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MCAT
Course: MCAT > Unit 9
Lesson 7: Separations and purifications- Separations and purifications questions
- Simple and fractional distillations
- Extractions
- Principles of chromatography
- Basics of chromatography
- Thin layer chromatography (TLC)
- Calculating retention factors for TLC
- Column chromatography
- Gas chromatography
- Gel electrophoresis
- Resolution of enantiomers
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Extractions
Learn about how chemicals can be separated through acid-base extraction. By Angela Guerrero. . Created by Angela Guerrero.
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- Around, would the aqueous layer always be on the bottom? I understand that a lower density of the organic layer means it floats on top, but are there common extractions where the aqueous layer is less dense? Thanks! 2:00(12 votes)
- Water is pretty dense (1 g/cm3) relative to most commonly used organic solvents. It would be a good bet on an MCAT question where density is uncertain.(13 votes)
- The end of the video makes me laugh... the real answer... you start your reaction over again to make more product!(18 votes)
- how do you do a liquid solid extraction?(5 votes)
- you would use filtration for that, like sand in a water and sand mixture.(7 votes)
- Atwhen the first extraction is done. What happens to the Chloride? Which lay is it in? 4:18(5 votes)
- @, wouldn't the acetic acid already be in the aqueous phase? I understand what you mean by the principle, but could you clarify? 7:00(3 votes)
- Good point! Actually, acetic acid is miscible with (infinitely soluble with) water, so if you were to add neutral water to that three-compound mixture, you would be able to extract the acetic acid in the resulting aqueous layer. This is because the acetic acid would, being an acid, self dissociate in water to become acetate; using a base would not be necessary but would likely increase the amount of acetic acid removed in each extraction :)(5 votes)
- How would this flow chart change if you were doing a wash instead of an extraction?(2 votes)
- When we add a base to deprotonate the acid in the mixture and make an ion which would go down to the aqueous layer, what happens to the added base? Where does it go?(1 vote)
- The added base would still be dissolved in the aqueous solution, but with new associations between the base and the acid. Remember the products of acid / base neutralisation are water and a salt, both of which would be found in the aqueous layer. The salt would be dissolved and form associated ions.(3 votes)
- Please explain how for the final extraction, NaOH can be used to separate the first and second compounds, but is too strong to be used to separate the AA from the the other two. Are they not all going to mix together?(2 votes)
- I think the idea is to separate the three layers into three separate flasks, and using NaOH at the start would just result in a mixture of acetic acid and phenol. By using two base strengths, you can have specific flasks for each compound.(1 vote)
- When we are doing solvent extractions, we know what the components of our mixture is, right? (Please correct me if I am wrong.) If we don't know what our components, for example if they give you a random solution and tell you to find out the components, can we apply this extraction process?(1 vote)
- You can use this process to separate a mixture into its components and eventually identify the compound, but it would not be your primary method of identification. However, if you were given a pure solution of a compound you would not use this as the method to solely identify the compound.(3 votes)
- At, in the molecular drawing of phenol, what does the circle in the middle of cyclohexane represent? 4:33(1 vote)
- The circle is just another way of showing the resonance structure, and acts the same as if there were 3 double bonds drawn instead.(2 votes)
Video transcript
Walking into a
department store, you're often hit by a variety
of smells coming from the perfume department. How do they make all
these different scents? Well, one way they do
that is by separating out chemical compounds and
mixing them in different ways to get their own
brand-specific formula. But in order to isolate
the chemical compounds, one of the techniques
that's frequently used is known as extraction. Extraction can be done in
your organic chemistry lab using these two main
pieces of glassware. The first which is
shown in blue is known as the separatory funnel,
or the sep funnel for short. At the top of this,
this has an opening through which you
can pour in liquid. At the bottom, you have
the stopcock, currently shown in the closed
position, which prevents liquid from flowing
freely from the sep funnel and into the flask,
shown here in pink. This happens after you shake
the sep funnel and allow the mixture to settle. When you pour in a mixture
of liquids, what you'll often notice is that you get
two different layers, one on the bottom and one on top. This happens after you shake
the separatory funnel and allow the mixture to settle down. But what does this mean? This means that these
liquids or solvents have different densities,
with a higher density being on the bottom and the lower
density floating on top. Often, these will
represent the organic phase and the aqueous phase. The aqueous phase contains
the water and other charged species, or ions, whereas
the organic phase contains uncharged species or
neutral compounds. So how exactly do
we do an extraction? Well, first of all, you
need to open the stopcock. And when you do that, you'll
see that although you still have your organic phase up
here, the aqueous phase is now able to flow down the sep
funnel and into the flask. And once you've collected
all of your aqueous layer in the flask, you can
close the stopcock again, which will leave you with just
the organic phase in the sep funnel. Now that we know how to do this
in lab, let's look at compounds we'd be able to separate. For example, take
hexane and propanamine. What do we know about
these two compounds? Well, hexane is pretty
neutral and propanamine is basic, because
of its amino group. So in order to get one of these
into the organic layer and one of them into the
aqueous layer, we'd want to make sure that
we can produce an ion and leave one other
compound uncharged. To do that, we can add an acid. By adding something
like hydrochloric acid, what you'll find is that this
amino group, the nitrogen, these electrons can
deprotonate the HCl and leave behind just
a chloride anion, which means that in your
aqueous layer you would have a
charged amino group, whereas in your organic layer
you would just have hexane. There, you've done
your first extraction. So next, in this example,
you have hexane and phenol. Phenol is weakly acidic. This means that in order to get
phenol into the aqueous layer, you would want to add to a base. We can use a base
like sodium hydroxide, which is a strong base. And what you'd find is
that these electrons would be able to
deprotonate the phenol, meaning that in
the aqueous layer this time you would be left
with the phenolate anion. And in the organic phase, you
would just have the hexane. Let's do something a little
bit more complicated. What if instead you had three
compounds this time, hexane, phenol, and acetic acid? Well, in this case, if
we took the approach we took previously,
which was trying to add a base to get these
acids into the aqueous layer. If we tried adding
NaOH, we'd find that since this
is a strong base, you would easily deprotonate not
only the acetic acid but even the phenol. So let's try using
something else instead. If instead we had a somewhat
milder base like sodium bicarbonate-- and I'll
show its structure here-- you have this anion
with a sodium cation. And since this is
a mild base, it won't be able to
deprotonate very weak acids, meaning it won't be able
to deprotonate the phenol, but it will be able to
deprotonate the acetic acid. This means that in your
aqueous layer this time, you would have
the acetate anion, while in your organic
layer, you have hexane and you have phenol. How do we separate
these last two? Well, we've already done
that in the previous example. So to get just one compound
into the aqueous layer, again we would want to add
a base, but this time we can use sodium hydroxide. The hydroxide, these
electrons here, would deprotonate the
phenol, giving you again the phenolate anion,
along with the hexane in your organic layer. And now we've learned
how to do extractions. So when you're actually
doing this in the lab, you'd want to make sure that
you save each and every layer. What happens if you accidentally
toss one out into the waste container? Well actually, using a
multi-step extraction, you'd be able to recover
your product from the waste container.