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2. Extraction

Pre-lab questions

Complete the following questions and submit before beginning the
experiment.

If you had 3 grams of 4-chlorobenzoic acid dissolved in ether, how many
milliliters of a 4M NaOH solution would be needed to react with it all?

To recover all of the 4-chlorobenzoic acid that had reacted with the
NaOH in the previous problem, how many milliliters of a concentrated HCl
(44 g/100 mL) solution would be needed to react with it all?

The partition coefficient of benzoic acid is 3 in methylene chloride
with respect to water. If 6 grams of benzoic acid was dissolved into
600 mL of water, how much benzoic acid can be extracted with four 150 mL
portions of methylene chloride?

Using Schemes 1 and 2 in your handout as references, fill in the
flowchart on the following page for the separation of an organic acid
(R-COOH), a base (R-NH2) and a neutral organic compound.

Pre-lab Notebook preparation

Prepare the following in your lab notebook before beginning the
experiment.

Start a new page by entering the title of this experiment.

Start a section with the subtitle “Part 1 - Procedure” and provide
an outline of the experimental procedure to be carried out. Do the same
for Part 2.

2. Extraction

Zubrick: Chapters 15 and 16

127-130 (general introduction)

133-136 (how to extract, sample extraction, washing)

138-139 (extraction hints)

141-143 (microscale extraction information)

Vocabulary: Miscible Two or more compounds which dissolve in one
another to

form a homogenous solution



Aqueous contains water

Organic non-aqueous (does not contain water),

usually hydrophobic

Emulsion No separation of the organic and aqueous layers (usually
cloudy).

Partition coefficient The same as the distribution coefficient. It is
the relationship between the solubility ratios of two different

layers for a compound which is soluble in both solvents.

This procedure has been adapted from the microscale procedure described
in Macroscale and Microscale Organic Chemistry Experiments by Kenneth L.
Williamson.

Background



What is extraction? There are several listings for extraction in the
American Heritage Dictionary (p. 466) but in its simplest form,
extraction is the separation of something from another. There are
numerous examples in our everyday lives. Vanilla extract is used in
cooking. The essence of perfume is extracted from flowers, plants, etc.
For the organic chemist, extraction is part of the normal work up of a
reaction. Sometimes, it is referred to as aqueous workup in the text or
manuscript. We wash the organic reaction mixture with water (acidic,
neutral and/or basic) to remove any byproducts or inorganic material.

There are three basic types of extraction: liquid/liquid, acid/base and
solid/liquid. We will be performing the first two types of extraction
in this experiment. First, we have to understand how to choose a
solvent for extraction. A couple of things we have to consider are
polarity and density of the solvent. The polarity of the solvent is
important because “like dissolves like.” So if the extraction is
from an aqueous solution, we do not want a solvent that is miscible with
water. Thus during the extraction two layers are formed, aqueous and
organic. It is important to know the density of the solvent so that we
can predict which one is the organic layer. Most chlorinated solvents
have densities greater than water (>1 g/mL). Aromatic (toluene, benzene
- any compound which contains a benzene ring), aliphatic compounds
(hexanes), and ethers have densities less than one. They would float on
the top of the water layer. When in doubt, add a drop of water and
observe which layer it joins.

Sometimes we run into compounds which are soluble in organic as well as
aqueous solvents. Such compounds partition between layers of these
solvents placed in contact with each other. A measure of this
equilibrium is the partition or distribution coefficient, k, which is
defined for a compound (A), a pair of solvents (including water (W) and
a solvent (S)), and a certain temperature. It can be calculated as
follows (where C stands for concentration):

k = CA in S ( CA in W

This relation can be approximated with (where Sol is the solubility in
g/100mL):

k = SolA in S ( SolA in W

For example, compound A has a solubility of 6 g/ 100 mL in diethyl ether
and a solubility of 3 g/ 100 mL in water, therefore k is equal to

k = 6 g/100 mL ( 3 g/100 mL = 2

Let’s assume that you have prepared a solution of 3 g of A in 100 mL
of water and then extracted (washed) the water layer with 100 mL of
diethyl ether. How much material remains in the water layer? Let x be
the grams of A in the ether layer. Then:

k = 2 = x g /100 mL ether ( (3 - x) g /100 mL water

2(3 - x) = x

6 - 2x = x

6 = 3x

x = 2

3 - x = 3 - 2 = 1 g (which is the mass of A left in the water layer)

Now, let’s assume that the extraction was carried out with two 50 mL
portions of diethyl ether instead of a single 100 mL portion. The
calculation is in two steps:

k = 2 = x g /50 mL ether ( (3 - x) g /100 mL water

2(3 - x) = 2x

3 - x = x

3 = 2x

x = 1.5 g

3 - x = 3 - 1.5 = 1.5 g (which is lthe mass of A left in the water
layer)

For the next aliquot of 50 mL:

k = 2 = x g /50 mL ether ( (1.5 - x) g /100 mL water

2(1.5 - x) = 2x

1.5 - x = x

1.5 = 2x

x = 0.75 g

1.5 - x = 1.5 – 0.75 = 0.75 g (which is the mass of A left in the
water layer)

It is clear that multiple extractions are more efficient than single
extractions. When working up a reaction, organic chemists usually
extract/wash three times.

Another typical extraction is acid/base. A Brшnsted acid is a proton
(H+) donor; whereas a Brшnsted base is a proton (H+) acceptor. Let's
consider the separation of a strong organic acid, an organic base, and a
neutral organic compound. From general chemistry you know about strong
and weak mineral acids, but in organic chemistry we consider carboxylic
acids as strong organic acids. These carboxylic acids have pKas in the
range from 1.83-6.0. Remember there is a tenfold increase in acidity
for every pKa unit.



Extraction of a strong organic acid from a neutral organic compound
(Scheme 1)

The separation of benzoic acid from 1,2,4,5-tetramethylbenzene is
depicted in Scheme 1.

The first step is to dissolve the two compounds in an organic solvent
like dichloromethane (CH2Cl2). Diethyl ether is also used but the
densities of the two different organic solvents are different. The
densities of chlorinated hydrocarbons are greater than that of water (d
> 1 g/mL), whereas the densities of ethers are less than that of water
(d < 1 g/mL).

In the second step, a base, sodium hydroxide solution, is added to the
solution. Since benzoic acid is a "strong" organic acid, a weak base,
such as sodium bicarbonate, can also be used. To insure that the
organic acid is completely transferred into the aqueous layer, this
layer is tested using pH paper. A drop of the aqueous layer is dropped
onto a piece of pH paper until the paper tests positive for base. It is
important to mix the solution after each addition of base.

There are two layers formed upon addition of the basic solution to the
organic solution, because the organic solvent (CH2Cl2) is not miscible
in the basic aqueous layer. In step three, the different layers are
separated from one another. Since CH2Cl2 is denser than the aqueous
layer, this layer is removed and treated further in steps five and six.


In step four, the organic layer containing the compound
(1,2,4,5-tetramethylbenzene) is dried using calcium chloride pellets.
Any time an organic layer is washed or touches water, it must be dried
using a drying agent. The solvent is then removed with a pipette to
another tube and isolated after removal of CH2Cl2 via evaporation with
heat.

In step five, the sodium benzoate is neutralized and acidified
(protonated) to regenerate benzoic acid, which precipitates from the
aqueous solution. The pH of the solution is tested with pH paper.

In step six, the benzoic acid can be isolated by vacuum filtration.

Extraction of an organic base from a neutral organic compound (Scheme
2)

The separation of 4-chloroaniline from 1,2,4,5-tetramethylbenzene is
depicted in Scheme 2.

The first step is to dissolve the two compounds in an organic solvent
like dichloromethane (CH2Cl2). Diethyl ether is also used but the
densities of the two different organic solvents are different. The
densities of chlorinated hydrocarbons are greater than that of water (d
> 1), whereas the densities of ethers are less than that of water (d <
1).

In the second step, an acid, hydrochloric acid, is added to the
solution. To insure that the organic base is completely transferred
into the aqueous layer, this layer is tested using pH paper. A drop of
the aqueous layer is dropped onto a piece of pH paper until the paper
tests positive for acid. It is important to mix the solution after each
addition of acid.

There are two layers formed upon addition of the basic solution to the
organic solution, because the organic solvent (CH2Cl2) is not miscible
in the acidic aqueous layer. In step three, the different layers are
separated from one another. Since CH2Cl2 is higher in density than the
aqueous layer, this layer is removed and treated further in steps five
and six.

In step four, the organic layer containing the compound
(1,2,4,5-tetramethylbenzene) is dried using calcium chloride pellets.
The solvent is then removed with a pipette to another tube and isolated
after removal of CH2Cl2 via evaporation with heat.

In step five, the 4-chloroaniline hydrochloride is neutralized
(deprotonated) to regenerate organic base, which precipitates from the
aqueous solution. The pH of the solution is tested with pH paper.

In step six, the 4-chloroaniline can be isolated by vacuum filtration.

Experiment



A. Partition Coefficient of an Organic Acid

Weigh approximately 0.25 g of the unknown and place into a small
reaction tube. Dissolve the solid in exactly 3 mL of dichloromethane.
Add exactly 6 mL of water, cap the tube and shake. Let the mixture
stand until two layers form. Remove the aqueous layer. Add calcium
chloride pellets to the organic layer. Remove the dichloromethane with
a pipette and filter through a funnel plugged with cotton into a
pre-weighed 50 mL beaker including a boiling chip. Wash the pellets
with a full pipette of dichloromethane. Remove the dichloromethane with
a pipette and filter through the funnel plugged with cotton into the 50
mL beaker. Evaporate off the dichloromethane in the hood on a hot plate
under low heat. When the beaker has cooled to room temperature, weigh
the beaker.

B. Multiple Extractions of an Organic Acid

Weigh approximately 0.25 g of the unknown and transfer it into a small
reaction tube. Dissolve the solid in exactly 3 mL of dichloromethane.
Add exactly 2 mL of water, cap the tube and shake. Let the mixture stand
until two layers are formed. Remove the aqueous layer. Wash the
organic layer two more times with exactly 2 mL of water (each time).
After the final wash, dry the organic layer with calcium chloride
pellets. Remove the dichloromethane with a pipette and filter through a
funnel plugged with cotton into a pre-weighed 50 mL beaker including a
boiling chip. Wash the pellets with a full pipette of dichloromethane.
Remove the dichloromethane with a pipette and filter through the funnel
plugged with cotton into the 50 mL beaker. Evaporate off the
dichloromethane in the hood on a hot plate under low heat. When the
beaker has cooled to room temperature, weigh the beaker.

C. Extraction of an Organic Acid with a Basic Solution

Weigh approximately 0.25 g of the unknown and transfer the unknown into
a small reaction tube. Dissolve the solid in exactly 3 mL of
dichloromethane. Add exactly 2 mL of a saturated sodium bicarbonate
solution, cap the tube and shake. Let the mixture stand until two layers
are formed. Remove the aqueous layer. Wash the organic layer twice
more with exactly 2 mL of a saturated sodium bicarbonate solution.
After the final wash, dry the organic layer with calcium chloride
pellets. Remove the dichloromethane with a pipette and filter through a
funnel plugged with cotton into a pre-weighed 50 mL beaker including a
boiling chip. Wash the pellets with a full pipette of dichloromethane.
Remove the dichloromethane with a pipette and filter through the funnel
plugged with cotton into the 50 mL beaker. Evaporate off the
dichloromethane in the hood on a hot plate under low heat. When the
beaker has cooled to room temperature, weigh the beaker.

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Extraction