So now I want to talk about a mechanism that competes directly with the SN1 mechanism and
that's called the E1 elimination mechanism.
If I were to sum up the entire mechanism into one sentence, what I would say is this E1
occurs when a weak nucleophile reacts with an inaccessible leaving group.
So let's just start right there.
Do you guys remember another mechanism that we talked about that had similar conditions?
Yeah, right?
The SN1.
Let's talk about the nucleophile part first.
We said that SN1 is favored with what type of nucleophiles, strong or weak?
Well, remember that strong favored the SN2 and E2, so weak is going to favor SN1 and
E1.
Why?
Because remember that SN1 – I'm sorry, remember that a weak nucleophile would not start the
back side attack first.
It would wait for something to happen.
Remember that the weak nucleophile just kind of was chilling and sitting back and waiting
for something to happen.
Well, then what's the other condition?
The other condition says that it's with an inaccessible leaving group.
Inaccessible really is just another word for tertiary.
It's basically, there's many R groups as possible.
Why is that good?
Well, the reason that's good is because remember that tertiary alkyl halides would make really
good carbocations.
And remember that carbocation was the rate determining step for SN1.
Is this all kind of coming back to you guys?
Well, it turns out that the same condition is also good for E1.
E1 also like to have an inaccessible leaving group.
So if we have the same exact conditions that are good for both mechanisms, how do we know
when one is going to happen or the other is going to happen?
It turns out, we don't.
So it turns out that these same exact conditions are going to also produce beta-elimination
in two steps.
What that means is that what we're going to find out is that SN1 and E1 always compete
with each other.
And they're almost impossible to separate.
What that means is that every time you see an SN1 reaction, remember I taught you SN1,
E1 is also occurring and you really can't do much to stop it.
That's one of the messy things about these reactions is that they're often going to lead
to a mixture of products and synthetically that's not very useful for us.
When I say synthetically, that's a word I'll use a lot more later as we get into orgo two.
Synthesis is a huge end goal of organic chemistry.
Eventually, what you're going to be doing, get excited, you're going to be trying to
synthesize a target compound.
You're professor's just going to be like, “Hey, make this.”
and you're just going to have a list of reagents and you're going to have to put them together.
SN1 and E1 aren't very helpful for synthesis.
The reason is because you get a mixture of products and that's bad because then you don't
know exactly what you're getting at the end.
Let's just go ahead and talk about this mechanism and see how you could get the SN1 or the E1.
Here I've got, as you'll see, a tertiary alkyl halide.
And the first step is going to be the same identical for both of these reactions.
So what's that first step?
Can you guys tell me?
Is it back side attack?
No.
Why?
Because my nucleophile isn't strong, so it can't initiate anything.
Remember it's not one of those arrows.
It doesn't do that.
So it's just going to wait around.
What actually happens is that the leaving group leaves by itself.
Remember that part?
It's kind of weird, but it dissociates.
What that's going to do is it's going to give me an intermediate that looks like this.
Single bond.
H's are exactly the same, H, H, H. But now what I have is a carbocation.
That carbocation came from the fact that the X just left by itself without getting anything
in return, so electrons are leaving.
Then finally, see how I have R1 and R2, those would still be here as well.
So I'd have R1 and R2.
Cool?
So it doesn't matter which one you put where because this is a carbocation, so you don't
actually know what you would get.
So now what?
Well, it turns out that this carbocation is going to do what next?
The nucleophile is going to react with it, right?
Cool.
So my nucleophile would want to react with it.
And the first thing you would think is that the nucleophile has electrons to give away.
Remember it doesn't have a negative charge, but it does have a lone pair, so it could
attack directly my carbocation.
And it turns out that that is going to happen.
That nucleophile is going to directly attack that carbocation and that's called the SN1
mechanism.
So what we would do here is we would draw the SN1 product and the SN1 product would
just be that I have now two R groups again and I would have these H's are exactly the
same, but then I would get a nucleophile attached to wherever that carbon was.
Now notice that I had R1 and R2, so those are different.
Does this make a chiral center?
Yes, most likely it does.
So what that means is that remember that I would actually get two different products.
I would get one enantiomer and then I would also get another enantiomer.
Well, let's say that this R1 was faced towards the front and this R2 is faced towards the
back.
I would also get the other enantiomer where I have the nucleophile facing the same way,
but now I have R2 in the front and R1 in the back.
Is that making sense?
I'm just going to leave this as CH3.
So remember that whenever you make a chiral center, you have racemization.
That means you get 50% of each and that's exactly what you would get for an SN1.
But now we know that's not the only thing that's going to take place.
There's also another mechanism that's going to compete with this and that's the E1.
What the E1's going to do is it's going to say, you know what, I could attack the carbocation
directly, but that's too boring.
Why would I want to do that?
Actually, I don't know exactly why it doesn't.
But it's also going to say, you know what, I am a pretty good nucleophile, but I could
also act like a base.
And what bases do is they pull off protons.
Remember that nucleophiles and bases many times are similar.
So it's going to say well, let me just pull off a beta-proton and then once I pull off
that beta proton, I can put these electrons into that bond there and satisfy the carbocation.
This is also going to happen and that's called the E1 mechanism.
So it's going to be competing with the SN1 at the same time.
So what that means is that now what we're going to get is a double bond that has H's.
Now only two H's, not three, only two.
But now it has a double bond.
Does that make sense guys?
So we would not only have two different substitution products because of the racemic part, but
we would also have an elimination product as well.
On top of that, remember that elimination products may yield several products.
In this case, I would only have one because I only drew one set of beta-hydrogens.
But if I had more beta-hydrogens, I could react with all of those.
So one of the reasons that this is so messy is because you have the possibility for up
to five different products at times because you could sometimes get two different substitutions
and then three different eliminations depending on what your leaving group looks like.
And for synthesis that's really bad because the last thing a researcher wants to do is
to separate a bunch of molecules that are very similar.
They're going to be very difficult to separate from each other.
And the really last thing that you want to do is try to make a target drug or a target
molecule and then have other things mixed inside.
Imagine if you were buying some pills from Walgreen's and the chemist was lazy and he
decided to include enantiomers in there or decided to include other types of double bonds,
you could die.
So that's why you have to be very careful when you're doing synthesis to only get one
product and that's not what we're doing here.