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- 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.