Solvent Selection for Separation

SOLVENT SELECTION FOR SEPARATION

Processes Requiring Solventsy y y y y y

Extraction Partition Fractional Crystallization Extractive Distillation Liquid Chromatography Gas-Liquid Chromatography

SOLVENT SELECTION FOR SEPARATION

Peripheral Properties of the SolventFactors that usually dont affect efficiency of the separation but are of interest y Synder defines a solvent as either a pure compound or a mixture of pure solvents.y

Binary and ternary solvent mixtures afford a wider range of solvents to choose than pure solvents

PERIPHERAL PROPERTIES

Boiling Point 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?

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

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

Peripheral PropertiesViscosity of Mixtures L =(La)xa (Lb)xb

Take home it is possible to use a viscous solvent when in a mixture

PERIPHERAL PROPERTIESViscosity of Water-Organic Solvent Mixtures

PERIPHERAL PROPERTIES

Solvent Properties Affecting DetectionUV 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?

PERIPHERAL PROPERTIES

Solvent Properties Affecting Detection Refractive IndexMaximize differences in refractive index between sample and solvent (Appendix A)

Note the relatively small differences What does this mean in relation to detection?

PERIPHERAL PROPERTIES

Solvent Properties Affecting DetectionSpecific Element Content Common Gas Chromatography Detectors: Method Element Electron Capture Cl Flame Thermionic N Flame Photometric S Solvent Chloroform Acetonitrile Dimethyl Sulfoxide

Peripheral PropertiesSOLVENT MISCIBILITY CHART Appendix C in your notes

PERIPHERAL PROPERTIES Toxicity Flammability Reactivity Cost Disposal

FACTORS AFFECTING SOLUBILITY AND SEPARATIONIf 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?

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);x x x x x x x x x x x

(G, free energy for transfer of 1 mole solute x from solvent B to A.

A

A

Bx x

x

B

SOLVENT SELECTION FOR SEPARATION(H (enthalpy) change for transfer of 1 mole solute x from solvent B to A. If (H positive, interaction with solvent B is stronger, the quantity on the right will be Cx,a)

In most solutions entropy effects are negligible: replace (G with (H

Solvent Selection for SeparationCx,b Cx,a100 y = ex 10 1 0.1 0.01 0.001 -4 -2 0 2 4

= ?

SOLVENT SELECTION FOR SEPARATIONSeparation y If solute x has a high solubility for the extracting solvent while solute y has a low solubility, solute x will separate from solute yThe 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?

SOLVENT SELECTION FOR SEPARATION

Solubility and Separationy

Visualize transfer of a molecule x from solvent B to A, (H = 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.

y y

SOLVENT SELECTION FOR SEPARATION Figure

(b) i is removed from solvent B leaving a cavity. (c ) The cavity collapses and B - i interactions are replaced with B - B interactions

Figure

SOLVENT SELECTION FOR SEPARATIONy

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)

y

y

WHAT GOVERNS THE STRENGTH OF THESE BONDS?

Intermolecular Interactions Dispersion Forces Dipole-Dipole Induced-Dipole Hydrogen Bonding Covalent Bonds

DISPERSION FORCESDispersion 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.

DISPERSION FORCES - SUMMARYPolarizability(larger atoms) High Polarizability = High intermolecular attraction

Molecular Size- Larger Size = More surface area and greaterintermolecular attraction

Molecular Shape - More branching or compact shape has lesssurface area and lower intermolecular attraction.

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.

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.

INDUCED DIPOLEA 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).

HYDROGEN BONDINGHydrogen 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.

INTERMOLECULAR FORCESIts time to playWho wants to be a millionaire?FASTEST FINGER QUESTION

Place these molecules in order from lowest to highest intermolecular forces

Intermolecular Forces

Boiling Pt. oC1) neopentane 2) 2,3-dimethyl butane 3) n-hexane 4) 2-methyl-2-butanol 5) 1-pentanol 10 58 69 102 138

POLARITYAbility 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.

EXAMPLE:Two Immiscible Solvents: Solvent A Solvent B Analyte: Acetone (i) Water Hexane

How will the acetone partition between Solvent A and Solvent B?

EXAMPLE: USE THIS EQUATION

Cx,a } Cx,bAx x x x

-(H/RT

e

What is the value of (H?x x

Bx

ESTIMATING THE VALUE OF (H2Hi,b = -Hb,b = Ha,a = -2Hi,a = B i bonds broken; B B bonds formed; A A bonds broken; 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)

CALCULATING (HAll 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)

CALCULATING THE RATIO

(H = ?

Cx,a } Cx,b

-(H/RT

e

R, gas constant (1.99 cal/oK); T, temperature (oK); = 298

HYDROPHOBICITYHydrophobic 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.

SELECTIVITY(H = (Pa2 - Pb2) + 2Pi.(Pb Pa)If there were only one type of interaction between molecules, the above equation for (H 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.

PART 2 SOLVENT CLASSIFICATION AND SELECTION

Outline:Solvent Classification Schemes Summary of Solvent Selection Extraction Efficiencies

HILDEBRAND SOLUBILITY PARAMETER

(H = Vx [(Ha2 - Hb2) + 2Hx (Hb - Ha)]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

ROHRSCHNEIDER POLARITY SCALEThe 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 Dioxane Nitromethane proton donor interaction proton acceptor interaction dipole interaction

ROHRSCHNEIDER POLARITY SCALEThe 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

SOLVENT SELECTIVITY TRIANGLE

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.

EXTRACTION OF COMPOUND X FROM A SAMPLE MATRIX

P1100 80 60 40 20

Px P2 Py

x P

y

EXTRACTION OF COMPOUND Y FROM A SAMPLE MATRIX

P1100 80 60 40 20

Px

P2 Py

x P

y

PARTITION COEFFICIENTSimplest 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/CwCo is concentration in the organic phase (solvent) Cw is the concentration in the aqueous phase (water)

PARTITION COEFFICIENTIf DV u100 then a single batch extraction can work:

D = Co/CwCo is concentration in the organic phase (solvent)

Assume: V = Vo/Vw = 10 D = Co/Cw = 5 Cw is the concentration in the aqueous phase (water)

Assume equal volumes

Then: U = (5)(10)/1+(5)(10) U = 98%

For unequal volumes, fraction extracted U

Co Vo DV U! ! Co Vo Cw Vw 1 DV

V = Vo/Vw

Fraction remaining in aqueous phase following n extractions: (1 - U)n = Xn

EXTRACTION EFFICIENCY

Wr = Wo (Vw/(KVo+ Vw))NWhere W r = 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