Acid/Base Extraction
Theory:
Extraction is one of the most frequently used separation techniques in
organic chemistry. An extraction essentially separates water-insoluble
organic compounds from water-soluble organic and inorganic
compounds. The organic compounds isolated may be solids, oils, or
liquids. The isolated material is a crude product that usually requires
further purification by recrystallization or another method. One almost
always performs an extraction at the end of a chemical reaction as a way
to separate the organic product(s) formed in the reaction from everything
else in the reaction mixture. You will be performing many more
extractions next semester when you begin to run more chemical
reactions.
_
One variation of an extraction is an acid/base extraction, which is a
very effective separation technique for organic compounds containing
specific functional groups: a carboxylic acid (RCO2H), an amine (RNH2),
or a phenol (RC6H4OH).
Before describing the theory behind an
acid/base extraction, we will first describe how a generic extraction
works.
The typical extraction is also known as a liquid-liquid extraction, and
as such involves a mixture of two immiscible liquids—two solvents that
do not dissolve in one another. The immiscible mixture you may be most
familiar with is oil and water; note how oil floats on water. In a liquidliquid extraction one of the solvents is always aqueous—water (either
neutral, acidic, basic, or salt), while the second solvent is an organic
solvent such as diethyl ether or dichloromethane, which is immiscible
with water. When the crude mixture and the two solvents are mixed
vigorously in a separatory funnel—a piece of glassware specifically
designed for this purpose, shown in figure 3-1, the water-insoluble
neutral organic components migrate toward the organic solvent, while
the water-soluble components dissolve in the aqueous solvent.
Figure 3-1
Separatory funnels
This is known as partitioning between the two layers. The physical
principle operating here is one we have seen before: "like dissolves like."
In other words, a substance will dissolve best in the solvent with the
most similar polarity characteristics to its own. As a general rule, water
will dissolve salts, including organic ions, and substances with strong
hydrogen bonding characteristics and very little hydrocarbon structure;
whereas, organic solvents will dissolve neutral organic substances with
weak intermolecular forces, i.e. compounds containing significant
hydrocarbon structure and few hydrogen bonding interactions relative to
hydrocarbon character.
Once the separatory funnel has been shaken vigorously, it is placed in a
ring clamp and the two immiscible solvents quickly separate into two
distinct layers, with the less dense solvent layer residing on top. Novices
often find it difficult to determine which layer is which. If you always
remember that the less dense layer is on top, this step will be easier.
You can look up the density of solvents in the Aldrich catalog or another
reference text; however, as a general rule of thumb, chlorinated organic
solvents are more dense than water, whereas n^~chloriiial£d^org^ul^
solvents are less dense than water.
One then simply separates the two layers by removing the stopper and
opening the stopcock at the bottom of the separatory funnel and allowing
each la}^er to carefully drain into two separate labeled Erlenmeyer flasks.
These layers are then referred to as the aqueous extract and the organic
extract.
Since partitioning of substances into their respective solvent layers is
never perfect, i.e. some organic material remains in the aqueous extract,
and visa versa, one typically returns the aqueous extract to the
separatory funnel and adds fresh organic solvent and repeats the
extraction process a couple more times. Calculations show that an
extraction is more effective when performed three times with 20 mL of
solvent than one time with 60 mL of solvent. This is the case anytime a
solute is distributed between two immiscible solvents. The amount of
solute in g/mL in each solvent phase is expressed quantitatively in terms
of the partition coefficient, K:
K = [solute A in solvent 1]/[solute A in solvent 2]
At equilibrium, at a given temperature, the ratio of concentrations
of A in the two phases will always be constant. For example, if the
partition coefficient, K, for a compound partitioned between diethyl ether
and water is 3.0; this means the compound preferentially partitions into
ether over water by a factor of 3. So if you have 12 g of compound being
partitioned between 60 mL of water and 60 mL of ether, the ether layer
will contain 9g of material and the water layer will contain 3 g of material
after one extraction. If you do the extraction 3 times with 20 mL of
solvent, then the 3g in the water layer after the first extraction will now
partion in a way to give K=3 again, so 2.25 g will be in the organic layer
and .75 g will be in the water layer (2.25/.75=3). Returning the water
layer to the separatory funnel and extracting for a third time with 20 mL
of fresh ether will cause the .75 g to partition between the layers so that
K=3, again. Thus, .56 g of compound is in the organic extract and . 19 g
is in the water layer (.56/. 19=3). Now when you combine the three
organic extracts you get 11.8 g (.56 + 2.25 + 9 g) of the original 12 g,
whereas if you did only one extraction you would recover only 9 g of
material. Notice also that with each additional extraction a smaller
percentage of the solute is obtained, so doing 4 or 5 extractions would
not increase your yield significantly, and would not be worth the extra
"effort. On the other hand the second and third extractions do contain
significant amounts of solute, and even more so the lower the value of K.
This calculation was somewhat simplified, since the partition coefficient
is probably somewhat smaller when the lesser volume of organic solvent
is used. It is for this reason that whenever an extraction is performed it
is done multiple times. The value of the partition coefficient depends on
the identity of the solute and the solvents. The final step of an extraction
concludes with removal of the solvent by evaporation or distillation to
yield the pure solute.
In the variation of an extraction known as an acid/base extraction the
same principles as those described above apply. The most important
difference is that the aqueous solvent- water, is not neutral; it is either
acidic or basic. For most organic functional groups this has no impact,
but for carboxylic acids, amines, and phenols, an acid base reaction
^^
of the mixture that reverses jj__ly
jtj_s_olubility ~
diaracjteristicsTThis is^e^nnu^FaTe^^yTHFacid/base reaction that
occurs with a carboxylic acid when extracted with aqueous base. The
reaction is shown below:
RCOOH + NaOH (aq)
water-insoluble
->
RCOO Na+ (aq) +
water-soluble
H2O
As the equation shows, most carboxylic acids are insoluble in water (the
exceptions are low molecular weight carboxylic acids), because they are
neutral and contain a fair amount of hydrocarbon character relative to
hydrogen bonding of the carboxylic acid functional group. However,
when base (hydroxide, carbonate, or bicarbonate) removes a proton from
the carboxylic acid, a carboxylate salt is produced, which is an organic
ion, and like any salt it will dissolve more readily in aqueous solvent than
organic solvent. So, now when the two layers are separated in the usual
fashion, the aqueous layer, not the organic layer, contains one of the
compounds of interest.
The carboxylate ion is then restored to its original carboxylic acid form by
another acid base reaction as shown below.
RCOO- Na+ (aq) +
water-soluble
HC1 (aq)
-> RCOOH + NaCl (aq)
water-insoluble
This acid base reaction is usually performed directly in the Erlenmeyer
flask. Since the carboxylic acid becomes neutral again, it also becomes
insoluble in water and precipitates from solution. Suction filtration can
then be used to recover the pure carboxylic acid crystals. Note,
bicarbonate and carbonate bases work equally well but also produce
carbon dioxide gas in the process.
A similar reaction occurs with phenols, except that since they are less
acidic than carboxylic acids, they can only be deprotonated with the
stronger hydroxide bases and not the carbonate and bicarbonate bases.
This can be advantageous when separating phenols from carboxylic
acids. Note that an acid/base reaction does not occur with ordinary
alcohols; they are not sufficiently acidic.
ArOH +
NaOH (aq)
-» ArO Na+ (aq) +
H2O
Compounds with an amine functional group can be converted into the
amine hydrochloride by treatment with aqueous HC1, and in an
analogous manner to the carboxylic acids, are converted from an organic
soluble substance to an organic insoluble salt that becomes soluble in
water as shown in the acid/base reaction below:
RNH2 + HC1 (aq)
water insoluble
->
RNH3+ Cl- (aq)
water soluble
Obviously, the difference from the carboxylic acid case is that addition of
acid not base achieves salt formation and the salt is a cation instead of
an anion. Either way it is a charge that causes it to be soluble in water.
The aqueous extract containing the amine hydrochloride is drained into
an Erlenmeyer flask, and is returned to the neutral amine form by
treatment with sodium hydroxide. Since this acid/base reaction causes
the organic substance to go from the water-soluble salt form into the
water-insoluble neutral form, it precipitates from solution. The reaction
is shown below. Once again, isolation is accomplished by a suction
filtration.
RNH3+
NaOH (aq)
—--»
water soluble
RNH-2
+
NaCl (aq) + H2O
water insoluble
In summary, in a standard extraction carboxylic acids, amines, phenols
and other organic compounds would all dissolve in the organic extract;
however, when an acid/base extraction is performed, one can selectively
remove the phenol with sodium hydroxide, the carboxylic acid with
bicarbonate or sodium hydroxide, and the amine with HC1. All other
organic substances remain in the organic extract. Incidentally, all
inorganic substances are removed in the aqueous extract. This a
powerful and simple technique for isolating certain organic compounds.
Experimental
Acid Base Extraction of a Mixture of Benzoic Acid, 4-nitroaniline,
and Naphthalene.
You will be provided with an equal mixture of benzoic acid, a carboxylic
acid; 4-nitroaniline, an amine; and naphthalene, a neutral organic
compound. Using the technique of acid/base extraction your goal is to
separate these three compounds and isolate them each in their pure
form. Calculate a total % recovery and perform a melting point
determination on each of the three solids to analyse the effectiveness of
your separation.
O2N
Benzoic Acid
4-Nitroaniline
Naphthalene
Begin by dissolving your mixture in 50 mL of dichloromethane and
transfer the solution to a 125 mL separatory funnel (remember to close
the stopcock first and double check that it doesn't leak). Next you will
extract the organic solvent three times with 6M HC1. To do this add 25
mL of 6M HC1 to the separatory funnel. Place the stopcock on the
separatoiy funnel and shake vigorously with frequent venting. Your
instructor will show you how to do this properly. It is very important
that you always remember to vent a separatory funnel frequently to avoid
solvent vapors from building-up and blowing the stopcock off. It is also a
good idea to wear gloves when handling a separatory funnel, in the event
of any spillage. Place the separatory funnel in its ring stand and let the
two layers separate. Open the stopcock slowly and allow each layer to
drain into two separate labeled 250 mL Erlenmeyer flasks, being
especially observant that you make the cut between the layers at the
right moment. Return the organic layer to the separator/ funnel and
repeat the process two more times, each time with 25 mL of fresh 6M
HC1. You may combine the aqueous extracts when draining the
separatory funnel each time. Label the flask "HC1 extract," and set it
aside. This flask contains the aqueous 4-nitroaniline hydrochloride salt.
Return the organic extract to the separatory funnel and extract it three
times with 3M NaOH. This should be done in the same way as described
above but using 25 mL of 3M NaOH each time. Set aside the combined
aqueous extracts and label it "NaOH extract." This flask contains the
aqueous sodium benzoate salt.
The remaining organic extract now contains only naphthalene, because
this substance is unchanged in both the HC1 and NaOH washes;
remaining neutral at all times. In order to isolate the pure naphthalene,
one first typically removes any residual water that is present as a result
of the numerous aqueous extractions. This is done by washing the
organic extract with an aqueous saturated sodium chloride solution. The
salt draws the water out of the organic layer and into the aqueous. To do
this, return the organic extract to the separatory funnel and add about
25 mL of saturated aqueous sodium chloride solution. Shake the
separatory funnel and separate the layers in the usual fashion. Add
approximately .5 g (estimate do not weigh) of sodium sulfate, a drying
agent, to the organic extract to remove the last traces of water. After
about 5 minutes, gravity filter the drying agent from the organic
solution.
A gravity filtration is set up as shown in figure 3-2. Place a powder
funnel in a clean dry tared Erlenmeyer flask and place a fluted filter
paper in the funnel so that it fits the entire funnel.
_Ring
Fluted filter paper
support
Shortstemmed
funnel
or a powder
funnel
Erlenmeyer
flask
Figure 3-2
Your instructor will show you how to flute a piece of filter paper.
Now
carefully pour the organic solution containing the drying agent into the
filter paper and allow it to drain through the paper by gravity, to remove
the drying agent.
Place the flask containing the filtrate (naphthalene and
dichloromethane) in the hood to evaporate the dichloromethane solvent.
When you return the following week your flask should contain pure
naphthalene. This is a good stopping point.
Be sure to label your flasks
with your name and place them in the "work in progress hood."
Weigh the flask and determine the mass of recovered naphthalene.
Scrape the naphthalene out of the flask with a spatula and transfer it to
a labeled vial and turn it in to your instructor.
Save a small amount for
a melting point determination.
Place the "HC1 extract" in an ice-cooled water bath. Then carefully add
3M aqueous NaOH until the solution is basic. Test the pH of the solution
by placing a drop of solution on pH indicator paper; do not dip the
indicator paper into the solution. -You should observe crystals
precipitating once your solution becomes neutral. These crystals are 4nitroaniline, which can be isolated by suction filtration. Remember to
rinse the crystals in the Buchner funnel with a minimum amount of cold
water and allow the solid to air dry for several minutes before
transferring to a beaker for further drying. You may store the solid in
the dessicator to dry.
Place the "NaOH extract" in an ice-cooled water bath and in a similar
manner add 6M HC1 to the solution until it is acidic (use pH indicator
paper). You should see benzoic acid crystals forming as you reach
neutrality. Isolate the solid by suction filtration in the same manner as
performed with the amine.
Obtain melting points on all three products and record these in your
notebook.
Since there is often some cross contamination in even the
best of extractions, the purity of these solids will not be as good as you
would observe in a recrystallization; therefore, you may see some melting
point depression, nevertheless you should be able to assess whether you
were able to successfully separate the three organic compounds from one
another. Describe the physical appearance of these solids in your
notebook. Turn all three samples into your instructor in vials labeled
with both your name and the identity of the sample.
Points will be
deducted for unlabeled samples. Calculate a total % recovery for the
combined samples.