Voiceover: 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 meth eth 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 bromo, 2 methyl,
2 methyl, methyl propane.
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 1 water molecule drawn right over here.
Let's imagine that we have this bromine right over here.
It's pretty neutral.
And it's neutral because it has-
or it is neutral because it has 7 valence electrons.
1, 2, or you could say- or it has the charge equivalent
as 1, 2, 3, 4, 5, 6.
And then it has half of this bond
that's composed of 2 electrons.
So that's what's keeping this neutral right over here.
But we do know that bromine,
well you know, it's not completely unstable.
If it were to take an electron,
because that would get it to 8 valence electrons,
we know that it's more electro-negative 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 2 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 methyl propane
is actually reasonably stable,
so I'm not saying that 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
you know, only one-way reaction,
I'll show that it's an equilibrium.
So the equilibrium, I'll draw 2 arrows like that.
So, once this happens, the bromine
would turn into a bromide anion.
So let me draw that right over here.
So the bromide anion.
So it's going to have 1, 2, 3, 4, 5, 6,
and then both electrons from that shared bond.
Both electrons from that shared bond.
It now has a negative 1 charge.
Now if that happened,
what would happen to this leftover,
to I guess what's now going to be
the thing that bromine
was connected to I guess you could say.
Let's draw that.
So we have this carbon, our number 2 carbon.
We have CH3 right over there.
Then we go back to a CH3
and then we had this blue methyl group.
So, I'll add this blue methyl group right over here
and the way I'm drawing it,
it kind of pops out of the page.
My best 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.
This is going to have a positive charge.
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 that it is connected to 3 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 tertiary, a tertiary carbocation is more stable
than a secondary and definitely
more stable than a primary.
A secondary would be
if it was only connected to 2 carbons,
but this is tertiary so let me write this down.
This is a tertiary carbocation.
This carbon is connected to 3 other carbons.
That's where the tertiary comes from.
So once again, it can happen.
I'm not saying that this reaction's
just going to go super fast in that direction.
That's why we did these equilibriums.
These equilibrium arrows.
Now let's bring this oxygen into play.
So let's bring- now let's bring-
sorry not the oxygen, let's bring this water into play.
So this water, as you can imagine, the water's neutral
and it seems pretty reasonable in this state,
but the oxygen end of the water does have
a partial negative charge.
We've seen in the past.
We've seen in the past that
because oxygen is
a lot more electronegative than the hydrogens,
that you have a partial negative side
on the side of the oxygens
and you have a partial positive charge
on the side of the hydrogens.
On the side of the hydrogens.
And so you have this positive carbocation.
This has a partially negative charge.
So you can imagine that this molecule
might be attracted to this nucleus.
Now it's not going to be super attracted
the way that an hydroxide anion would.
A hydroxide anion has a full negative charge.
This only 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 could imagine that one of these pairs
of this oxygen would then go-
one of these pairs on this oxygen would then go
and essentially form a bond with this carbon,
making this carbon right over here neutral.
So what is that going to look like?
And this might happen, this would happen faster
so I'm not going to do this as an equilibrium.
This could happen like this.
And let me- 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.
Actually let me erase that little smudge
and make sure it still has a negative charge.
So, negative charge.
It's a bromide anion.
Now let me draw everything else.
So I have this carbon,
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.
I have the carbon in front like that.
CH3.
And now my oxygen,
or I should say this molecule right over here,
the water molecule has attached to the carbon.
So these 2 electrons,
you could say they've attacked the carbon.
They now form a bond with the carbon.
So these are those 2 electrons
in this covalent bond.
And now let's draw the rest of this.
So you have the oxygen.
You have these 2 electrons right over here.
These 2 valence electrons.
Then you have 1 hydrogen and then another hydrogen.
Now what is going to be the charge,
what is going to be the charge
right over here on the oxygen?
Well it was, it had a partial negative charge
because it's more electronegative
but it just gave, it just gave the
essentially half of a bond,
half of a pair.
It's now sharing a pair with this carbon.
So it's equivalent to giving up 1 electron's
equivalent of charge so you would now say
that this thing right now has a positive,
a positive charge.
So you could imagine,
you could imagine,
maybe another water molecule comes along
maybe another water molecule comes along.
Let me draw a 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, okay well this thing right over here
is positive but the oxygen is hogging these bonds
a little bit more than the hydrogen,
especially now that it's become positive.
So this might give its pair to a hydrogen proton
and then this hydrogen-
and then this bond with the hydrogen
will go entirely, entirely to the oxygen.
Let me draw that in a different color.
So this bond, this bond right over here,
goes entirely, goes entirely to the oxygen.
Electronegativity means that if this is more
electronegative than that then if a bond breaks
that pair is more likely to go to the more electronegative
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 molecule.
So I have the carbon.
It had bonded to another carbon right up there,
which bonded to another carbon right over there.
We have our carbon in front.
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.
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 1 electron charge.
Now it has the whole bond.
So now it has the equivalent,
it literally has both electrons.
Now this thing is neutral
and now this molecule right over here
becomes a hydronium cation.
Let me draw that.
So this water, let me see.
That's the 2 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 the hydrogen, with this hydrogen right over here.
With this hydrogen right over here
and now this will have a positive charge.
This is a hydronium, this is a hydronium cation.
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,
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 a little bit of,
there's this hydronium cation that has a positive charge.
We 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-
and instead of just saying propane,
it's propanol.
So let's write that down.
This is prop- prop- prop- an
and some people say just propanol generally
but if want to be a little bit particular,
this is 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, 2 methyl,
actually let me make this
2 methyl propan 2 ol is what we are left with.
Now let's think about what we can name
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.
Substitution.
And we substituted with a-
it was a weak nucleophile,
but it was a neucleophil none-the-less.
So it involved a neucleophile.
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 an equilibrium.
This is the rate determining step.
Only one reactant is involved
in the rate determining step.
One reactant in rate determining step.
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.