E2 reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

November 17, 2019 0 By Kody Olson


Let’s try to come up with the
reaction of when we have this molecule right here reacting
with sodium methoxide in a methanol solution, or with
methanol as the solvent. Just so we get a little practice
with naming, let’s see, this is one, two,
three, four carbons. So it has but- as a prefix and
no double bond or triple bonds, so it’s butane. And we have a chloro
group here. So if we start numbering
at the side closest to it, one, two. So it’s 2-chlorobutane. Let’s think about what
might happen here. The first thing– let
me just redraw the molecule right there. The first thing you need to
realize is this sodium methoxide is a salt. When it’s not dissolved,
it’s made up of a positive sodium cation. Let me draw them right here. It’s a positive sodium
cation and a negative methoxide anion. Let me draw the methoxide
part right here. It normally would have just
two pairs, but now we have three pairs here. The oxygen has an
extra electron. Actually, let me draw
another electron as the extra electron. This’ll be useful for
our mechanism. It could be any of them. The oxygen has an
extra electron. It has a negative charge. In solid form, when they’re not
dissolved, they had formed an ionic bond and they form
a crystal-like structure. It’s a salt, but when you
dissolve it in something, in a solution, in this case, we have
methanol as the solution, they will disassociate
from each other. And what you have right here is
this methoxide right here. This is a very, very,
very strong base. This right here is
a strong base. And if you use the Lewis
definition of a base, that means it really, really, really
wants to give away this electron to something else. If you use the Bronsted-Lowry
definition, this means that it really, really, really wants
to take a proton off of something else. In this situation, that is
exactly what it will do. I’ll actually give you the most
likely reaction to occur here, and we’ll talk about other
reactions, and why this is the most likely reaction
in future videos. So it wants to nab a proton. It is a strong base. It wants to give away
this electron. Let’s say that it gives away
this electron to this hydrogen right over here. Now, this hydrogen already
had an electron. If it’s getting an electron from
the methoxide anion, the methoxide base, then it can give
away its electron to the rest of the molecule. It can give away this electron
to the rest of the molecule. Now, carbon won’t need
the electron. Carbon doesn’t want to have
a negative charge. Maybe simultaneously that
electron goes to that carbon right over there. But once again, this carbon
doesn’t want it. It already has four bonds, but
what we see is we have this chloro here. This is a highly electronegative
group. The chlorine is very
electronegative, so the whole time, the chlorine was already
tugging on this electron right here. Now, all of a sudden, it’s all
happening simultaneously, when an electron becomes available
to this carbon, this carbon doesn’t need this electron
anymore. The chlorine already wanted it,
so now the chlorine can take the electron. And just like that, in exactly
one step, let’s think about what happened. If all of these simultaneous
reactions occurred, what are we left with? Let me redraw my two
chlorobutanes. I’m going to have to
change it now. So now, the chlorine
has disappeared. So we clear it. That’s disappeared. The chlorine has now left. This chlorine is now up here. It has left. It had this electron right over
there, which is right over there. The other electron it was paired
with that was forming a bond with is now also
on the chlorine. It becomes a chloride anion. Let me draw the rest of
the valence electrons. One, two, three, four, five,
six, and it has the seventh one right here. One, two, three, four, five,
six, seven, eight, so it now has a negative charge. This electron up here– let me
clear this part out as well. Now, this electron right here,
this magenta electron, this is now given to this carbon. Let me draw it here. That magenta electron is now
given to that carbon. And if we look at the other
end, the one that it was paired with, that it was bonded
with, it still is bonded with it, so that green
electron is now still on that carbon and now they
are bonded. Now, they’re forming
a double bond. This all of the sudden has
become an alkene and now the methoxide took the hydrogen. Let me redraw the methoxide. OCH3, the oxygen had one,
two, three, four, five. It had six. Actually, it had seven valence
electrons, one with the carbon and then all of these
six unpaired ones. Neutral oxygen has six. But now it gave one of them
away to a hydrogen. It gave that green electron
there to the hydrogen– I’ll make this hydrogen the same
color– to this hydrogen right over here, so now
this hydrogen is now bonded with it. So what are we left with? We have a chloride anion,
so this is chloride. We now have methanol. This was a strong base. Now it has become its
conjugate acid. So now we have methanol, which
is the same as the solvent, so it’s now mixed in. And now we’re left off
with one, two, three, four, still four. We still have four carbons,
but now it’s an alkene. We have a double bond, so we
could call this but-2-ene, or sometimes called 2-butene. Let’s think about what
happened here. It all happened simultaneously. Both reactants were present. There was actually only one
step so this is the rate-determining step. If we would try to name
it, it would probably have a 2 in it someplace. And something got eliminated. The chloride, or I guess you
could call it the chloro group, got eliminated. It left. It was a leaving group
in this situation. So this was eliminated, and this
type of reaction where something is eliminated and
both of the reactants are participating in the
rate-determining step, and we only had one step here so that
was the rate-determining step, is called an E2 reaction. And once again, E stands
for elimination. And the 2 stands for that both
reactants were involved in the rate-determining step. We had one reactant
and two reactants. They were both involved with
the rate-determining step. E2, just like Sn2. It’s obviously not the same
reaction, but in Sn2, we had substitution. The leaving group left,
nucleophile came. It was substitution using
a nucleophile so that’s the S and the n. And then the 2 was we
had both reactants. It was all happening
simultaneously. I’ll leave you there. We’re going to go into more
detail on the different types of each reactions, which ones
are favored, and then we’ll talk a little bit about E1. You could maybe guess what
that’s going to be based on our experience with
Sn2 and Sn1.