Solvent Effects on Sn1 and Sn2 Reactions - Organic Chemistry, Chemistry Video Lecture - JEE

Let's talk a little bit more
about the comparison between
Sn1 and Sn2 reactions, and maybe
we can also talk about
what type of conditions make
one or the other more
favorable, especially
the solvent.
We've talked about it a little
bit, but we'll go into a
little bit more detail.
So just as a bit of a review,
an Sn1 reaction, you might
have some type of a-- let me
just draw it like this.
So you'll have some
carbon that's
attached to three things.
I'm making these three things
just arbitrary colors.
They could be other
carbons, CH3.
It could be hydrogens.
It could be other things.
And then it's also attached
to a leaving group.
Let me do this in a color that
I haven't used yet: in pink.
And then you also have a
nucleophile floating around.
That'll be our nucleophile.
It has one excess electron.
It has a negative charge.
And in an Sn1 reaction, the
first step only involves one
of the reactants, one
of the reagents.
That's why it's called Sn1.
So in the very first step, the
nucleophile just waits
passively while the leaving
group leaves.
So maybe there's an electron
right there, the leaving group
already has an electron.
It swipes one, so it has to be
reasonably electronegative.
It swipes one from that carbon,
and so once it does
that, our situation will
look like this.
You have your substrate.
It now just lost an electron.
It lost that electron, so it
now has a positive charge.
Let me add that little
trace of something.
Let me erase this right here
so you don't think that
electron is still there.
So it now has a positive
charge.
The leaving group, which was
neutral, took an electron.
So it out now has a
negative charge.
The leaving group now took an
electron, so it now has at
least two valence
electrons there.
It has a negative charge,
probably a lot more.
And then in the next step, this
nucleophile-- so let me
copy and paste this.
That's not what I
wanted to do.
Let me copy and paste
it again.
So copy and paste.
OK.
So that thing that was waiting
patiently during the
rate-determining steps
says, hey, there's
a carbocation there.
I'm a nucleophile.
I'm not a strong nucleophile in
the case of an Sn1, but I
still like giving away electrons
to nucleuses.
I love nucleases.
I see a positive carbocation
there.
Actually, a carbocation is
positive by definition, so let
me give my electrons to him.
And in the very last step, we
have a situation where this
guy-- let me copy and
paste this again.
Let me copy and let me paste.
That guy is bonded to
the nucleophile.
Since the nucleophile gave the
electron to this carbon, the
carbon doesn't have
a charge anymore.
Since the nucleophile lost that
electron, it's now neutral.
So you have it bonded to the
nucleophile, and, of course,
you have the leaving
group still here
with a negative charge.
We've hinted at this a little
bit, but in order for this
step to happen on its own with
only this reagent happening
spontaneously to some degree,
or for it to just happen.
And we won't talk
about the rates.
And it won't necessarily happen
fast, but in order for
this to happen at all, this
has to be a stable
carbocation.
And we've seen before that in
order for this to be a stable
carbocation, it has to be bonded
to things that are
willing to lend it some
electrons so it's not super
hungry for electrons.
And in order for that to happen,
this carbocation right
here has to be tertiary.
Tertiary is ideal.
Tertiary carbocation means that
these three things over
here are all carbons
or carbon chains.
So ideally, it would be tertiary
and secondary will
also work if at least two of
these guys are carbon.
That'll also work if the solvent
is favorable, if the
environment we're dealing with
is a favorable environment.
If only one of these were
carbon, so if this was a
primary carbocation or this was
a methyl carbocation where
nothing here is a carbon,
then you're unlikely
to have an Sn1 reaction.
Now, on the other side of the
picture, Sn2 reaction, this is
the one where you have a super
strong nucleophile, or at
least a strong nucleophile, and
the whole reaction happens
in one step involving both
of the reagents.
So you start here with
your leaving group.
The way you can think about it,
Sn1, the leaving group is
the thing that does something
first. In Sn2, it all happens
at once, but the nucleophile's
kind of the aggressor.
It really wants to give that
electron away badly.
So in the very first step, it
comes in here and gives it to
that carbon, even though that
carbon didn't necessarily want
it that badly.
It would have a partial positive
charge if this thing
is really electronegative and
might be hogging this electron
already, but that thing that it
was already hogging, now it
could say, hey, you already
got an electron.
I can take it from you.
And so in literally one
step, we end up with
something like this.
Actually, let me just
redraw that.
So we have our leaving
group has left.
It's taken the electron
that the
carbon didn't need anymore.
So it now has this electron and
that one and maybe more,
and it has a negative charge.
And now this carbon is attached
to the blue thing.
I'll just draw the purple thing,
and maybe in back, it's
attached to the orange thing.
Let me draw it so you can see
what I'm talking about, so the
blue thing and then
the purple thing.
And then the nucleophile had
come from the other direction,
so now it's also bonded
to the nucleophile.
The nucleophile is giving an
electron, so it loses its
negative charge.
And essentially, the negative
charge gets transferred to the
leaving group, which is able to
take an electron, and then
that carbon in the middle
stays neutral.
And we saw before that this
can only reasonably happen
with a strong nucleophile.
That has to be a strong
nucleophile, and there can't
be a lot of steric hindrance.
These things can't
really block this
nucleophile from attacking.
So this will work best if this
is a methyl or primary carbon.
And if the conditions are just
right and if the nucleophile
is strong enough,
maybe secondary,
maybe a secondary carbon.
Because once you have too many
carbons or big carbon chains
over here, then it blocks
the nucleophile.
It cannot attack.
If these are all hydrogens,
there's a lot of space for
this nucleophile to attack.
If only one of these are carbon,
still a lot of space.
Now, that's just kind of a
general comparison, and it
gives you some clues.
Weak nucleophile, you're
dealing with Sn1.
Strong nucleophile, Sn2.
If this is a tertiary or
secondary carbon, so if it's
bonded to a lot of other
carbons, Sn1
should be in your brain.
If this isn't bonded to a lot
a carbons and you have a
strong nucleophile, Sn2 should
be in your brain.
And now the last thing you
should have to think about is
the solvent itself.
What are these reagents
actually dissolved in.
So let's think about it.
So in either an Sn1 or an Sn2
reaction, we're dealing with
things that have charge.
And in general, if you're
dealing with ions or things
that have charge, they're going
to be more soluble in
polar solvents because the
charge likes to be around
other things with charge so that
they can mix and match.
So in both situations, you want
to have polar solvents.
And you are familiar with
polar solvents.
Examples of polar solvents
are, well,
water is a polar solvent.
H2O, and it's polar because
oxygen is so electronegative
it hogs electrons.
So you have a partial negative
charge on the oxygen side and
partial positive charges
on the hydrogen side.
Another example of a polar
solvent that you may have not
seen before is something
like diethyl ether.
So diethyl ether would
look like this.
CH2, CH3, CH2, CH3.
And I know we havn't
named ethers yet.
Let me write this.
This is diethyl-- let me make
sure I get that "y" right.
This is diethyl ether.
And this is a bit of a review.
This is probably the best
way to learn things, as
we learn as we go.
An ether is something
that has a form.
You have some carbon chain or
some-- you could call it a
radical-- some group attached to
a carbon chain, I guess is
the way my brain processes it,
to an oxygen which is attached
to another carbon chain.
It doesn't have to be
the exact same one.
And diethyl ether is the same.
And you're literally just
saying diethyl.
You have to ethyl groups
attached to an ether.
Ethyl is two carbons.
So two carbons, oxygen,
two more carbons.
Diethyl ether.
And this would also be a polar
solvent because oxygen is more
electronegative then carbon, so
you still have this partial
negative charge-- let me do this
in a different color-- on
the oxygen end and you still
have a partial positive charge
on your carbon end.
So this is also a
polar solvent.
Both of these are
polar solvents.
Now, the difference between
these, and this will probably
be a new term for you, is this
water and most types of
alcohol, something that has
hydrogen that could be knocked
off, this is a protic sovent.
And protic just comes
from proton.
And that means that if something
is dissolved in
this, or if you have a solution
of this, then these
hydrogens might be able to get
knocked off and you might have
free protons lying around.
Because let's say this
oxygen swipes this
electron from this hydrogen.
If it swipes that electron from
that hydrogen, then you
have a situation where you
have OH minus, right?
The oxygen now has a
negative charge.
And this hydrogen, all it had to
its name was that electron.
If you take that away, all
it is is a proton.
So when someone draws H plus,
that literally is a proton.
That literally is a proton.
And this can occur in water,
that you have some protons
floating around, so water
is fundamentally a
protic polar solvent.
Diethyl ether, on the other
hand, it doesn't have
hydrogens where oxygens can
take its hydrogens away.
The hydrogens here are all
bonded to carbon, which is not
that electronegative, so this
is an aprotic solvent.
Let me draw this.
Now, let's think about whether
an Sn1 or Sn2 is better and--
well, I already said the
solvents have to be polar.
That's because we're dealing
with charged
things in both reactions.
But which one will be better?
So let me write this down.
So in both situations, we want
a polar solvent, so let me
write that down.
And I'll give you a hint: one
of them will be good in a
protic environment and one of
them will not be good in a
protic environment.
So in an Sn2, if you had-- let's
just think it through.
Let's just not memorize this.
If we had protons floating
around in an Sn2 situation,
what is likely to happen?
We have this strong
nucleophile.
We have this thing that wants
to give away electrons.
Now, it could give it away to a
carbon that might maybe have
a partial positive charge and
then the leaving group could
take it away, but what
wants electrons more?
A carbon that has a partial
negative charge or a proton,
something that has a full
positive charge?
Well, a proton would
want it more.
So if you had protons floating
around, the nucleophile would
just give its electrons to the
proton and the Sn2 reaction
wouldn't actually occur.
The protons would take
all the electrons.
So you can't have protons
floating around.
You cannot have protons
floating
around in an Sn2 reaction.
Then this carbon will never
get the electron.
The proton will take it.
So you need an aprotic
solvent.
So Sn2 would work well
in diethyl ether.
If this is a strong nucleophile,
it's a solution
in diethyl ether, and these were
all hydrogens, then you'd
be just ready to go for
an Sn2 reaction.
Now, on the other end of the
spectrum, the key in an Sn1
reaction is that this
carbocation be stable.
And for a carbocation to be
stable-- or actually even more
important, instead of the
carbocation being stable,
that's dependent on it being a
tertiary or secondary carbon.
But order for the leaving group
to be stable when it's
negative, it actually would
be nice if there was
some protons available.
Maybe it could give electrons
to the proton.
Maybe it could bond
with the proton.
So in this situation, the
protons will actually help
stabilize the situation.
So in an Sn1 reaction, the
ideal is actually a polar
protic solvent.
So if you see a situation where
you have a secondary or
tertiary carbon, it is a weak
nucleophile, and you're
dealing with a polar protic
solvent, the most famous of
which is water, but an alcohol
would also work, then you
should be thinking Sn1.
If you see a strong nucleophile,
there's not much
steric hindrance.
There's space that
it can attack.
This is a methyl or a primary,
maybe a secondary carbon, and
it's happening in an aprotic
polar solvent like diethyl
ether, then you know you're
ready for an Sn2.