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8/10/2019 pemisahan kimia

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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.


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.


What will you do?

You may use an aqueous solution ofeither 5% HCl, to extract the amine asa salt form and benzoic acid has

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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

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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


ammonium chloridesalt


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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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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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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F l E P d


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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


Partitioning of organic acids between two

phases:

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When the solute (acid/base) can exist in

different forms, D (distribution coefficient) is

used instead of K (partition coefficient)

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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+]


Substitute for [A-] in D eq. and rearrange

D = [HA]

[HA]

2

1 K HA

H

a[ ]

[ ]

1

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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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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


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


In general:

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