Transcript Chemistry of Separation - Food Science & Human Nutrition

Solvent Selection
for Separation
SOLVENT SELECTION

FOR
Processes Requiring Solvents






Extraction
Partition
Fractional Crystallization
Extractive Distillation
Liquid Chromatography
Gas-Liquid Chromatography
SEPARATION
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
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 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
Peripheral Properties
Viscosity of Mixtures h =(ha)xa (hb)xb
Take home – it is
possible to use a
viscous solvent when
in a mixture
PERIPHERAL PROPERTIES
Viscosity of Water-Organic Solvent Mixtures
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?
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?
PERIPHERAL PROPERTIES
Solvent Properties Affecting Detection
Specific Element Content
Common Gas Chromatography Detectors:
Method
Element
Electron Capture
Cl
Flame Thermionic
N
Flame Photometric
S
Solvent
Chloroform
Acetonitrile
Dimethyl Sulfoxide
Peripheral Properties
SOLVENT MISCIBILITY
CHART
Appendix C in your
notes
PERIPHERAL PROPERTIES
– Toxicity
– Flammability
– Reactivity
– Cost
– Disposal
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?
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
A
x
A
x
x
x
x
x
B
x
x
x
x
x
x
B
DG, free energy for transfer of 1 mole
solute x from solvent B to A.
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
Solvent Selection for
Separation
Cx,b
Cx,a
= ?
100
y = ex
10
1
0.1
0.01
0.001
-4
-2
0
2
4
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?
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.

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
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)
WHAT GOVERNS
THESE BONDS?
THE
STRENGTH
OF
• Intermolecular Interactions
– Dispersion Forces
– Dipole-Dipole
– Induced-Dipole
– Hydrogen Bonding
– Covalent Bonds
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.
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.
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 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).
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.
INTERMOLECULAR FORCES
Its 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. oC
1) neopentane
2) 2,3-dimethyl butane
3) n-hexane
4) 2-methyl-2-butanol
5) 1-pentanol
10
58
69
102
138
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.
EXAMPLE:
Two Immiscible Solvents:
Solvent A
Solvent B
Analyte:
Water
Hexane
Acetone (i)
How will the acetone partition between Solvent A and
Solvent B?
EXAMPLE: USE
THIS EQUATION
Cx,a

Cx,b
What is the value of DH?
A
x
x
x
x
B
x
x
e
-DH/RT
x
ESTIMATING
2Hi,b
-Hb,b
Ha,a
-2Hi,a
=
=
=
=
THE
VALUE
OF
B – i bonds broken; B – B bonds formed;
A – A bonds broken; i – A bonds formed
DH = (Ha,a – Hb,b) + 2(Hi,b – Hi,a)
Approximation:
The interaction between molecules
is based on the product of their
polarities. Thus:
DH = (Pa2 - Pb2) + 2Pi.(Pb – Pa)
DH
CALCULATING DH
All we need now are the polarities of A, B, and i
to substitute in this equation:
DH = (Pa2 - Pb2) + 2Pi.(Pb – Pa)
From Appendix A:
i = Acetone P’ = 5.1
B = Hexane P’ = 0.1
A = Water
P’ = 10.2
DH = (10.22 - 0.12) + 10.2(0.1 – 10.2)
CALCULATING
DH = ?
THE
RATIO
Cx,a

Cx,b
e
-DH/RT
R, gas constant (1.99 cal/oK);
T, temperature (oK); = 298
HYDROPHOBICITY
Hydrophobic interactions - associated with "nonpolar"
solutes in "polar" solvents
assume: B - polar
A - non-polar
i – small (non polar)
DH = (Pa2 - Pb2) + 2Pi.(Pb – Pa)
the polar solvent "squeezes" out the nonpolar solute into phase A.
SELECTIVITY
DH = (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.
PART 2 SOLVENT CLASSIFICATION
SELECTION
Outline:
Solvent Classification Schemes
Summary of Solvent Selection
Extraction Efficiencies
AND
HILDEBRAND SOLUBILITY PARAMETER
DH = 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
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
Dioxane
proton donor interaction
proton acceptor interaction
Nitromethane
dipole interaction
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’
SOLVENT SELECTIVITY TRIANGLE
EXTRACTION OF COMPOUND X FROM
SAMPLE MATRIX CONTAINING Y
A
Begin by studying the extraction of x and y as a
function of solvent polarity.
EXTRACTION OF COMPOUND X
SAMPLE MATRIX
P’1
P’x P’2 P’y
100
80
60
40
20
FROM A
x
y
P’
EXTRACTION OF COMPOUND Y
SAMPLE MATRIX
P’1
P’x
P’2 P’y
100
80
60
40
20
x
y
P’
FROM A
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)
PARTITION COEFFICIENT
If DV 100 then a single
batch extraction can
work:
D = Co/Cw
Co
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:
 = (5)(10)/1+(5)(10)
 = 98%
For unequal volumes, fraction extracted 
Co Vo
DV


Co Vo  Cw Vw 1  DV
V = Vo/Vw
Fraction remaining in aqueous phase following n extractions:
(1 - )n = Xn
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