In the last video we looked at Strecker synthesis for creating an alpha amino acid.
We started at an aldehyde, added the alpha amine and alpha carboxy.
In this video we're going to look at the Gabriel Malonic Ester Synthesis which is another method
for synthesizing an alpha amino acid but this time we're starting with the backbone and
adding first the nitrogen then the R-group.
As the name implies, Gabriel Malonic Ester Synthesis is a combination of the Gabriel
Amine Synthesis followed by malonic ester synthesis and together, we're going to form
that amino acid.
Let's do a quick overview and then we'll dive into the step-by-step reaction and mechanism.
The reaction starts out with a malonic ester that has a leaving group at the alpha position.
We need this leaving group so that we can add a nitrogen through an SN2 reaction.
Gabriel Amine synthesis is an ideal way for forming a primary or secondary amine while
protecting the nitrogen every step of the way so that we don't have multiple alkylation.
Gabriel Amine synthesis uses nitrogen in the form of potassium phthalimide.
With a negative nitrogen that�s protected and its reactivity slowed down so that we
only have one addition and it will attack the alpha carbon through an SN2 reaction,
giving us N-phthalimido malonic ester, where N is the nitrogen phthalimido, refers to the
phthalimide group, and malonic ester refers to this entire group.
This gives us the nitrogen that'll ultimately be the alpha nitrogen of our amino acid, but
it's protected so that the nitrogen will not react in the next step when we first add a
base and then an R with a leaving group.
The R used in step 2 is specifically the variable group of the amino acid, this is how we add
the amino acid side chain to the reaction.
The base will activate the alpha carbon on the ester allowing an SN2 reaction to add
the R-group, and now that we're done adding all of the groups we'll start kicking off
protecting groups one at a time.
Carried out in acid or base, hydrolysis will turn the ester into a carboxylic acid and
unprotect the nitrogen giving us the alpha amine.
And in the final step, we add heat for decarboxylation kicking out that second carboxylic acid and
ultimately giving us our alpha amino acid product.
Biological systems prefer to have L-amino acids but in this reaction since the alpha
amino acid can attack the R-group from the top or the bottom, we're going to get a racemic
mixture of D and L amino acids.
The important thing in this reaction is to recognize where all of your components come
from.
Let's look at our final product and break it down.
We have the alpha carbon, this is that central carbon in the amino acid attached to four
groups.
The alpha hydrogen, the variable side chain, this is what gives us the 20 common amino
acids.
The alpha amine here, protonated because we're at a low pH and therefore acidic conditions.
And the alpha carboxy which is also protonated at these acidic conditions.
Where do each of these atoms come from?
The alpha carbon in the malonic ester is the one that starts out most reactive and winds
up the alpha carbon in the final product.
This is where we're slowly adding the groups to.
We use malonic ester instead of malonic acid because a carboxylic acid would not undergo
the malonic ester reaction.
As you can see in the malonic ester video reference below.
So we start out with the ester and later convert it back to the acid.
The alpha nitrogen comes from the potassium phthalimide, but it comes protected until
we know there will be no further reactions so that we don't wind up with random side
products.
And finally the alpha hydrogen comes from the last step when the reaction undergoes
decarboxylation and then keto-enol tautomerization which is part of the carboxylation, we're
going to get a hydrogen take the position where the second carboxylic acid used to be,
and this gives us our final product.
Now let's take a look at key components of this mechanism, keeping in mind that for the
MCAT you simply want to understand why and how each step happens but you don't have to
worry about being able to repeat and write it out step-by-step.
If you�re required to know this for your class, obviously go through and draw it out
again and again until you have it memorized.
The reaction starts out with the Gabriel portion of the Gabriel Malonic Mechanism, but we still
need to have that malonic ester, in this case we're using ethyl as the ester R-group and
we're starting out with a bromine at the alpha position.
This makes the alpha carbon very partially positive and therefore susceptible to attack,
and that�s exactly what we�ll see in this step.
Potassium phthalamide has potassium as a counter-ion.
It's K+ it doesn't do anything except balance nitrogen's charge.
It's the negative nitrogen that we're interested in.
One electron pair is used for resonance, the other one is capable of doing a backside attack
at the alpha position, kicking out the bromine and forming a new bond between nitrogen and
carbon.
For the simplicity, I added the nitrogen and drew the rest of the group in tiny so that
it doesn't get in the way of the mechanism itself.
In fact, the molecule likes the fact that it's big and bulky because this will make
sure we don't have a second attack and it'll help control the direction of the reaction.
With the nitrogen in place, the next step is to add the R-group and this is where we
transition into the malonic ester mechanism.
The first step in Malonic Ester Synthesis is deprotonating the alpha position to get
a reactive negative carbon.
To do this we're going to use an OR-group ethoxide Eto- because we want the same R as
we have on the ester itself.
Ethoxide is a strong base, will grab that alpha proton, collapsing the electrons back
onto carbon.
I'm showing the electrons on the alpha carbon, but don't forget that we can have resonance
to the right and left, which is the secret behind the stability of this molecule.
In addition to resonating, the electrons can also attack, which is exactly what we'll see
in the next step.
In the next step we add the R-group or the amino side chain.
For this reaction, we're going to use a chlorine bound to a two carbon chain, bound to an SCH3.
If chlorine is the leaving group for this SN2 reaction, recognize that the rest of it
is the R-group for methionine, which is -CH2CH2SCH3.
The alpha carbon electrons will attack carbon, kicking out chlorine, the leaving group, giving
us the methionine side chain attached to the alpha carbon.
At this point we have everything that we need.
We have the alpha carbon, we have the carbonyl which will be the alpha carboxy but it's an
ester, we have to change that.
We have the nitrogen but again, it has too much on it, we have to change that and we
have the R-group.
And let's not forget we have this instead of the alpha hydrogen so we have to find way
to get this on as well.
How do we get rid of all these protecting groups?
Base or acid catalyzed hydrolysis.
For this reaction we'll focus on the acid catalyzed hydrolysis which is carried out
using acid dissolved in water for H3O+.
When you dissolve reactive groups in water, the water will cut the groups in half, for
example an ester will get cut so that the carbonyl gets part of the water OH and the
OR gets the other hydrogen.
This is hydro-lysis, add water and lyse.
The same thing happens with the phthalimide group because if you think about it, the reactive
portion on the phthalimide simply looks like two amides fused together where we have a
carbonyl, a nitrogen, another carbonyl, and R, which is just the rest of the chain.
Acid catalyzed hydrolysis will cut between the carbonyl and the nitrogen, giving us an
OH carboxylic acid and NH.
This will also happen on the other side so once again we have OH carboxylic acid and
NH.
I'm not gonna go through either mechanism right now because they're tedious and I cover
them in the carboxylic acid derivative reactions.
Instead we'll simply apply the trick and see what happened.
Each ester is turned into a carboxylic acid, nitrogen is now not just an NH2, it's NH3+
because at low pH it will be protonated giving us a positive nitrogen, and let's not forget
our side chain.
We are nearly where we want to be.
In the next step, when we add heat, the molecule will rearrange itself to give us a decarboxylation.
To show you this reaction we have to get creative and flip this carboxylic acid so the OH and
the carbonyl switch positions.
I want you to pay particular attention to the colors because that way you'll be able
to follow where every electron and bond moves to and winds up.
This reaction happens through a concerted mechanism where all of the electrons move
simultaneously in a cyclic direction.
The electronegative oxygen will reach for the hydrogen atom forming a new bond but hydrogen
can only have one bond so its electrons instead of collapsing onto the green oxygen, will
move down and form a pi bond between oxygen and the carbonyl carbon.
Carbon can only have four bonds so it kicks out that sigma bond between itself and the
alpha carbon and moves it over to form a pi bond between the alpha carbon and the other
carbonyl carbon.
But that other carbon can only have four bonds, it has five, so the pi electrons between itself
and oxygen will be kicked up onto the oxygen atom.
Let's redraw this skeleton exactly as we see the atoms and then fill in the bonds.
Notice the atoms haven't moved but now let's see what happened to all of the electrons.
Oxygen had two red electrons pairs, one is still sitting as a lone pair on oxygen, the
second is sitting as a single bond between oxygen and hydrogen.
The blue electrons between oxygen and hydrogen are now sitting as a pi bond between carbon
and oxygen, and oxygen hasn't lost any of its lone pairs.
The bond that used to bind the carbonyl to the alpha carbon is now sitting on the other
side as a pi bond between the alpha carbon and the former carbonyl carbon on the right.
And the former pi bond on the right carbonyl, is now sitting as a lone pair of electrons
on oxygen.
This is what we want to see but it's not very pretty so let's play with it some more.
Notice that this molecule here completely broke away from the rest and has a central
sp hybridized carbon atom.
That means it shouldn't be bent like that, it should be redrawn as a linear form, and
notice this is CO2 gas which bubbles out of the solution.
What do we have on the right?
Let's start with the alpha carbon, bound to a protonated alpha amine, and the methionine
side chain.
The alpha carbon is double bound to another carbon atom, which is bound to two hydroxy
groups.
Does this look like an amino acid to you?
I didn't think so.
But recognize that this portion right here is an enol, and enols can undergo keto-enol
tautomerization which is exactly what happens here.
Carbon is still bound to two oxygen atoms but one is double bound as a carbonyl and
the second is the OH of the carboxylic acid.
And we have a hydrogen bound to the alpha carbon as the alpha hydrogen of our amino
acid.
Let's make sure everything is there.
Alpha carbon bound to a protonated alpha amine, alpha carboxy, alpha hydrogen, and the thianine
side chain.
Be sure to join me in the next video where we look at amino acid hydrolysis and dehydration
peptide synthesis.
You can find it along with this entire amino acid video series, practice quiz and cheat
sheet on my website leah4sci.com/aminoacids.
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of tutorials and that I should keep making more.