Professor Dave here, let’s talk about extraction.
If you have a mixture of some kind, sometimes
it can be very easy to separate the components.
With a mixture of water and sand, you can
just use some filter paper, and the smaller
water molecules will pass through the tiny
pores of the paper, while the larger sand
particles will not.
But what if we have a mixture with several
substances that are all small molecules?
Particle size will not be of any help here.
Instead we can make use of certain physical
properties, like solubility, or even chemical
properties, like some specific type of reactivity.
A technique we can use to do this is called
extraction, so let’s learn how to do this now.
Let’s say we have a mixture of sodium chloride
These are both white, crystalline solids,
so if mixed together, it may seem impossible
to get them separated again.
But in fact, we can use their physical properties
to separate them with ease.
Sodium chloride is an ionic compound that
is water soluble.
Cholesterol, on the other hand, is a steroid,
which is a type of lipid, and is therefore
nonpolar, and largely water insoluble.
However, cholesterol will easily dissolve
in a nonpolar solvent, like ether.
So to separate this mixture, we would just
toss in some water, and then some ether.
We swirl it around for a while, and then put
everything in something called a separatory funnel.
Since these two solvents are immiscible, meaning
they do not mix, we will get two layers.
On the bottom will be an aqueous layer, which
contains all the water and all the sodium
chloride, which has dissolved and dissociated
in the water.
On top will be an organic layer, which contains
all the ether, and all the cholesterol, which
has dissolved in the ether.
The water is on the bottom because it is more
dense than the ether.
So we just open up the funnel and drain the
aqueous layer until we just barely get to
the organic layer, and then we pour that out
the top of the funnel into something else.
Evaporate the solvents, and there you have
your substances, nice and separated.
That one was pretty easy because the substances
had differing solubilities.
Now let’s look at a trickier example.
Say we have a solid mixture of 4-chloroaniline,
benzoic acid, and 1,4-dibromobenzene.
These are structurally similar, and therefore
have very similar solubilities, so we can’t
just use two different solvents and expect
much to happen.
Instead, we have to do some chemistry.
Let’s toss the mixture in a flask and dissolve
it in ether, and then transfer over to our
Now let’s think, is there some type of reaction
that only one of these compounds might undergo
that the others would not?
Well 4-chloroaniline is somewhat basic, because
of this amino group.
Benzoic acid, as the name suggests, is acidic.
And dibromobenzene is neither.
So what if we do some acid-base reactions?
Starting with the aniline, this will react
with a strong acid, like hydrochloric acid,
while neither of the other two compounds will
react, since they are not at all basic.
So let’s toss in some of an aqueous HCl solution.
Since this is aqueous, it will be more dense
than the ether, and it will pass right through
to collect on the bottom, forming an aqueous layer.
But as it passes through, it will perform
acid-base reactions, transferring protons
to all the aniline molecules.
Once aniline is protonated, the nitrogen atom
will bear a formal positive charge.
And whereas the neutral aniline was not particularly
water soluble, this aniline salt will be,
because the formal charge can make ion-dipole
interactions with water that are favorable.
So it’s as though the acid yanks the aniline
into the aqueous layer as it passes through.
Just to make sure everything reacts properly,
we will put a stopper over the funnel, pick
it up, and shake it around a bit.
This will generate some pressure inside the
funnel, so we will have to periodically vent
by holding it upside down with a finger on
the stopper, and turning the stopcock to release
the gas that builds up.
This will produce an audible sound.
Then close the stopcock and shake some more,
and then vent again.
You can keep doing this until the venting
doesn’t make any more noise, then you know
Then just drain the aqueous layer into a new,
To be thorough, we will typically wash the
organic layer again with another portion of
the acid, and repeat the whole process, and
then maybe even one more time with just deionized
water, collecting all three aqueous extracts
in the same flask.
Now we’ve got our aniline salt sitting in
water, and we can put it aside for a moment.
So what’s left in the ether?
There is benzoic acid, and there is dibromobenzene.
As you might guess, the next step will be
to add some base, since the acid will react,
while the other compound will not, since it
is not remotely acidic.
So we can add some aqueous sodium bicarbonate,
which will pass through the organic layer
to form an aqueous layer at the bottom, but
as it goes it will accept a proton from benzoic acid.
This leaves the benzoate ion, which will be
very water soluble, since the negative charge
can interact with water molecules in solution.
So just like the previous step, the base drags
benzoic acid into the aqueous layer.
We perform the same actions as before, shaking,
venting, collecting the aqueous layer in another
labeled flask, washing again with base, and
then again with just water.
And now we have our benzoic acid salt sitting
in water, which we can put aside.
So what’s left in the separatory funnel?
That will just be the dibromobenzene in ether,
and we can pour that out the top to avoid
contamination, into yet another new labeled
Wash the empty funnel with a little ether
to make sure we get it all, and add that to
Now toss in a little bit of a drying agent
like sodium sulfate to suck up any water that
may have made it in there.
Then use a funnel with some filter paper to
transfer the solution into yet another flask,
this time pre-weighed if you intend to measure
the mass of product.
So now we have successfully separated our
mixture into its three components.
So how do we get the solids back?
Well the dibromobenzene is easy.
That’s just sitting in ether, and ether
is quite volatile, so let’s just put it
on a hot plate and use a low setting to gently
evaporate the ether away.
The dibromobenzene will remain as a residue,
and you can weigh the flask again to get its mass.
The other two substances will be a little
tricky, because they are now salts, so we
have to do a little more chemistry.
The aniline salt needs to be deprotonated,
so we need a strong base, like sodium hydroxide.
Let’s add some of it to the flask, immerse
it in a cold water bath, and watch the 4-chloroaniline
We can be sure that this is finished by checking the pH.
Once all the salt has reacted then no more
hydroxide can react, and it will thus remain
in solution, and the pH should reflect that,
reaching around 11 or 12.
Now that we have our solid sitting at the
bottom, we can just set up a Buchner funnel,
hook it up to a vacuum pump, pour the solution
onto the filter paper sitting in the funnel
and wait for it to dry.
Then we can do the same thing with the benzoic
acid salt, except here we need a strong acid
to protonate it.
So to this flask, let’s add some HCl, watch
benzoic acid precipitate, check for a pH of
around 2 to ensure that this is complete,
and then collect that in a Buchner funnel
And there you have your three solids, completely separated.
So hopefully we now understand what extraction
is and how it is performed.
This technique is ubiquitous in the organic
Almost every single time a reaction is performed,
it will end with some kind aqueous workup,
to neutralize any ionic products, after which
the contents of the reaction flask will be
transferred to a separatory funnel for extraction.
So if you are interested in doing a lot of
organic chemistry, make good friends with
your separatory funnel now.
And with this technique understood, let’s
move forward and learn about some others.