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Steps of separation of natural
products
General
isolation
strategy of
natural
products
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Crude aqueous
extracts of certain
plants (and animals)
provided pigments,such as indigo and
alizarin.
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Other examples of naturalproducts:
ephedrine
from Ephedra sinica(for respiratoryailments) tetrahydrocannabinol (marijuana)geraniol (rose oil)
cinnamaldehyde(cinnamon) diallyldisulfide (garlic)
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morphine
(narcotic analgesic)
1817
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strychnine
(poison) 1818
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cocaine
(narcotic stimulant)
1859
nicotine(toxic) 1828
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Liquid-liquid extraction is a useful method to separate components(compounds) of a mixture
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Let's see an example.
Suppose that you have a mixture of sugar in vegetable oil (it tastes
sweet!) and you want to separate the sugar from the oil. Youobserve that the sugar particles are too tiny to filter and yoususpect that the sugar is partially dissolved in the vegetable oil.
What will you do?
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How about shaking the mixturewith water
Will it separate the sugar from the
oil? Sugar is much more soluble inwater than in vegetable oil, and,as you know, water is immiscible(=not soluble) with oil.
Did you see the result?The waterphase is the bottom layer andthe oilphase is the top layer, because
water is denserthan oil.
*You have not shaken the mixtureyet, so sugar is still in the oil phase.
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By shaking the layers (phases) well, youincrease the contact area between thetwo phases.The sugar will move to thephase in which it is most soluble: the
water layer
Now the water phase tastessweet,because the sugar is moved tothe water phase upon shaking.**Youextracted sugar from the oil withwater.**In this example,water wasthe extraction solvent ;the originaloil-sugar mixture was the solution to
be extracted; and sugar was thecompound extracted from one phaseto another. Separating the two layersaccomplishes the separation of thesugar from the vegetable oil
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Did you get it? .....the concept of liquid-liquid extraction?
Liquid-liquid extraction is based on the transfer of a solute
substance from one liquid phase into another liquid phase accordingto the solubility.Extraction becomes a very useful tool if you choosea suitable extraction solvent.You can use extraction to separate asubstance selectively from a mixture, or to remove unwantedimpurities from a solution.In the practical use, usually one phase is awater or water-based (aqueous) solution and the other an organicsolvent which is immiscible with water.
The success of this method depends upon the difference in solubilityof a compound in various solvents. For a given compound, solubilitydifferences between solvents is quantified as the "distribution
coefficient"
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Partition Coefficient Kp (Distribution Coefficient Kd)
When a compound is shaken in a separatory funnel with two immiscible
solvents, the compound will distribute itself between the two solvents.
Normally one solvent is waterand the other solvent is awater-immiscible organic
solvent.
Most organic compounds aremore soluble in organic solvents,while some organic compounds
are more soluble in water.
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(1) If there are 30 particlesof compound , these aredistributed between equalvolumes of solvent1 and solvent2..
(2) If there are 300particles of compound , the
same distribution ratio isobserved in solvents 1 and 2
(3) When you double the
volume of solvent2 (i.e., 200mL of solvent2 and 100 mL ofsolvent1),the 300 particles ofcompound distribute as shown
If you use a larger amount of extraction solvent, more solute isextracted 15
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What happens if you extract twice with 100 mL of solvent2 ?In this case, the amount of extraction solvent is the same volume as wasused in Figure 3, but the total volume is divided into two portions and
you extract with each.
As seen previously, with 200 mLof solvent2you extracted 240particles of compound . Oneextraction with 200 mL gave a
TOTAL of 240 particlesYou still have 100 mL of solvent1,containing 100 particles. Now youadd a second 100 mL volume offresh solvent2. According to the
distribution coefficient K=2, youcan extract 67 more particlesfrom the remaining solution
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An additional 67 particles areextracted with the second portionof extraction solvent(solvent2).The total number of
particles extracted from the first(200 particles) and second (67particles) volumes of extractionsolvent is 267.This is a greaternumber of particles than the
single extraction (240 particles)using one 200 mL portion ofsolvent2!
It is more efficient to carry outtwo extractions with 1/2 volumeof extraction solvent than onelarge volume!
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If you extract twice with 1/2 the volume, the extraction is moreefficient than if you extract once with a full volume. Likewise,extraction three times with 1/3 the volume is even more efficient.
four times with 1/4 the volume is more efficient.five times with 1/5the volume is more efficientad infinitum
The greater the number of small extractions, the greater thequantity of solute removed. However for maximum efficiency the
rule of thumb is to extract three times with 1/3 volume
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Chemically active (acid-base) extraction
Can you change the solubility property of a compound? How?
Most organic compounds are moresoluble in organic solvents than inwater,usually by the distributioncoefficient K > 4
However, specific classes oforganic compounds can bereversibly altered chemically tobecome more water-soluble.
This is a powerful technique and allows you to separate organiccompounds from a mixture -- if they belong to different solubilityclasses
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What type of organic compounds can be made water-soluble?Compounds belonging to the following solubility classes can beconverted to their water-soluble salt form
(1) Organic acids include carboxylic acids (strong organic acids)and phenols (weak organic acids).
(2) Organic bases includes amines
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How can organic acids or bases be converted to a water-solubleform?1. Organic Acids can be converted to their salt form when treated with an
aqueous solution of inorganic base (e.g., NaOH (sodium hydroxide) andNaHCO3 (sodium bicarbonate)).Salts are ionic, and in general, ions aresoluble in water but not soluble in water-immiscible organicsolvents.Remember: water is a verypolarsolvent thus salts (i.e., ionic
species) are well dissolved in it.
A. Carboxylic Acids areconverted to the salt formwith 5% NaOH aqueoussolution. NaOH is a stronginorganic base.
Carboxylic acids are strong
organic acids (pKa = 3 to 4),
so they can also be ionized
with weak inorganic bases
(e.g.,NaHCO3(sodium
bicarbonate)) aqueous
solution.
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Let's try a sample problem.Here is a mixture of naphthalene and benzoic acid, dissolved in
dichloromethane.
You want to separatethese two compounds.
What will you do?
You may use an aqueous solution ofeither 5% NaOH or sat. NaHCO3, to
extract benzoic acid as a salt form22
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B. Phenols are considered to be weak organic acids. Phenol, the parentcompound, is partially water-soluble (1 g will dissolve in 15 mL of water),whereas substituted phenols are not.Sodium bicarbonate (NaHCO3) aqueoussolution, a weak inorganic base, will not deprotonate phenols to make itionic, because it is not strong enough.However, treatment with NaOH, a
strong inorganic base, can change phenol to its ionic (salt) form.
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Let's try a another sample problem.Here is a mixture of benzoic acid and p-methoxyphenol, dissolved in
dichloromethane.
You want to separatethese two compounds.
What will you do?
You cannot use 5% NaOH to separate these twocompounds. NaOH will react with both benzoic acid and p-methoxyphenol, thus both compounds will be extracted into the
aqueous layer.24
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Let's try this problem again.Here is another mixture of benzoic acid and p-methoxyphenol, dissolved in
dichloromethane.
Strong organic acids such as benzoic acid would bedeprotonated and ionized, while weak organic acids such
as phenols would NOT be deprotonated
NaOH was too strong a base,
thus it does not differentiatethe strong and weak organicacids. Use of weak inorganicbase such as NaHCO3 willdifferentiate between the
compounds
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Let's try a third sample problem.Here is a mixture of benzoic acid and p-chloroaniline, dissolved in
dichloromethane.
You want to separatethese two compounds.
What will you do?
You may use an aqueous solution ofeither 5% HCl, to extract the amine asa salt form and benzoic acid has
remained in the organic layer 27
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You can separate four different classes of compounds from a mixture based ondiffering solubility properties. The four classes are:1.Amines (organicbase)2.Carboxylic acids (strong acid)3.Phenols (weak acid)4.Neutral
compounds.
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After the separation of the mixture of four components, we will have foursolutions: each solution contains one component.
The first three compounds are chemically altered, existing in their salt formdissolved in aqueous solution. The fourth compound is not chemically altered,but it is dissolved in an organic solvent.We now want to recover each compound
in its original state (i.e., in the non-ionic form) to complete the experiment. Wecall this step isolation or recovery.
Let's see, one by one, how to recover each compound obtained from the
separation process
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I l ti (R ) f i
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Isolation (Recovery) of aminesAn amine is a basic compound. It is protonated in the presence of excess HClforming a salt that is soluble in aqueous solution. This is how you separated theamine from the original mixture containing it.
An amine is soluble in acidic aqueoussolution because it forms a salt, an
ionic form.
However, if you change the pHof the solution to basic theamine can no longer staydissolved because it is no longerionic! This process is calledbasification.
Basification is done by carefully
adding concentrated NaOH
solution to the solution containing
the amine salt until it becomes
basic. 30
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In the basification step, you useconcentrated NaOH solution tominimize the volume of the finalsolution. Recall that a dilute
solution of HCl was used toextract the amine as its water-soluble salt (see the picture onthe right side).
Basification must be donecarefully, portion by portion, withswirling each time because theacid-base neutralization reactionis exothermic.
Check the pH of the solution to
ensure that it is basic. (~pH 10)
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l (R ) f d
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Isolation (Recovery) of Acids
There are two different groups of organic acids: carboxylic acids (strong acids) andphenols (weak acids).In the separation procedure, acids were extracted using (weak orstrong) basic aqueous solutions
Both acids can be returned to the original form in the samemanner! Organic acids are currently dissolved in a basic aqueoussolution, because the acid forms a salt, an ionic form. When youmake the aqueous solution acidic, the organic acids no longerremain dissolved because they are no longer ionic and usuallyprecipitate out of solution. This process is called acidification.
Acidification is done by carefully addingconcentrated HCl solution until the mixturebecomes acidic,
When the weak base, NaHCO3, was theextracting solution, CO2 gas will evolveduring acidification.
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The recovery of organic acidsrequires acidification withconcentrated HCl solution. Recallthat in the extraction step for
the separation of an organic acideither dilute NaHCO3 or NaOHwas used. Concentrated HCl willnow help minimize the volume ofthe final aqueous solution
Acidification must be donecarefully, portion by portion, withswirling each time because theacid-base neutralization reactionis exothermic.
Check the pH of the solution toensure that it is acidic. (~pH 3)
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RCO2H / ArOH / RNH2/ RH General Acid/Base Extraction
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Aqueous Phase #1RCO2Na dissolved in aqueous base
Organic Phase #1ArOH / RNH2/ RHdissolved in Et2O
Organic Phase #2 RNH2/ RHdissolved in Et2O
Aqueous Phase #2ArONa dissolved in aqueous base
extract with5% NaHCO3
acidifywith HCl
dry, filter &concentrate
extract with5% NaOH
acidifywith HCl
make basicwith NaOH
2 2
dissolved in diethyl ether
RCO2H
ArOH
extract with 5% HCl
RHRNH2
Organic Phase #3RH dissolved in Et2O
Aqueous Phase #3dissolved RNH3Cl
General Acid/Base Extraction
Flow Diagram
separation of organic acids, bases, and neutrals
RCO2H = carboxylic acids
ArOH = phenols
RNH2 = amines
RH = others (neutrals)
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A t id h A i i C ff i
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Acetamidophen Aspirin Caffeine
O CH3
O
OHO
NO
CH3
NCH
3
O
N
N
CH3
ACE(phenol)[weak organic acid]
ASP(carboxylic acid)[strong organic acid]
CAF(amine)[organic base]
OH
N CH3
O
H
phenoxidesalt
polar, water soluble
carboxylatesalt
polar, water soluble
ammonium chloridesalt
polar, water soluble
extract withaqueousNaOH
extract withaqueous
NaHCO3
extract withaqueous HCl
NO
CH3
NCH3
O
N
N
CH3
H+
Cl-N CH3
O
H
Na+ -O
O CH3
O
O O-Na+
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example
extraction
flow diagram
ACE / ASP / CAFmethylene chloride
solid #2
solid #3
aspirin
solid #1
acetaminophen
lower layer in theseparatory funnel}
density of CH2Cl2 = 1.325 g/mL( )
gravity filter
solution #1 {separatory funnel}
solution #2
(organic phase)CAF dissolved in CH2Cl2
solution #3
(aqueous phase)ASP salt dissolved in aqueous base
extract with10 mL NaHCO3
acidify with2 mL 6N HCl
dry withanh. MgSO4
separates insoluble
solid from solution
filter with glass fritted funnel
concentrate byremoving solvent
ASP / CAF in CH2Cl2
separates acidsfrom neutrals
filter ppt.
for separation of three components:
acetaminophen, aspirin & caffeine
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Separatory Funnel Extraction Procedure
Separatory funnels are designed to facilitate the mixing of immiscible liquids
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Separatory Funnel Extraction Procedure
1. Support the
separatory funnel in aring on a ringstand.Make sure stopcock isclosed
2. Pour in liquid to
be extracted
3. Add extractionsolvent
4. Add ground glass
Stopper (well greased)
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Separatory Funnel Extraction Procedure
Pick up the separatoryfunnel with the stopper inpalce and the stopcockclosed, and rock it once
gently.
Then, point the stem up and slowly open thestopcock to release excess pressure. Close thestopcock. Repeat this procedure until only a smallamount of pressure is released when it is vented
Shake the separatory funnel.
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Separatory Funnel Extraction ProcedureShake the separatory funnel vigorously.
Now, shake the funnel vigorously for a few seconds. Release
the pressure, then again shake vigorously. About 30 sectotal vigorous shaking is usually sufficient to allow solutes tocome to equilibrium between the two solvents.
Vent frequently to prevent pressure buildup,which can cause the stopcock and perhapshazardous chemicals from blowing out. Takespecial care when washing acidic solutions withbicarbonate or carbonate since this produces alarge volume of CO2 gas
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Separatory Funnel Extraction ProcedureSeparate the layers.
Let the funnel restundisturbed until the layersare clearly separated
While waiting, remove thestopper and place a beakeror flask under the sepfunnel.
Carefully open the stopcock andallow the lower layer to draininto the flask. Drain just to thepoint that the upper liquidbarely reaches the stopcock
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Solvent Extraction
Equilibrium constant for this partitioning is
K (partition coefficient)
K=[S]2[S]
1
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Solvent Extraction
Determination of solute concentration in
each phase
Define some variables:
V1 & V2 are volumes of solvents 1&2
m = total # of moles of solute (S) present
q = fraction of solute remaining in phase 1 at
equilibrium
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Solvent Extraction
[S]1 = qm/V1
[S]2 = (1-q)m/V2
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K=[S]2
[S]1
qm/V1
(1-q)m/V2= =q/V1
(1-q) /V2
q =KV2 + V1
V1(1-q) =
KV2 + V1
KV2
fraction of S in: phase 1 phase 2
Rearrange:
Solvent Extraction
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Solvent Extraction
If remove V2 and extract V1 with fresh
layer of V2, what fraction remains in V1?
Initial moles = m
after first extraction - qm
after second extraction - q(qm)=q2m
q(2) =KV
2+ V
1
V12
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Solvent Extraction
q(n) =KV2 + V1
V1n
Fraction in V1 after n extractions:
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Solvent Extraction
Example: Solute A has a partition
coefficient of 4.000 between hexane and
water. (K = [S]hexane/[S]water= 4) If 150.0
ml of 0.03000 M aqueous A is extractedwith hexane, what fraction of A remains if:
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Solvent Extraction
a) one 600.0 ml aliquot of hexane is used?
q =4(600ml) + 150ml
150ml= 0.05882 = 5.882%
# moles remaining
0.05882 (0.03M0.150L) = 2.647x10-4 moles
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q =4(100ml) + 150ml
150ml= 0.0004115
6
# moles remaining
4.115 x 10-4 (0.03M0.150L) = 1.852x10-6moles
Solvent Extraction
b) 6 successive 100.0 ml aliquots of
hexane are used?
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Solvent Extraction
Although same volume of hexane is used,
it is more efficient to do several small
extractions than one big one!
1 600 ml extraction extracts 94.12%
6 100 ml extractions extract 99.96%
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B + H2O BH+ + OH-
Kb
HA H+ + A-Ka
Generally, neutral species are more soluble
in an organic solvent and charged species
are more soluble in aqueous solution
Solvent Extraction (pH effects)
with organic acids/bases:
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organic
HA H+ + A-Kaaqueous
HA H+ + A-
very little here, ions
havepoor solubility
Solvent Extraction (pH effects)
Partitioning of organic acids between two
phases:
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Solvent Extraction (pH effects)
When the solute (acid/base) can exist in
different forms, D (distribution coefficient) is
used instead of K (partition coefficient)
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Solvent Extraction (pH effects)
D =total conc. in phase 2
total conc. in phase 1
D =[HA]org
[HA]aq + [A
-
]aq
HA
HA H+ + A-Ka
K
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Ka =[H+][A-]
[HA][A-] =
Ka [HA]
[H+]
Solvent Extraction (pH effects)
Substitute for [A-] in D eq. and rearrange
D = [HA]
[HA]
2
1 K HA
H
a[ ]
[ ]
1
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Solvent Extraction (pH effects)
D =[HA]
[HA]
[HA]
[HA]
2
1
2
1
K HA
H
K
H
a a[ ]
[ ] [ ]
1
1
D = [HA][HA]
2
1
1
KH
a
[ ]= K
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Solvent Extraction (pH effects)
D = K1
K
H
a
[ ]
D =
K
1
K
H
K H
H Ka a
[ ]
[ ]
[ ]
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D
pH
[H+]=Ka
pH=pKaK
[H+
]>>Ka
mainly
HA
[H+]
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K1
4 8
K2
D
pH
Solvent Extraction (pH effects)
Example problem: Want to separate two organic
acids using a scheme based on pH. Acid 1 (pKa =
4), Acid 2 (pKa = 8)
Acid 2 stays in
organic phase,
acid 1 is extracted
into aqueous phase
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[H+] + KaD =
K Ka
D
pH
K[H
+
]=KapH=pKa
[H+]>>Ka
mainly
BH+
[H+]
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D
pH
Kacid base
Solvent Extraction (pH effects)
In general: