move faster?

18. Solvent Selection for Separation • Solubility and Separation • Visualize transfer of a molecule x from solvent B to A, DH = heat of transfer, and determines the relevant solvency of B vs. A for solute x. • Figure (a) portrays a part of molecule x • (i , functional group) with surrounding molecules of solvent B.

19. Solvent Selection for Separation • Figure (b) i is removed from solvent B leaving a cavity. • Figure (c ) The cavity collapses and B - i interactions are replaced with B - B interactions

20. Solvent Selection for Separation • Figure (d) original A- A interactions in Solvent A • Figure (e) the i group is added, breaking bonds between A molecules and forming a cavity • Figure (f) the i group is inserted into the cavity (dissolved)

21. What Governs the Strength of these Bonds? • Intermolecular Interactions • Dispersion Forces • Dipole-Dipole • Induced-Dipole • Hydrogen Bonding • Covalent Bonds

22. Dispersion Forces Dispersion forces arise from the temporary variations in electron density around atoms and molecules. At any instant the electron distribution around an atom or molecule will likely produce a dipole moment, which can induce a (temporary) dipole moment in any nearby molecules. It is the Polarizability of the molecules, which determines the size of the induced dipole moments and thus the strength of the dispersion forces. Molecules containing large atoms (e.g. bromine or iodine) have large polarizability and so give rise to large dispersion forces. This explains the increasing melting and boiling points of the halogens going down that group of the periodic table.

23. Dispersion Forces - Summary Polarizability-High Polarizability = High intermolecular attraction (larger atoms) Molecular Size- Larger Size = More surface area and greater intermolecular attraction Molecular Shape -More branching or compact shape has less surface area and lower intermolecular attraction.

24. Dipole - Dipole If two neutral molecules, each having a permanent dipole moment, come together such that their oppositely charged ends align, they will be attracted to each other.

25. Dipole - Dipole Interactions? Orthodinitrobenzene has a high overall dipole moment because of the 2 nitro groups, but the overall dipole moment of the para compound is 0 because of the cancellation of group dipoles. However, both molecules have 2 nitro groups and the interactions of these two compounds with surrounding molecules are similar.

26. Induced Dipole A polar molecule (lower left) carries with it an electric field and this can induce a dipole moment in a nearby non-polar molecule (lower right). This will cause the attraction between the molecules. This type of force is responsible for the solubility of oxygen (a non-polar molecule) in water (polar).

27. Hydrogen Bonding Hydrogen bonds are usually listed as a type of dipole-dipole force, but the details of hydrogen bonding are subtle and these bonds have some partial covalent bond character. If a hydrogen bond can form between a pair of molecules it will be stronger than other intermolecular forces between the molecules.

28. FASTEST FINGER QUESTION Place these molecules in order from lowest to highest intermolecular forces Intermolecular Forces Its time to play- Who wants to be a millionaire?

29. Intermolecular Forces Boiling Pt. oC 1) neopentane 10 2) 2,3-dimethyl butane 58 3) n-hexane 69 4) 2-methyl-2-butanol 102 5) 1-pentanol 138

30. Polarity • Ability of a molecule to engage in strong interactions with other polar molecules. Thus, it describes the ability of the molecule to enter into many different interactions (dispersion, dipole, hydrogen bonding, etc.). • Relative polarity – sum of all these interactions.

31. Example: Two Immiscible Solvents: Solvent A Water Solvent B Hexane Analyte: Acetone (i) How will the acetone partition between Solvent A and Solvent B?

32. Cx,a Cx,b A e  -DH/RT x x x B x x x x Example: Use this equation What is the value of DH?

33. Estimating the Value of DH 2Hi,b = B – i bonds broken; - -Hb,b = B – B bonds formed; Ha,a = A – A bonds broken; - -2Hi,a = i – A bonds formed H = (Ha,a – Hb,b) + 2(Hi,b – Hi,a) Approximation: The interaction between molecules is based on the product of their polarities. Thus: H = (Pa2 - Pb2) + 2Pi.(Pb – Pa)

34. Calculating DH All we need now are the polarities of A, B, and i to substitute in this equation: H = (Pa2 - Pb2) + 2Pi.(Pb – Pa) From Appendix A: i = Acetone P’ = 5.1 B = Hexane P’ = 0.1 A = Water P’ = 10.2 H = (10.22 - 0.12) + 10.2(0.1 – 10.2)

35. Cx,a Cx,b e  -DH/RT Calculating the Ratio H = ? R, gas constant (1.99 cal/oK); T, temperature (oK); = 298

36. Hydrophobicity Hydrophobic interactions ‑ associated with "nonpolar" solutes in "polar" solvents assume: B ‑ polar A ‑ non‑polar i – small (non polar) H = (Pa2 - Pb2) + 2Pi.(Pb – Pa) the polar solvent "squeezes" out the nonpolar solute into phase A.

37. Selectivity H = (Pa2 - Pb2) + 2Pi.(Pb – Pa) If there were only one type of interaction between molecules, the above equation for DH would be valid. However, in reality there are usually several types of interactions. These differences in interaction makes it possible to separate analytes of similar polarity. This ability is known as solvent selectivity.

38. Part 2 Solvent Classification and Selection Outline: Solvent Classification Schemes Summary of Solvent Selection Extraction Efficiencies

39. Hildebrand Solubility Parameter H = Vx [(a2 ‑ b2) + 2x (b ‑ a)] Vx (molar volume) of solute x affects its relative solubility The larger is Vx more affected will be the solubility of x by a change in solvent polarity

40. Rohrschneider Polarity Scale The Rohrschneider polarity scale is based on experimental data. This method estimates the polarity of a solvent based on the solubility of three reference solutes below: Ethanol proton donor interaction Dioxane proton acceptor interaction Nitromethane dipole interaction

41. Rohrschneider Polarity Scale The three values can be plotted as a Selectivity Triangle, with the 3 legs of the triangle calculated as the ratios of each individual term to the total polarity of the solvent as follows. • Xe = log(K"g) ethanol • P’ • Xd = log (K"g) dioxane • P’ • Xn = log (K"g) nitromethane • P’

42. Solvent Selectivity Triangle

43. Extraction of Compound X from a Sample Matrix Containing Y Begin by studying the extraction of x and y as a function of solvent polarity.

44. P’1 P’x P’2 P’y 100 80 % Extracted 60 40 20 x y P’ Extraction of Compound X from a Sample Matrix

45. P’1 P’x P’2 P’y 100 80 % Extracted 60 40 20 x y P’ Extraction of Compound Y from a Sample Matrix

46. Partition Coefficient • Simplest form of batch extraction • Complete extraction not possible; greater than 99% extraction can occur • Extraction efficiency by this method is based on Partition Coefficient (K) or Distribution ratio (D) D or K = Co/Cw Co is concentration in the organic phase (solvent) Cw is the concentration in the aqueous phase (water)

47. Partition Coefficient • If DV 100 then a single batch extraction can work: • Assume: V = Vo/Vw = 10 • D = Co/Cw = 5 • Then: • = (5)(10)/1+(5)(10) • = 98% • D = Co/Cw • Co is concentration in the organic phase (solvent) • Cw is the concentration in the aqueous phase (water) • Assume equal volumes • For unequal volumes, fraction extracted  V = Vo/Vw Fraction remaining in aqueous phase following n extractions: (1 - )n = Xn

48. Extraction Efficiency Wr = Wo (Vw/(KVo+ Vw))N Where Wr = weight of solute remaining following extraction, Wo = weight of solute in original solution, Vw = volume of aqueous phase, Vo = volume of extracting solvent, K = partition coefficient, N = number of extractions. Example: K = 2, Vw = 60 mL, Wo = 1 g Calculate Wr for 1 extraction with 60 mL solvent 2 extractions with 30 mL each 3 extractions with 20 mL each