So now what I want to show you guys is how does this relate to substitution.
Well, remember that we also had the Lewis acid and base definition and that didn't have
to do with protons.
What that means is that sometimes you're going to have electrophiles and nucleophiles that
want to react together, but there's no acidic hydrogens that they can react with.
Does that mean that you give up?
No.
You still react.
This is an example that I also used when we were talking about acids and bases.
We said like in this compound, which one would be the Lewis acid, which one would be the
Lewis base.
Do you remember?
Well, remember that Lewis acid, so I'm just going to write here, actually the definition
of a Lewis acid is that it's an electrophile.
Those two words are actually synonymous with each other.
Lewis acid means it's a good electron acceptor.
Remember that a Lewis base is synonymous with nucleophile.
It means that it's good at giving away electrons.
So in this case which one would be good at giving away electrons?
The double bond.
I've mentioned this several times during the span of this course, but I keep saying, double
bonds are really good sources of electrophiles, of electrons, not electrophiles, of electrons
because they have these two free electrons in the pi bond.
So I know that I'm going to start from there.
On top of that, is boron a good electrophile?
Actually, yeah.
Remember that boron and aluminum were two special atoms that I keep pointing out that
happen to have an incomplete octet or basically they don't have 8 electrons, they only have
6, and they have an empty p orbital.
Now if this is the first time that you're hearing me say that, that's okay.
That just means that maybe you didn't get to watch the old reviews.
But from now on this is going to be a very important fact for the rest of orgo you need
to remember that aluminum and boron are very good at accepting electrons because they just
have this empty p orbital that's just waiting to have some electrons in it.
So I'm going to go ahead and draw the rest of my mechanism.
My electrons would go straight into that orbital.
So my end products here – when it's a Lewis acid/Lewis base, we actually don't use the
equilibrium arrows, we use just a forward arrow.
The reason is because what we're going to get is a new covalent bond without the exchange
of hydrogens.
When you have an exchange of hydrogens, you use an equilibrium sign because the hydrogen
can go from one place to another and then it could go back.
But with Lewis acid/Lewis base there's no exchange
As you can see from my description I didn't read it but that's because I wanted to show
you guys.
When a nucleophile and electrophile react with an empty orbital, that's called a Lewis
acid/Lewis base.
So this is what we were used to doing in the acid/base chapter.
When we had an empty orbital we would just draw this.
And what I would get now is a square again.
That's cyclobutane.
Now you guys know how to name that.
And I would get now BH3.
Remember that basically ever arrow always has to turn into a sigma bond or a single
bond.
Now I have that and I just have to figure out the formal charges.
Are there any formal charges here that I have to worry about.
Yes, there are because let's look at the double bond for a second.
Anytime you break a double bond, what that means is that you are going to be removing
electrons for two atoms, not just one.
So this top carbon would have had a hydrogen.
This bottom carbon would have also had a hydrogen.
Why?
Because that carbon wants to have four bonds, so obviously according to bondline they need
one hydrogen each.
After the reaction, does that change?
Absolutely not, they still have one hydrogen each, there and there.
The only difference is that now one of the carbons is happy.
Its octet is filled because it has four bonds.
The other one is not happy because it only has three.
What are the formal charges that I'm going to have to put here?
There's going to be a positive charge here because that carbon is missing electrons.
There's going to be a negative charge here because boron wants to have three; boron now
has four.
Easy.
Now we're going to leave this right here.
Later on, in future chapters, we're actually going to continue.
I'm just going to put a question mark because we don't know what that is yet.
But in the addition chapter, once we get there, oops, addition, what we're going to find is
that this is the precursor to a very important reaction.
But we're not there yet.
But I just want to show you guys that this is another example of an acid/base but this
is the Lewis definition.
So once again you're like, “Okay, Johnny, I get it.
What does this have to do with substitution.”
Finally, let's get to it.