Tip:
Highlight text to annotate it
X
In the video on E2 reactions, I showed you how a strong
base-- and all of this happened simultaneously-- a
strong base can nab a hydrogen off of this carbon right here.
And it's just nabbing the proton itself.
It's not grabbing the hydrogen and the electron.
And then that electron goes to this carbon right over there,
and then that allows that carbon to give away the
electron that was forming a bond with the chlorine to go
with the chlorine.
This all happened at once.
The chloride got eliminated.
Now, one thing that might pop out in your brain is why did I
pick this hydrogen?
Why did I pick the hydrogen right over here?
Why couldn't I have picked that hydrogen over there?
And I'm going to introduce you to a little bit of terminology
and then I'll introduce you to a rule.
And then I'll tell you a little bit about why people
think this rule works.
So in general, the carbon that has the functional group on
it, that's the alpha carbon.
So let me label it.
This carbon right here is the alpha carbon.
And in order to have an E2 reaction in this case, when we
did the video on E2 reactions, but actually, the rule will
hold as well for E1 reactions.
But in order to have the E2 reaction, the hydrogen has to
get swiped off of a beta carbon.
And a beta carbon is just a carbon that's one away from
the alpha carbon.
So this is a beta carbon and this is also a beta carbon.
And so it's completely reasonable one would think
that, well, I could swipe it from there or I could swipe it
from there.
And let's think about this reaction.
Let's just draw it out, so you can visualize it
a little bit better.
So here we swiped it from this beta carbon.
Let me redraw the reaction where we're swiping it from
the other beta carbons.
I want both reactions on the screen at the same time.
Let me draw our methoxide.
So we have our oxygen bonded to a CH3 to a methyl group.
The oxygen has seven valence electrons: one, two, three,
four, five, six, seven.
I'll do the seventh one.
The one that will bond with the hydrogen,
I'll do it in green.
Instead of attacking that hydrogen or taking that
proton, I should say, because it's not taking the electron
with it, it does it to this one.
So what it does is it takes this proton or it bonds with
that hydrogen proton, and then the hydrogen protons or that
hydrogen's electron can then be taken by this molecule, so
then it goes to the alpha carbon to form a double bond
between this beta carbon and the alpha carbon.
And so now, this alpha carbon got that electron.
It doesn't need the electron that's bonded with the chloro
group anymore, and so that goes to the
chlorine to form chloride.
Chlorine was already way more electronegative.
It was already hogging it.
Now it gets to go there.
And now, when all is said and done, our
products look like this.
So we still have the methanol just like we had in the
original reaction because this grabbed this hydrogen.
Let me draw it.
So you have your OCH3.
Let me draw all of this in.
You have that pair right over there.
You have this pair right over here.
And then you have this purple electron and now it's bonded
with this green electron, which is now on the hydrogen.
So it is now bonded with this green electron that has been
given to the hydrogen.
Oh, and make sure we don't forget.
This oxygen over here had seven valence electrons, so it
had a negative charge.
Neutral oxygen would have six.
So this had a negative charge.
But now that it gave its electron to this hydrogen, it
now has a neutral charge.
It is now methanol.
Exactly what we saw when we first
learned about E1 reactions.
We also know the chloro group, it took that electron.
It's now a chloride anion.
So let me draw that.
And that's exactly the same as in the
first video on E1 reactions.
So it's now a chloride anion.
It has grabbed this orange electron.
Let me do it in orange.
It has grabbed this orange electron, so it now has a
negative charge.
You can imagine that the negative charge has been
transferred from the methoxide to the chloride anion.
And now what's different this time is that the double bond
is now between the one and the two carbon and not between the
two and the three carbon.
So now it's going to look like this.
So now, the result of this product, if the reaction went
this way would look like this.
We have this carbon right here.
It is bonded to two hydrogens.
Now, it has a double bond with this carbon.
It has a double bond, and I'll do the second bond.
I'll do the pi bond of the double bond in purple right
over there.
I'll assume that that's the pi bond.
That's the new double bond form.
And now this carbon, which was the alpha carbon,
is right over here.
This was the alpha carbon.
It's bonded to one hydrogen.
And then let me draw everything else.
So then you have a carbon, a carbon.
This guy is bonded to three hydrogens.
I could have just written it as CH3 if I wanted.
This guy's bonded to two hydrogens.
And we're done.
So instead of, as we saw on the first video on E2
reactions, instead of forming but-2-ene, we now have-- it's
still one, two, three four carbons, so it's still but-,
but the double bond is on the one carbon.
We'd start one, two, three, four, So we could call this
but-1-ene or 1-butene, either way.
So let's call this but-1-ene.
So the question is, which is more likely to happen?
Do both happen?
Does one happen disproportionately?
And the answer is, is that, yes, one happens
disproportionately.
This one, this is the dominant product.
If you were to perform this reaction and you were to
analyze in your beaker or wherever you're performing the
reaction what you see most of, the majority product, the
great majority, is going to be the
but-2-ene, not the but-1-ene.
Maybe you see very, very little of this.
And the question is why?
Or how would you even be able to determine that?
Or how could you have predicted that?
And to predict it, there's something
called Zaitsev's rule.
And I'm sure I mispronouncing it, but let me write it down.
So Zaitsev's rule.
And it's kind of analogous to Markovnikov's rule, but for
elimination reactions.
If you think about it, the addition reactions that we did
many videos ago are the opposite of
the elimination reactions.
In the addition reactions, we're adding the chloro group,
and in the elimination, we're taking it off.
And so Zaitsev's rule is kind of analogous to
Markovnikov's rule.
Now first, I'll just tell you the rule, then we can think a
little bit about why it works.
The jury's not out on this.
They think they know why it works, but
they're not 100% sure.
So Zaitsev's rules says the carbon that is going to lose
the hydrogen is the one that has fewer hydrogens.
So let me write it down over here.
Carbon more likely to lose hydrogen is-- I should say the
hydrogen proton because it keeps the electron still-- is
the one with fewer hydrogens.
So if you were to look at this reaction right here, we have
our alpha carbon.
Either this beta carbon or this beta carbon could lose
its hydrogen.
This one has three hydrogens on it.
This one only has two.
So Zaitsev's rule tells us that this is the hydrogen, or
actually the proton, that is more likely to be
reacted with the base.
You could almost view it as it is the more acidic proton.
It is a lower-hanging fruit for this
strong base to capture.
Now, a more interesting question-- and that's a pretty
easy rule to follow.
And if they both have the same number, then you'd see equal
products depending on which side it gets.
Now the question is why is this happening?
And here, something called
hyperconjugation comes into effect.
I'm not going to go into details in it and to the
quantum mechanics of it.
And hyperconjugation is the notion that the fact-- so we
said that the one with the fewer hydrogens is the one
that's less likely to lose.
Or the one with fewer hydrogens is the one more
likely to lose the hydrogen proton.
But the one with fewer hydrogens is also bonded to
more carbons.
This guy's bonded to one carbon outside
of the alpha carbon.
He's actually bonded to two: the alpha and this carbon.
This guy right here is only bonded to the alpha carbon.
And hyperconjugation is the notion that not the beta
carbon, but the carbons one over from that help stabilize
the double bond that eventually forms. I almost
think of it you have more electrons over here because
carbons have more electrons to offer than hydrogens.
At the end of the day, this guy is more likely able to
donate electrons to form from the right-hand side to make a
double bond than from the left-hand side.
I won't go into the detail of hyperconjugation, but it's all
based on the notion that the more stable double bond will
be formed if we have other carbons near the double bond.
Now, another way to think about is
to look at the products.
So we saw or Zaitsev's rule tells us that but-2-ene is a
more likely product than but-1-ene.
And if you look at but-1-ene, we could rewrite it like this.
We could draw the double bond like this.
This carbon is what was the alpha carbon.
We could draw a carbon right here.
And then it is bonded to a hydrogen.
It's bonded to this hydrogen.
And then it's bonded to just a chain of carbons.
We'll just write R on that.
And then this guy is just bonded to two hydrogens.
Well, this isn't necessarily but-1-ene.
I just put an R here, but this is how it could be
represented.
Now, the but-2-ene, if we wanted to draw it like this
would look like this.
We could call this right here R prime.
It's just a chain of carbons.
And then we could call this R prime prime.
It's not even a chain.
It's just one carbon.
But if we call it that, then the but-2-ene-- let me draw it
down here where I have more real estate.
The but-2-ene would look like this.
You have your carbon-carbon double bond.
Now, the left-hand carbon is bonded to a hydrogen, that
hydrogen right there, and to R prime.
And the right-hand carbon is bonded to
hydrogen and R prime prime.
So it's bonded to a hydrogen and R prime prime.
All I did here-- let me see if I can fit it all on the same
screen-- is I just redrew this and I just abstracted away the
chain as it goes away from the double bond.
And I did that so that we can look at it this way.
We can just have the double bond kind of as our focus of
attention and think about what's going on around it.
Over here, for the but-1-ene, and we already said this is
the lesser product, so this is the greater product.
This is the greater, or the dominant product.
In the lesser product, if we go off of the double bond, we
only have one alkyl group right there.
That R right here.
Over here, we have two.
And we say that this is more substituted.
And when w say substituted, you're imagining that you're
substituting hydrogens with carbon
chains, with alkyl groups.
So this one right here is more substituted.
And hyperconjugation, so the idea would have it, is that
these carbon chains that are near the double bond help
stabilize it.
Some of their sigma electrons and sigma orbitals are there
to somehow help stabilize the pi orbitals.
And now this is getting into quantum mechanics and all of
that, so it's a little bit-- you know, the world isn't 100%
clear whether that's definitely the mechanism,
although people have run the experiment, and they know that
the more substituted product is what you're going to see
more of as opposed to the less substituted, and that all
comes from Zaitsev's rule.
So, hopefully, you at least get Zaitsev's rule.
The hyperconjugation, that's kind of a deeper concept.
You know, the jury's not even out on it.
That's a belief of why Zaitsev's rule works.
But the rule itself is pretty straightforward.
If you're trying to pick between two beta carbons, the
one that's going to lose the hydrogen is the one that
already has pure hydrogens or the one that's
bonded to more carbons.
And this is true.
I drew it.
Everything we focused on right now was in an E2 reaction.
But it's just as true in an E1 reaction.