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- Let's think about what'll happen if
- we have this molecule.
- Let's name it.
- We have one, two, three, four, five carbons.
- No double bond.
- Five tells us pent.
- It's pentane, and it has two groups on the number three
- carbon, one, two, three.
- It doesn't matter which side we start counting from.
- We have a bromo group, and we have an ethyl group, two
- carbons right there.
- On the three carbon, we have three bromo, three ethyl
- pentane right here.
- So we have 3-bromo 3-ethyl pentane dissolved in a
- solvent, in this right here.
- It's an alcohol and it has two carbons right there.
- Meth eth, so it is ethanol.
- This right there is ethanol.
- Let's think about what might happen if we have 3-bromo
- 3-ethyl pentane dissolved in some ethanol.
- Now ethanol already has a hydrogen.
- It's not super eager to get another proton, although it
- does have a partial negative charge.
- It is polar.
- Oxygen is very electronegative.
- It has a partial negative charge, so maybe it might be
- willing to take on another proton, but doesn't want to do
- so very badly.
- It's actually a weak base.
- Ethanol right here is a weak base.
- It's not strong enough to just go nabbing hydrogens off of
- carbons, like we saw in an E2 reaction.
- It's just going to sit passively here and maybe wait
- for something to happen.
- What might happen?
- Well, we have this bromo group right here.
- We have this bromine and the bromide anion is actually a
- pretty good leaving group.
- It's a fairly large molecule.
- It's able to keep that charge because it's spread out over a
- large electronic cloud, and it's connected
- to a tertiary carbon.
- This carbon right here.
- This carbon right here is connected to
- one, two, three carbons.
- So if it were to lose its electron, that electron right
- there, it would be-- it might not like to do it-- but it
- would be reasonably stable.
- It's within the realm of possibilities.
- It could occur.
- Maybe in this first step since bromine is a good leaving
- group, and this carbon can be stable as a carbocation, and
- bromine is already more electronegative-- it's already
- hogging this electron-- maybe it takes it all together.
- Let me draw.
- Neutral bromine has one, two, three, four, five, six, seven
- valence electrons.
- Maybe it swipes this electron from the carbon, and now it'll
- have eight valence electrons and become bromide.
- What happens now?
- And of course, the ethanol did nothing.
- It's a weak base.
- It wasn't strong enough to react with this just yet.
- What is happening now?
- This is going to be the slow reaction.
- What I said was that this isn't going to happen super
- fast but it could happen.
- This is actually the rate-determining step.
- What happens after that?
- Let me just paste everything again so this is our set up to
- begin with.
- But now that this little reaction occurred, what will
- it look like?
- The bromine has left so let me clear that out.
- We clear out the bromine.
- It actually took an electron with it so it's bromide.
- Let me draw it like this.
- I'll do it in blue.
- This is the bromine.
- The bromine is right over here.
- It had one, two, three, four, five, six,
- seven valence electrons.
- It swiped this magenta electron from the carbon, now
- it has eight valence electrons.
- It has a negative charge.
- The carbon lost an electron, so it has a positive charge
- and it's somewhat stable because it's a tertiary
- carbocation.
- Now let's think about what's happening.
- And I want to point out one thing.
- In this first step of a reaction, only one of the
- reactants was involved.
- This rate-determining, the slow step of reaction, if this
- doesn't occur nothing else will.
- But now that this does occur everything
- else will happen quickly.
- In our rate-determining step, we only had one of the
- reactants involved.
- It's analogous to the SN1 reaction but what we're going
- to see here is that we're actually eliminating.
- We're going to call this an E1 reaction.
- We're going to see that in a second.
- Actually, elimination is already occurred.
- The bromide has already left so hopefully you see why this
- is called an E1 reaction.
- It's elimination.
- E for elimination and the rate-determining step only
- involves one of the reactants right here.
- It didn't involve in this case the weak base.
- Now that the bromide has left, let's think about whether this
- weak base, this ethanol, can actually do anything.
- It does have a partial negative charge over here.
- It does have a partial negative charge and on these
- ends it has partial positive charges, so it is somewhat
- attracted to hydrogen, or to protons I should say, to
- positive charges.
- But not so much that it can swipe it off of things that
- aren't reasonably acidic.
- Now that this guy's a carbocation, this entire
- molecule actually now becomes pretty acidic, which means it
- wants to give away protons.
- Another way you could view it is it wants to take electrons,
- depending on whether you want to use the Bronsted-Lowry
- definition of acid, or the Lewis definition.
- Either way, it wants to give away a proton.
- It could be this one.
- It could be that one.
- It has excess positive charge.
- It wants to get rid of its excess positive charge.
- So it's reasonably acidic, enough so that it can react
- with this weak base.
- What you have now is the situation, where on this
- partial negative charge of this oxygen-- let me pick a
- nice color here-- let's say this purple electron right
- here, it can be donated, or it will swipe
- the hydrogen proton.
- Then hydrogen's electron will be taken
- by the larger molecule.
- In fact, it'll be attracted to the carbocation.
- So it will go to the carbocation just like that.
- Now in that situation, what occurs?
- What's our final product?
- Let me draw it here.
- This part of the reaction is going to happen fast. The
- rate-determining step happened slow.
- The leaving group had to leave. The
- carbocation had to form.
- That's not going to happen super fast but once that
- forms, it's not that stable and then
- this thing will happen.
- This is fast.
- Let me paste everything again.
- So now we already had the bromide.
- It had left.
- Now the hydrogen is gone.
- The hydrogen from that carbon right there is gone.
- This electron is still on this carbon but the electron that
- was with this hydrogen is now on what was the carbocation.
- That electron right here is now over here, and now this
- bond right over here, is this bond.
- We formed an alkene and now, what was an ethanol took a
- hydrogen proton and now becomes a positive cation.
- Let me draw that.
- So this electron ends up being given.
- It's no longer with the ethanol.
- It gets given to this hydrogen right here.
- That hydrogen right there.
- And now they have formed a new bond and since this oxygen
- gave away an electron, it now has a positive charge.
- And all along, the bromide anion had left in
- the previous step.
- The bromide anion is floating around with its eight valence
- electrons, one, two, three, four, five, six, seven, and
- then it has this one right over here.
- That makes it negative.
- Then our reaction is done.
- We had a weak base and a good leaving group, a tertiary
- carbon, and the leaving group left.
- We only had one of the reactants involved.
- It was eliminated.
- And then once it was eliminated, then the weak base
- was then able to take a hydrogen off of this molecule,
- and that allowed this molecule to become an alkene, formed a
- double bond.
- This is called, and I already told you, an E1 reaction.
- E for elimination, in this case of the halide.
- One, because the rate-determining step only
- involved one of the molecules.
- It did not involve the weak base.
- We'll talk more about this, and especially different
- circumstances where you might have the different types of E1
- reactions you could see, which hydrogen is going to be picked
- off, and all the things like that.
- And we're going to see with E1, E2, SN1, and SN2, what
- kind of environments or reactants need to be there for
- each one of those to occur in different circumstances.