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What is a chemical separation? Examples: Filtration Precipitations Crystallizations Distillation HPLC GC Solvent Extraction Zone Melting Electrophoresis Mass Spectroscopy. Chemical Separations. What is the object of the separation. Collection of a pure product

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Chemical separations l.jpg

What is a chemical separation?

Examples:

Filtration

Precipitations

Crystallizations

Distillation

HPLC

GC

Solvent Extraction

Zone Melting

Electrophoresis

Mass Spectroscopy

Chemical Separations


Chemical separations2 l.jpg

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?

Chemical Separations


Chemical separations3 l.jpg

Major Industries

Petroleum Distillation

Distilled Spirits

Chemical Separations


Chemical separations5 l.jpg

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

Chemical Separations



Distillation l.jpg

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.

Distillation


Distillation8 l.jpg

What is whiskey? compounds will boil away and can be collected.

What is brandy?

Distillation


Interesting facts l.jpg

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.

Interesting Facts


Interesting facts10 l.jpg

Malt whisky - Whisky made purely from malted barley. 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.

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.

http://www.whisky-world.com/words/index.php

Interesting Facts


Extraction l.jpg

Liquid – Solid 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.

Liquid - Liquid

Extraction


Liquid solid extraction soxhlet whale wheelock edu bwcontaminants analysis html l.jpg
Liquid Solid Extraction (Soxhlet) 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.whale.wheelock.edu/ bwcontaminants/analysis.html


Liquid solid extractor bunn l.jpg
Liquid Solid Extractor (Bunn) 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.


Solvent extraction liquid liquid l.jpg

We want to partition a solute between two immiscible phases. 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.

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.

Solvent Extraction (Liquid-Liquid)


Slide15 l.jpg

http://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.htmlhttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html


Solvent extraction l.jpg
Solvent Extractionhttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html


Slide17 l.jpg

Replace concentration with moles over volume and http://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.htmllet q equal the fraction in the aqueous phase.m will be the moles of solute in the entire system


Define a new term for the ratio of the volumes of the phases l.jpg
Define a new term for the ratio of the volumes of the phaseshttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html


We can do a little algebra and find an expression for q l.jpg
We can do a little algebra and find an expression for qhttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html


If it does not end up in the aqueous phase it must be in the organic phase l.jpg

p is the term for the fraction in the organichttp://oceanexplorer.noaa.gov/explorations/02sab/background/products/media/chroma1.html

p + q = 1

Giving

If it does not end up in the aqueous phase it must be in the organic phase.


Sample problem l.jpg

You have 100.0 mL of an aqueous solution that is 100.0 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.

Sample Problem


Solution l.jpg
Solution 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.


We can do multiple extraction from the aqueous phase l.jpg
We can do multiple extraction from the aqueous phase. 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.

  • We end up with the following expression for what is left in the aqueous phase.


Example l.jpg

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  0.9999.  This leaves 0.0001 in the aqueous phase.  Since we have equal volumes then Vr is 1.00.

We are able to find from the Interactive Analysis Web site that K for Aspirin is 35.5.  We plug these values into the q equation and the power is the unknown.

Example


Solution25 l.jpg
Solution aspirin from an aqueous solution with an equal volume of n-octanol?


What if our compound can dissociate or participate in some other equilibrium l.jpg

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???

What if our compound can dissociate or participate in some other equilibrium?


Dissociation l.jpg

So if we have dissociation then less will go into the organic phase.

Kp 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?

Dissociation


Distribution coefficient l.jpg
Distribution Coefficient organic phase.

  • Where C is the formal concentration of the species.

  • Ca = [HA] + [A-]

  • Dc will vary with conditions

  • For this compound what is that condition?


Slide29 l.jpg
D organic phase. c

  • 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.


Acid equilibria l.jpg
Acid Equilibria organic phase.

  • What is the equilibrium?

  • Ka


With a little algebra l.jpg
With a little algebra organic phase.

  • So if you know Kd and Ka then you can determine Dc as a function of H+ (pH)

  • However if [H+] is much larger than Ka then Dc will equal Kd. If the [H+] is close in value to Ka then D will be related to the pH

  • Plotting this we get.


So what why is this useful l.jpg

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.

So What, Why is this useful.


Separation l.jpg

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 Dc 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.

Separation


Example35 l.jpg

System I another. What if we are try to separate two compounds, A and B

Da = 32 Db = 0.032 (A ratio of 1000)

Vr= 1

Let's recall our equations

q (fraction in aqueous)  =  1 /  (DVr + 1)

p (fraction in organic)       =   DVr / (DVr + 1)

Vr (volume ratio)            =   Vo / Va

Example


Case i l.jpg

p another. What if we are try to separate two compounds, A and Ba  =  32*1 / (32*1 + 1)  =  0.97

pb  =  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 + 0.03)  =  0.97 or 97 %

Case I


Case ii l.jpg

D another. What if we are try to separate two compounds, A and Ba = 1000 Db = 1 VR = 1 (Ratio is still 1000)

pa  =  1000*1 / (1000*1 + 1*1)  =  1000/1001  =  0.999  

Aha! we got more a into the organic, as we would expect with a higher D value.

Now

pb =  1*1 / ( 1*1 +1) =  1/2  = 0.5

oh-oh

What do we get for purity of compound a now?

purity =  0.999 / (0.999 + 0.50)  =  0.666

Yuck!

Case II


How can we get around this issue l.jpg

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  =  DVr / (DVr + 1)

Not much here except Vr  and in fact that is the key to this problem.  Is there an optimum Vr value for the values of D that we have?  Yes!

Our equation for this is      V r(opt)  =  (Da*Db)-0.5

How can we get around this issue?


Revisit the two cases l.jpg

So let us look at our two cases and see which will give us the optimum values.

Case I

Da  =  32   and Db  =  0.032

V r(opt) =  (32 * 0.032)-0.5   =   ( 1 )-0.5  = 1

So we were already at the optimum.

Revisit the two cases


Case ii revisited l.jpg

Case II the optimum values.

Da = 1000 and Db = 1

Vr (opt)  =  (1000*1)-0.5  = 1000-0.5  =  0.032

Which mean that when we do our extraction we will extract _______ mL of organic for each _______ mL of aqueous.

Case II Revisited


Purity for case ii l.jpg

What is our purity for this system? the optimum values.

pa  =  1000*0.032 / (1000*0.032 + 1)  =  32/33  =  0.97

and

pb  =  1*0.032 / (1*0.032 + 1)  =  0.032/1.032 = 0.03 

Purity of a then is 0.97/ (0.97 + 0.03)

Which will give us the 97% purity we had for Case I with with the Vr of 1.

Purity for Case II


Can we improve this purity l.jpg

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.

Can we improve this purity?



Back extraction case i example l.jpg

Let's look at the numbers. phase.

Da = 32 Db = 0.032 Vr = 1

pa = 0.97 pb = 0.03

qa = 0.03 qb = 0.97

Let’s prepare a table. 

Back Extraction Case I Example


Initial conditions prior to starting back extraction l.jpg

Before Shaking phase.

Amount A

Amount B

Organic Phase

0.97

0.03

Fresh Aqueous Phase

0

0

Initial conditionsprior to starting back extraction

.


Now we extract shake shake shake l.jpg

How much goes to the Aqueous phase 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 Shaking

Amount A

Amount B

Organic Phase

(0.97)(0.97)

(0.03)(0.03)

Aqueous Phase

(0.97)(0.03)

(0.03)(0.97)

Now we extract – shake shake shake


Now what is the purity for a in the organic phase l.jpg

Purity = Amount A / (Amount A + Amount B)  =  phase.

0.97*0.97 / (0.97*0.97 + 0.03*0.03) =

0.94/(0.94 + 0.0009) = 99.9%

What is the yield of A (fraction of the total amount that we started with)

Now what is the purity for A in the organic phase???


Let s do it again can we improve purity even more l.jpg

After second Back Extraction phase.

Amount A

Amount B

Organic Phase

0.94*0.97

0.0009*0.03

Aqueous Phase

0.94*0.03

0.0009*0.97

Let’s do it again – Can we improve purity even more?

Purity A   =  0.913 / (0.913 + 0.000027) =  99.997%

But our yield has dropped to 91.3%,    there is a price to pay for the added purity.


Can we expand this why would we want to l.jpg

Such multiple extraction systems have been developed. phase.

Still a viable option for preparative work.

For separations it has been replaced by HPLC

Called Craig Counter Current Extraction.

Special glassware is used.

Can We Expand This?Why Would We Want to?


Craig cce l.jpg
Craig CCE phase.

  • 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.


Craig cce52 l.jpg
Craig CCE phase.

  • 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 90o to 100o.


Craig cce53 l.jpg
Craig CCE phase.

  • Rotate Back to Horizontal


Slide54 l.jpg

Starting Conditions phase.

After One Equilibrium

Transfer Step 1



Slide56 l.jpg

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.

We transfer again.

Transfer Step 3


Slide57 l.jpg

Shake Again  Equilibrium 4 equilibrium.

Transfer 4

See a trend????


Craig cce58 l.jpg

How about a binomial expansion? equilibrium.

(q  +  p)n  =  1

Powers of the two terms in each tube will add up to n

Coefficients will be found from Pascal Triangle

1 1     1 1     2     1 1     3     3     1 1     4     6     4     1 1     5     10     10     5     1 1     6     15     20     15     6     1 1     7     21     35     35     21     7     1

Craig CCE


Craig cce59 l.jpg

Or the formula equilibrium.

Fr,n  =  n!/((n-r)!r!) pr q(n-r)

n is the number of transfer and r is the tube number. You start counting at zero!

Craig CCE


Craig cce60 l.jpg

Let's look at and example for a four tube system. equilibrium.

Da   = 3 p =  0.75 q = 0.25

Db  = 0.333 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               p4   or  0.754   =   0.3164 Amount of B               p4   or  0.254   =   0.0039

Purity of A         0.3164 / (0.3164 + 0.0039)  =  0.9878    or 98.78%

Yield of A          We collect a fraction of 0.3164  or 31.64%

Horrible Yield!

Craig CCE


Craig cce61 l.jpg

What if we collect the last two tubes?? equilibrium.

Amount of A      p4   and  4p3q  or  0.754 +  4*(0.75)3(0.25)  =   0.3164 + 0.4219 = 0.7383 Amount of B      p4   and  4p3q   or  0.254 +  4*(0.25)3(0.75)  =   0.0039 + 0.0469 = 0.0508

Purity of A    (0.3164 + 0.4219) / (0.3164 + 0.4219 + 0.0039 + 0.0469 )  =  0.9356 or 93.56%

Yield of A              We collect a fraction of 0.3164  +  0.4219  =  0.7383 or 73.83 %

Purity still ok and yield is much better.

Craig CCE


Craig system n 200 transfers d a of 2 0 and d b of 4 0 p a of 0 666 p b of 0 800 l.jpg
Craig system n= 200 transfers.  equilibrium.Da of 2.0 and Db of 4.0 pa of 0.666 pb of 0.800.


Final formulas 1 l.jpg

r equilibrium.max =   np   = nDVr/(DVr +1)

To find the separation between two peaks we would use.

Drmax = (rmax)a - (rmax)b  =  n(pa-pb)

The Gaussian distribution approximation for our binomial expansion would be (when n>24)

Fr,n =  (2p)-0.5(npq)-0.5 exp-[((np-r)^2)/2npq]

Final Formulas(1)


Final formulas 2 l.jpg

The width of the distribution through the system would be: equilibrium.

w = 4s = 4(npq)0.5

Resolution would be

R = Drmax/w = Drmax/4s

or

R = nDp/(4(npq)0.5)  =  n0.5Dp / 4(pq)0.5

Final Formulas(2)



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