E1 reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

November 13, 2019 0 By Kody Olson

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.