Leah here from Leah4Sci.com.
And in this video we'll continue our SN2 discussion by analyzing a number of SN2 reactions and
their mechanisms.
We'll start with the chiral 2-Iodobutane reacting with NaSCH3 in the presence of Acetone.
But first let's determine that an SN2 reaction can take place by analyzing the alkyl chain,
leaving group, attacking nucleophile and solvent.
You can find detailed videos showing you what to look for in each of this scenarios by visiting
my website: HTTP://LEAH4SCI.COM/SUBSTITUTION-ELIMINATION So let's begin.
We have a leaving group sitting on a secondary carbon which tells me that in the case of
an SN1 reaction, I can have a stable carbocation but in the case of an SN2 reaction I don't
have enough steric hindrance to prevent a nucleophile from attacking directly and this
tells me that I can have a one-type and a two-type reaction take place based on the
alkyl chain.
Iodine is a decent leaving group so we can have a one-type and a two-type reaction.
Our attacking molecules where it gets a little tricky, recorded in an SN1 reaction, you typically
have a weak nucleophile and because it is not strong enough to attack, it waits for
the leaving group to leave and then sweeps in on the carbocation.
In this case it appears that we have a neutral or weak attacking molecule but that's actually
not the case.
Anytime you see Na or some other group on metal at the start of your molecule recognize
it as a positive spectator ion secretly hiding a negative charge atom or molecule.
And in this case, we have a negative Sulfur which is charged and therefore strong enough
to initiate an attack meaning a two-type reaction.
But we will take this further than just a blanket to two which can include S into our
E2.
Sulfur, which sits directly under Oxygen on the periodic table, is a larger atom with
its charge distributed.
This makes it a weak base, because the charge isn't as concentrated or strong.
However, because the sulfur is so large, the atom is polarizable and when you can polarize
your attacking molecule it tends to form a better nucleophile.
So, in addition to writing 2, we'll say that SN2 is favored over something like E2 meaning
it is pointing towards an SN2 reaction.
And lastly, we'll analyze our solvent which is acetone or simply propanone, notice that
it is a polar aprotic solvent.
Polar because we have the carbonyl oxygen hugging electron density towards itself and
a partially positive carbonyl carbon.
But the Hydrogens on this molecule are located on Carbons and therefore are not polar enough
to participate in Hydrogen Bonding.
Since acetone doesn't have the capability of Hydrogen bonding or nucleophile, it tends
to be more reactive in this solution and therefore more likely to attack.
And Polar aprotic solvents tend to favor the SN2 reaction.
And so the clues justify that an SN2 reaction can take place.
Let's redraw the starting molecule and show the reaction.
I note to twist starting molecule before looking at the reaction.
Because I want to emphasize what happens to the leaving group when the molecule gets attacked.
Iodine is an electronegative atom and even though it's a covalently bound to Carbon,
will pull electron density towards itself giving it a partial negative charge.
This makes the attached Carbon partially positive.
We'll show the nucleophilic portion of our attacking molecule which is a CH3-S with a
negative charge and this negative Sulfur will polarize all its negativity towards the direction
of the Carbon atom to the point where it's actually attracted enough to form a bond between
itself and Carbon.
When Sulfur attacks, it creates a fifth bond to Carbon, but Carbon can only have four bonds
so Iodine is forced to take its bonding electrons and break away from the molecule.
Now, the key part of this reaction is paying attention to what happens in the intermediate
step.
We'll show the invisible Hydrogen to make the intermediate make a little more sense.
We'll show the central Carbon, so you get a better understanding of what goes on with
the hybridization and the way the atoms are attached.
At the top of the molecule we have the CH3 group.
It starts out facing left but as the Sulfur comes in and attacks, it gets pushed slightly
to the right and so the CH3 group in the transition state is pointing straight up.
The Hydrogen which was down and towards the back is still going to stay where it is.
And the two Carbon chain towards the left is going to come slightly forward.
The CH3S is partially attached during the transition state and at the same time the
Iodine leaving group has partly broken away.
So noticed that we had a molecule that started out chiral, sort of straighten out in the
middle, meaning the Carbon has the substituents facing straight up and down but you have the
leaving group and the nucleophile attached so this means that your chirality end product
is going to be controlled in terms of how it winds up.
We put brackets around the transition state because this is not something you have to
show in working through the reaction.
Instead I want you to understand what's happening.
And finally, to complete this reaction, the bond between Carbon and Sulfur will get stronger,
the bond between Iodine and Carbon will completely break and as this happens, everything attached
to Carbon will be slightly pushed over to the other side to make room for that incoming
nucleophile.
Let's quickly verify the stereochemistry of the starting and the ending molecule.
In the starting molecule, Iodine is my highest priority, then Ethyl, then Methyl and I already
have Hydrogen towards the back.
Tracing from 1 to 2 to 3, I get an R configuration from my starting molecule.
I did the same thing to my product with the SCH3 as priority one, Ethyl is 2, Methyl 3,
Hydrogen 4 is already in the back so ignore that.
And once again, we trace from one to two to three, in this case going counterclockwise
and giving me the S configuration.
As you can see, we started out with a pure R enantiomer and after undergoing an SN2 reaction,
we round up with a pure S enantiomer given that an SN2 reaction involves an inversion
of configuration.
The final thing we want to verify in any reaction is the conservation of charge.
We showed the starting molecules NaSCH3 which was neutral.
Here we're showing it as S minus, it doesn't matter which one you use as long as you're
consistent.
If we use just the S minus, then we can say the products have a net charge of negative
one.
The product appears to be neutral but do not forget that we had an i minus break off in
solution, that I minus has a negative charge giving the product a net charge of negative
one.
Now that we understand how the SN2 reaction takes place, I want to show you a quicker
way to draw this because the transition I used above is kind of tricky and will get
messy in your examination paper.
Given that these are our reacting molecules, we can show a lone of pair of electrons on
Sulfur attacked from the opposite side of Iodine.
But understand that because Iodine is drawn with the wedge, it is coming out of the paper.
And even though we're showing Sulfur attacked from the top it's attacking from the opposite
side of Iodine, in other words, it's attacking from beneath the molecule or beneath the paper.
When we show the product, we keep everything the same except for the leaving group and
the incoming group which switch places.
To show that Sulfur is beneath the paper, we draw the bond between Carbon and Sulfur
as dashes and then simply add SCH3.
On paper, it looks like the chirality inverted but let's quickly prove that by finding the
stereochemistry of the reactant and product.
The product appears to be R but because Hydrogen is forward and group number 1 is in the back,
R actually becomes S because we envision this backwards and low and behold, an R molecule
undergoes an SN2 reaction and gives me an S product.
This example will analyze what happens when a primary Chlorine on a substituted cyclopentane
is attacked by a strange molecule LICCCH3.
Let's prepare our checklist and see what we can make of this set of conditions.
Our alkyl chain has a leaving group on a primary Carbon which means we cannot form a stable
carbocation.
And since the SN1 and E1 reaction undergo carbocation intermediate, we're ruling out
the one-type leaving us with the two-type reactions, SN2 and E2.
Chlorine is an okay leaving group which means that it can be kicked out in a reaction and
therefore, a substitution or elimination can take place.
And we have our strange attacking molecule, at first glance it does not appear to be a
molecule that I recognized, but if you've analyzed the atoms carefully, you can actually
make sense of it.
"Li" at the start of the molecule is very likely a positive spectator ion, meaning it
has a negative charge on the attached molecule.
The fact that a molecule is negative means it is a strong attacker and will prefer a
two-type reaction meaning SN2 or E2.
If we then draw out the nucleophile to see the bonds carefully, notice that we have methyl
group in one end of the molecule, a negative charge on the other end, but we don't have
any Hydrogen on those first two Carbons and this tells me that I have to have a triple
bond between them.
Now we have a molecule that we recognize, an SP Carbon with a negative charge, is a
highly polarizable molecule due to the pi bonds and this makes it an okay base but a
very good nucleophile.
The fact that R molecule is a better nucleophile than it is a base; we're going to have the
SN2 reaction preferred over the E2 reaction.
And lastly we have the DMSO which is dimethyl sulfoxide, this is a polar aprotic solvent
which means it doesn't really stabilize any charge intermediates, anything charged in
solution will be considered stronger by default and this tends to favor an SN2 reaction.
Analyzing all the clues together, an SN2 reaction can indeed take place.
We will redraw the molecules so that we can show the mechanism and will only focus on
the negative portion of our attacker.
Chlorine, as a halogen, draws electron toward itself giving it a partially negative charge
causing the attached Carbon to become partially positive.
A nucleophile is attracted to this partially positive Carbon attacking with a lone pair
of electrons and forming a bond between itself and the Carbon holding the leaving group.
This causes Chlorine to grab the electrons that attach into Carbon and break away into
solution.
This one-step reaction results in a much larger chain where I first had my primary Carbon.
Notice that we didn't talk about stereochemistry in this reaction and that's because Chlorine
was not on a chiral molecule to begin with.
Let's not forget that the primary Carbon had two invisible Hydrogen atoms.
And the criteria for a Chiral Carbon is Four Unique Substituents, which means that we didn't
have a chiral-starting product nor do we have a chiral-ending product.
Be sure to join me in Part 3 of the SN2 Series where I show you how to do a reaction with
a leaving group that doesn't want to leave.
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