Now I've given you an example here of basically an alkyl halide with a nucleophile.
Let's use the Big Daddy flowchart to figure out what mechanism this would be.
Let's go ahead and ask our first question.
Notice that my nucleophile is NaOEt.
Is that negatively charged or is that neutral?
For those of you that said neutral, you're forgetting that sodium can dissociate.
So what it's actually going to look like is OEt- so that's going to be a negatively charged
nucleophile.
That's going to go down the left-hand side of the flowchart.
So let's go to step two.
Step two is is NaOEt one of my bulky bases.
No.
We have a list of bulky bases, NaOEt is not one of them.
So I'm just going to say no.
Let's go to my third question.
What type of alkyl halide do we have or what type of leaving group?
Well, this carbon right there is attached to two other carbons, one, two, so this would
be a secondary alkyl halide.
So do we know the mechanism now?
No, we have to ask one more question.
The last question is – I'm just going to put it down here.
Is my base a better nucleophile or a better base?
So for this, you have to remember what were the strong bases.
Is NaOEt one of those strong bases?
Yes, it is.
Remember that one of the strong bases was oxides.
Oxides have the general formula of OR-.
And that's exactly what we have.
We have OEt, which is an ethyl group, negative.
So this is an oxide, so this is going to favor E2.
Cool.
So now we're got E2.
Now we have to figure out how do we actually draw the mechanism for this and how do we
predict the products.
Do you guys remember what the first step of E2 is?
Figure out how many different beta-protons I have.
So, what have we got?
We've got two different beta-carbons.
Let's say this is beta one and let's say this is beta two.
Those are my two different options.
Do both of them have hydrogens on them?
Yes, they both do.
On the green one, I've got a hydrogen towards the front and a hydrogen towards the back.
On the red one, I just have a hydrogen towards the back.
So are we ready to eliminate yet?
We have to ask ourselves one more question.
Now that we know all of our beta-protons, which is three.
How many of them could actually react in an E2 reaction using the anti-coplanar rule?
Remember you always have to think of that rule.
The answer is two of these could react.
I could eliminate in the green direction with this one right there.
And I could also eliminate in the red direction using that one right there.
The reason is because my chlorine is facing towards the front, so I can only eliminate
with a hydrogen that's facing towards the back.
So now that I know that, let's go ahead and draw one of the mechanisms.
We don't have to draw both, but let's just draw one of them and then predict the products.
So the mechanism would be three arrows, just like it always is.
Let's say I'm taking off the red H. then what I'm going to do is I'm going to dump my electrons
into the bond between the alpha and the beta and then I'm going to kick out my Cl.
So what that means is I'm going to have two different products possible.
I'm going to have a product possible where I eliminate where I just did, so it would
be a double bond right there with now no Cl.
I don't have to draw that and a methyl group facing down like that.
Now, why is it important that I drew my methyl group on a stick?
Because remember now this is trigonal planar If I draw it on a wedge that would look like
I have no clue what I'm doing.
I would kind of look like an organic noob and you don't want to look like a noob when
you're trying to get points from your professor.
So that's the way we would draw that product.
Obviously, I would have my Cl negative leaving group.
But there's also another product I could have gotten.
The other product would have been if I eliminated up to the green hydrogen and that would have
given me a product that looks like this.
A double bond there and – let me just straighten that out a little bit.
And that methyl group would still be on the wedge.
Why am I drawing it on the wedge here?
Because that carbon is still tetrahedral because it still has an H on it.
So that is a carbon with four different groups around it so that one should be drawn as a
tetrahedral.
So now we've drawn our two different products.
Are they both going to be formed in equal amounts?
No.
It turns out that Zaitsev's rule explains – just look down here – that we will always
favor the more substituted thermodynamically stable product.
Now, what the hell does thermodynamically stable mean?
I know that whenever you bring thermodynamics or kinetics, things get confusing.
It just means overall which one is going to have the least energy.
Which one is going to be the overall most stable at the end?But guess what?
We know how to figure this out because we can just use the alkene stability rules to
determine that.
So out of these two double bonds which of them is going to be more stable and which
one of them is less stable.
What do you guys think?
This is like its own question now.
Well, this one here is trisubstituted.
This one here is disubstituted.
So which one going to be the overall more stable product?
Red.
Red is going to be more stable than green.
Which one is the Zaitsev product?
Red.
Because Zaitsev is the more stable one.
This one's the less stable, it's Hofmann.
So instead of saying less stable/more stable, from now on I'm just going to be using the
words Zaitsev and Hofmann because guess what?
Those are just synonyms of the less stable and more stable are just Zaitsev and Hofmann.
I can just use those words.
So now we have to figure out which one is major and which one is minor.
Which one am I going to form in higher amount than the other?
What Zaitsev's rule explains is that I'm going to favor the Zaitsev product.
That means that this is going to be my major product and this is going to be my minor product.
Does that mean that I only get one of these?
No, I still get both, but I'm going to get a lot more of the red and a lot less of the
green.
Now sometimes your professor may ask a question that just says give the major product for
the reaction.
If they're asking for major product, then you would just draw red.
But if your professor asks give all the products, then you would draw both of them.
Does that make sense?
But many times especially in multiple choice exams, if you have any multiple choice component,
many times your professor is just going to say draw the major product or select the major
product.
If you're doing online homework and it says select the major product, it would be this
one.
If you're just checking different resources out online, this would be the more stable,
more favored product.
Awesome guys.
So Zaitsev's rule's pretty easy, right?
We just use the number of R groups to figure out which one is more stable, which one's
less stable.
Select the more stable one.