Tip:
Highlight text to annotate it
X
So what do we have right here?
Well, this is just a water molecule.
And this right over here, let's see,
we have 1, 2, 3 carbons in our backbone.
So this would be ***, ethy, propane.
We all have single bonds right here.
So this is propane.
Let me write that out.
This is-- give myself some space-- propane.
And then on the number 2 carbon, I
have both a methyl group and a bromo group.
And bromo comes before methyl in alphabetical order.
So I could write this as 2-bromo-2-methylpropane.
So that's this molecule that we have right over here.
Now, what we're going to imagine--
it's sitting in some water, and I
have one water molecule drawn right over here.
Let's imagine that we have this bromine right over here.
It's pretty neutral.
It is neutral because it has seven valence electrons.
Or you could say, it has the charge equivalent.
It has 1, 2, 3, 4, 5, 6.
And then it has half of this bond that's
composed of two electrons, so that's
what's keeping this neutral right over here.
But we do know that bromine-- well,
it's not completely unstable if it were to take an electron,
because that would get it to eight valence electrons.
We know that it's more electronegative than carbon,
which means that, if it's sharing a bond,
it's more likely to take the entire bond than carbon is.
So let's imagine a scenario.
Let's imagine a scenario where bromine takes these two
carbons in this bond right over here.
And this isn't going to be a super-fast thing that happens,
because this molecule right over here,
the 2 bromo-2 methylpropane, is actually reasonably stable.
So I'm not saying this is going to just happen automatically.
It's going to happen super-fast.
But it could happen.
And so to show that this isn't going
to be only a one-way reaction, I'll
show that it's in equilibrium.
So the equilibrium, I'll draw two arrows like that.
And so once this happens, the bromine
would turn into a bromide anion.
So let me draw it right over.
So the bromide anion-- so it's going to have 1, 2, 3, 4, 5, 6.
And then both electrons from that shared bond,
it now has a negative 1 charge.
Now, if that happened, what would happen to this left over,
to what's now going to be the thing that the bromine was
connected to, I guess you could say?
Well, let's draw that.
So we have this carbon, our number 2 carbon.
We have the CH3 right over there.
And we go back to CH3.
And then I had this blue methyl group.
So I had this blue methyl group right here.
And the way I'm drawing, it kind of pops out of the page.
My attempt to draw it popping out of the page-- so this
is it popping out of the page.
And now, what's its charge going to be?
Well, it just gave up half of this bond,
the equivalent of an electron's charge.
So it just gave up-- you can kind of
think of it as giving up an electron.
So this is going to have a positive charge.
And a positive charged carbon like this,
we would call this a carbocation.
So this right over here is a carbocation, positive charge.
And the reason why this is not a crazy thing to happen
is that this is a tertiary carbon, which
means it is connected to three other carbons-- 1, 2, 3.
A tertiary carbon, one way to think about it
is that this positive charge can kind of
be shared with its brothers a little bit more.
So a tertiary carbocation is more stable than a secondary
and definitely more stable than a primary.
Secondary would be if it was only connected to two carbons.
But this is tertiary-- so let me write this down.
This is a tertiary carbocation.
This carbon is connected to three other carbons.
That's where the tertiary comes from.
So once again, it can happen.
I'm not saying that this reaction is just
going to go superfast in that direction.
that's why we did these equilibrium arrows.
Now, let's bring this oxygen into play.
Sorry, not the oxygen, let's bring this water into play.
So this water, as you can imagine, the water is neutral.
It seems pretty reasonable in the state.
But the oxygen end of the water does
have a partial negative charge.
We've seen in the past that, because oxygen is a lot more
electronegative than hydrogen, you
have a partial negative side on the side of the oxygens,
and you have a partial positive charge,
on the side the hydrogens.
And so you have this positive carbocation.
This has a partially negative charge.
So you could imagine that this molecule might
be attracted to this is nucleus.
Now, it's not going to be super attracted
the way that a hydroxide anion would.
A hydroxide anion has a full negative charge.
This only just has a little partial negative charge.
But it could still be attracted to something very positive
like this.
So in this situation, the water molecule is a weak nucleophile.
So this is a weak nucleophile, not as strong as a hydroxide
anion that has a full negative charge.
But if it sees something positive like this
and it bumps in the right way, you
can imagine that one of these pairs of this oxygen
would then go and essentially form a bond, making
this carbon right over here neutral.
So what is that going to look like?
And this would happen faster, so I'm not
going to do this as in equilibrium.
This could happen like this.
And let me just copy and paste the bromine,
so I don't have to redraw the whole thing.
So let me copy, and let me paste it.
So that's our bromine right over there, draw it right there.
And actually, let me erase this little smudge
and make sure it still has a negative charge.
So negative charge-- it's a bromide anion now.
And now, let me draw the everything else.
So I have this carbon.
Let me scroll down a little bit, so I have some space.
Carbon, I have the carbon in the back-- CH3.
I have the carbon on top-- CH3.
I have the carbon in front, like that-- CH3.
And now my oxygen, or I should say, this molecule right over,
the water molecule, has attached to the carbon.
So these two electrons, you could say,
they've attacked the carbon.
They now form a bond with the carbon.
So these are those two electrons in this covalent bond.
And now, let's draw the rest of this.
So you have oxygen.
You have these two electrons over here, these two
valence electrons.
And then you have one hydrogen and then another hydrogen.
Now, what is going to be the charge
right over here on the oxygen?
Well, it had a partial negative charge,
just because it's more electronegative.
But it just gave essentially half of a bond, half of a pair.
Or it's now sharing a pair with this carbon,
so it's equivalent of giving up one electron's equivalent
of charge.
So you would now say that this thing right now
has a positive charge.
So you could imagine maybe another water molecule
comes along.
Let me draw another water molecule right over here.
We're sitting in water, so there's
plenty of water molecules.
So another water molecule might come along and say, OK, well,
this thing right over here is positive.
But this oxygen is hogging these bonds a little bit more
than the hydrogens, especially now that it's become positive.
And so this might give its pair to a hydrogen proton.
And then this bond with the hydrogen
will go entirely to the oxygen.
Let me draw that in a different color.
So this bond right over here goes entirely to the oxygen.
Electronegativity means if this is more electronegative
than that, that if a bond breaks,
that pair is more likely to go to the more electronegative
molecule, or I should say, the more electronegative atom.
So what will this look like when all is said and done?
So let's draw an arrow like this.
So let me draw my main molecules.
So I have the carbon.
It bonded to another carbon right up there.
It bonded to another carbon right over there.
We have our carbon in front just like that.
We have the bond with our oxygen.
So we have our bond with our oxygen right over there.
So that's our oxygen.
We have this bond with a hydrogen.
We have this pair.
And now, it has gained this pair.
It has gained this pair that was with that hydrogen,
so just like this.
This is now neutral.
It went from being positive-- over here,
it only had half of this bond.
So it's kind of the equivalent of one electron charge.
Now, it has a whole bond.
So now it has the equivalent.
It literally has both electrons.
So now this thing is neutral.
And now this molecule right over here
becomes a hydronium cation.
Let me draw that.
So this is water.
So let me see.
So that's the two hydrogens right over there.
This is that pair that is on top.
And now it is sharing this pair with the hydrogen,
essentially with a proton, so with this hydrogen
right over here.
And now, this will have a positive charge.
This is a hydronium cation and maybe
this bumps into another water molecule
and does the same thing with it.
And it just keeps going on and on and on.
And of course, we can't forget our bromide anion friend.
So there's our bromide anion friend.
We've got a negative charge right over there.
And so what are we left with?
Well, we are left with-- well, there's this hydronium cation.
It's a positive charge.
You have this bromide anion.
And now, what is this molecule now called?
Well, we'll go into more detail in the future.
But when you have a hydroxyl group attached to the carbon
backbone, it is an alcohol.
And so our main backbone, once again, has 1, 2, 3 carbons.
So that tells us prop.
But instead of just saying propane, it's propanol.
So let's write that down.
This is propan-- and some people say just propanol, generally.
But if we want to be a little bit particular,
the hydroxyl group is attached to the number 2 carbon.
So we would say, propan-2-ol.
And of course, we still have this blue methyl group.
So we could say 2-methyl-- let me just make this--
2-methylpropan-2-ol is what we are left with.
Now, let's think about what we can
name this reaction we just saw.
Well, we've just substituted a bromo group
with a hydroxyl group.
So I think it's reasonable to say
that this is going to be a substitution reaction.
And we substituted with a-- it was a weak nucleophile,
but it was a nucleophile nonetheless.
So it involved a nucleophile.
But this one, unlike the Sn2 reaction
in the rate-determining step-- and this
is the rate-determining step.
This is the slowest step of the process.
Everything else will just kind of
start happening once this thing happens.
This thing kind of is in equilibrium.
This is the rate-determining step.
Only one reactant is involved in the rate-determining step.
In this case, it was the bromo group.
So you might guess what we're going to call it.
It's substitution involving a nucleophile, so nucleophilic
substitution.
You can imagine, this stands for nucleophilic.
And only one reactant is involved
in the rate-determining step.
So this right over here, this whole reaction that we've now
drawn is an Sn1 reaction.
We're now going to explore one of the most
fundamental reactions in organic chemistry.
And first, let's look at what are
the things that are going to react.
So over here, I have a hydroxide anion.
It has a negative 1 charge, and that negative charge
comes from the fact that oxygen is neutral
if it has six valence electrons.
But this has 1, 2, 3, 4, 5, 6 plus half
of this bond with hydrogen.
So you could think of that as another electron or it's
share-- it's got half of these two,
so that's another electron, which
gives it a negative 1 charge.
Now, this right over here, it has one carbon.
I guess, you could say, in its backbone.
So it's a methane.
And it has a group right over here consisting of bromine.
So this is bromomethane.
So let's think a little bit about what
might happen under the right conditions.
So this hydroxide anion, we've already gone over the fact
that you can kind of view it as having an extra electron, which
gives us its negative charge.
Well, if you have a negative charged anion like this,
it will be attracted to positive things.
And one example of a positive thing is a nucleus.
And so in this reaction, we are going
to refer to this hydroxide anion as the nucleophile.
Phile literally refers to things that something likes.
This thing likes-- it's a nucleus lover,
I guess you could say.
And so it might want to share maybe a pair of electrons
with a nucleus.
Now, over here, I have this bromo group
in the bromomethane.
And bromine is quite electronegative.
It is more electronegative than carbon.
So even in this bond-- we have two electrons
in this bond right over here-- bromine
is going to hog the electrons a little bit more than carbon.
And also, electronegativity says that if this bond were
to break, it's more likely that this pair, the pair that's
in this bond, is going to go to the bromine
than it's going to go to the carbon.
So the way that this reaction works-- and it's really
kind of happening in one step, and this
is key to the name of this reaction,
as we'll see in a second-- is that this nucleophile right
over here might say, hey, let me share.
I've got this extra charge.
Let me share some of that charge with a nucleus,
in this case, the carbon.
Let me draw it like this.
In this case, the carbon, so it would
share this pair, which would form a bond.
Carbon would have half of the pair.
Oxygen would have half of the pair.
So it would share this pair.
It would share this pair with carbon.
Now, if carbon is now sharing half of this,
it's essentially getting an electron.
It doesn't need this electron, which bromine is already
hogging a little bit.
And so bromine would take this entire pair back.
Whenever we draw full arrows like this, just normal arrows
that you're used to seeing, it's referring
to the action of pairs.
So this whole pair is going to bromine.
This pair right over here is now going
to be shared with the carbon.
It's going to form a bond with the carbon.
So when we do both of these steps,
after both of these steps, it would look like this.
Let me draw my bromine first, which
is now going to become a bromide anion.
So bromine 1, 2, 3, 4, 5, 6, and now it
gets both of these electrons from this bond.
So it has both of those electrons right over there.
Now, this will have a negative 1 charge.
How do we know that?
Well, bromine is neutral, when it has seven valence electrons.
This thing now has eight.
So this now has a negative charge.
Now, once again, let's think about what
happens to all of this business.
So let me draw my carbon first, my carbon.
Now, let me draw this hydrogen that's popping out.
So this hydrogen that's popping out, let me draw it like that.
So that's the bond, trying to show
that's popping out of the page.
So that's that hydrogen.
I just made them different colors,
so we can keep track of the different hydrogens.
Then the hydrogen that's going into the page, that's
that one right over there.
And you have this hydrogen that's going up,
so that hydrogen right over there.
And now the hydroxide anion has bonded to this carbon.
Now it will be a hydroxyl group.
So this thing has now bonded.
So let me draw the bond in the color of these two electrons.
It has now bonded.
So those two electrons now make up this blur,
this bond right over here.
And now we have our oxygen, but with those two, these two
right over here, and then the bond to the hydrogen.
Now, what just happened here?
The oxygen is now neutral.
Why is it neutral?
Well, it has 1, 2, 3, 4.
And then it has half of this bond,
so you could view that as an electron
or the equivalent of an electron charge.
And then it has half of this bond.
So it has the electron charge equivalent
of six valence electrons, which makes oxygen neutral.
And now bromine is negative.
So one way to think about is that the negative charge
has been transferred to the bromine in this reaction.
Now, in this reaction, bromine has left.
And that's why we call this the leaving group.
We call this right over here the leaving group of the reaction.
This thing right over here is the thing
that is going to leave.
That is the leaving group.
Now, the other thing that you might find interesting,
it looks like kind of this hydroxide anion--
some people even say-- attacked this bromomethane.
The bromomethane left.
The bromine left, not bromomethane left.
It attacked the bromomethane.
The bromine left.
So in some ways, you'd say that there's really
two reactants over here.
You have the hydroxide anion, and you have the bromide anion.
And this carbon right over here with the hydrogens,
this was kind of the thing that just to some degree
facilitated the reaction.
This thing bumped into it and attached,
and then this thing left.
And so in this context, we would call the carbon
with a three hydrogens-- we would call this the substrate.
Now, what should we call this reaction?
Well one way to think about it is, we substituted this bromo
group with a hydroxyl group.
So it might make sense to call this a substitution reaction.
Substitution has occurred.
Now, what did we substitute with?
Well, we substituted with a nucleophile.
And how many reactants were involved
in the rate-determining step?
And I know that might not make a lot of sense to you,
but this had essentially one determining step.
This had to want to bump in the right way.
And this would want to have to leave right at the same time,
or roughly the same time, in order
for this reaction to occur.
So this reactant and this reactant
were involved in figuring out, well,
how fast is this reaction going to happen?
And we're going to see future reactions, where you don't have
both reactants involved in the rate-determining step.
We'll see only one reactant involved
in the rate determining step.
But I'll just write this right over here-- two reactants
in rate determining step.
There's a garbage truck outside.
I think the garbage truck has left now, safe to resume.
So as we were saying, there are two reactants
involved in the rate determining step.
And so the shorthand for this reaction
is to call it an Sn2 reaction.
It's substitution with a nucleophile, where
both reactants are involved in the rate-determining step.