All right, so I know that I already told you guys that this was an E1 reaction, but how
would you know that if I wouldn't have told you?
Now just so you know, this entire question of what mechanism do I use is going to be
the very last topic that we end on for this lesson.
And I'm going to spend a lot of time explaining how to determine that.
But even before we have all the tools, we can already start to just use our logic of
knowing nucleophiles and leaving groups to figure this out.
So what I would say is what type of nucleophile do I have.
Is it strong or is it weak?
Well, in this case, MeOH stands for methanol.
This would just be CH3OH.
Is that negative charged or neutral?
That's neutral.
If that's neutral, that means that it's weak.
That means that is it going to start off by doing a back side attack or an E2 elimination?
No.
It's just going to chill.
It's just going to stay there for a while.
What type of leaving group do I have?
I have a tertiary.
What is tertiary really good at doing?
Making carbocations.
So I know that my first step is going to be to make a carbocation and that means that
my mechanism is going to be a combination of E1 and SN1.
Is that cool?
Awesome.
So let's go ahead and draw this mechanism.
The first part of the mechanism is going to be the same for all the products.
It's just going to be that my F leaves on its own.
What's that's going to do is it's going to give me a carbocation that looks like this.
And that was my – right here this is my rate – that's my rate-determining step.
Now what happens after this is what determines which path I take.
If my methanol pulls off a beta-proton, that is going to be an elimination.
So how many different types of beta-protons do I have?
That's my alpha.
By the way, my alpha is now whatever has the carbocation.
Well, I have three betas, beta, beta, beta.
Do all of them have at least one hydrogen on them?
Yes.
Do all of them – are they all the same, are they different?
They're all the same, so it doesn't matter which one I pick.
I'm just going to pick this one over here for my H. Remember that also anti-coplanar
doesn't matter because this is E1.
I'm going to take my CH3OH and I'm going to pull off that proton.
So what I'm going to get for the E1 mechanism is that and that E1.
And what I'm going to wind up getting is the double bond that looks like this.
Is that cool?
So that's my elimination product and that would have been – that's my only product.
But remember that I could have gotten a mixture of products if I would had more than one type
of beta-hydrogen.
Cool So that's the E1, so we're done with that.
But like I said, every time you get an E1, now you're going to have to be thinking about
SN1 as well, so let's just go ahead and draw that really quick.
So remember that for SN1, and this is also good practice of the SN1 mechanism.
So remember that for SN1, what I would do is I would take my CH3OH and I would directly
attack the carbocation.
What that's going to give me is a new compound that looks like this, where I have the three
methyl groups and now that's attached to O with an H in one place and with a CH3 on another
and a positive charge.
Remember that that SN1 was a little bit annoying because remember that SN1 always requires
a proton removal step or what we just called deprotonation.
This is the only mechanism that requires a deprotonation because you always get a formal
charge at the end.
So do you guys remember what I could use to take away that H?
I could just use another – I could use the thing that left which would just be my F-.
So I have my F- and I'm going to pull it off.
F to H and then these electrons down to O .
Now I have to ask myself, did I make a new chiral center.
Are there any chiral centers on that molecule, on that SN1 right here?
No, that's not chiral, so I'm only going to get one product and it would just look like
this OCH3.
This part was my SN1.
It's hard know where to put it.
I guess I'll just put it here.
This one was my SN1 where I attacked directly.
The red one was my E1 where I did a beta-elimination.
And in the end, I would up with two different products.
Now this is honestly one of your best case scenarios for E1 and SN1 where you just get
one product for each.
But like I said, sometimes you can get up to five products.
The reason five is because you could have three different double bonds because you have
three different types of eliminations and then you could get two enantiomers for the
SN1 and that's just crazy.
So I'm not going to make you draw all five, but we are going to draw now question (b)
and question (b) is going to have more products than the one I just drew.
So go ahead and try to get all of the products for (b) using both mechanisms and also the
logic of why we're going to use these mechanisms.
So go for it.