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- Let's think about what type of reaction might occur if we
- have this molecule right over here.
- I won't go through the trouble of naming it.
- It would take up too much time in this video.
- But it's dissolved in methanol.
- When we talk about what type of reactions, we're going to
- pick between Sn2, Sn1, E2 and E1 reactions.
- Now, maybe the first place to start or the place I like to
- start is to just look at the solvent itself.
- And when we're trying to decide what type of reaction
- will occur, the important thing to think about is, is
- the solvent protic or aprotic?
- And if you look at this solvent right
- here, this is methanol.
- It is protic.
- And in case you don't remember what protic means, it means
- that there are protons flying around in the solvent, that
- they can kind of go loose and jump around from one molecule
- to another.
- And the reason why I know that methanol is protic is because
- you have hydrogen bonded to a very
- electronegative atom in oxygen.
- So every now and then, in one of the methanol molecules, the
- oxygen can steal hydrogen's electron.
- And then the hydrogen itself, without the electron, the
- hydrogen proton will be flying around because it
- doesn't have a neutron.
- So this is a protic solvent.
- Now, you might say, well does anything with a hydrogen,
- would that be protic?
- And the answer is no.
- If you have a bunch of hydrogens bonded to just
- carbons, that is not protic.
- Carbon is not so electronegative that it could
- steal a hydrogen's electron and have the hydrogens
- floating around.
- So a big giveaway is hydrogen bonded to a very
- electronegative atom like oxygen.
- So this is protic.
- And when we think about protic out of all of the reactions we
- studied, that favors-- well, even a better way to think
- about it is it disfavors.
- So it tells us that it's unlikely to have an Sn2 or an
- E2 reaction.
- And the logic there is an Sn2 reaction needs a strong
- nucleophile.
- An E2 reaction needs a strong base.
- Now, if you have protons flying around, the nucleophile
- or the base is likely to react with the proton.
- It would not be likely to react with
- the substrate itself.
- So a protic solution, you're likely to have an Sn2 or E2.
- What you are likely to have is an Sn1 or an E1 reaction.
- Both of these need the leaving group to leave on its own, and
- actually, having protons around might help to stabilize
- the leaving group to some degree.
- So it makes Sn2, E2 unlikely, Sn1, E1 a little more likely.
- So far, these are our good candidates.
- Now, the next thing to think about is to just look at the
- leaving group itself, or see if there is
- even a leaving group.
- And over here, everything we see on this molecule is either
- a carbon or a hydrogen, except for this iodine right here.
- And we know that iodide is a good leaving group.
- Well, a good leaving group, it does not make it any less
- likely that you'd have Sn2 or E2.
- Both of those can do well with a good leaving group.
- But it's a necessary requirement for an Sn1 or an
- E1 reaction.
- Remember, an Sn1 and E1, in both of them, the first step
- is that the leaving group leaves on its own.
- That is the rate-determining step.
- So that's a requirement for Sn1 or E1, so
- it still looks likely.
- We haven't seen anything that would make us think that we
- wouldn't have an Sn1 or E1.
- This is a good leaving group.
- Let me write it here.
- Good leaving group.
- Now, the last thing that we can think about right here is
- the carbon that we might be leaving from.
- So far, everything is pointing in the Sn1 and E1 direction,
- and kind of the final thing is when this leaving group
- leaves, it's going to form a carbocation from the carbon
- that it's bonded to right now.
- And in order for that carbocation to be reasonably
- stable, at minimum it should be a secondary carbocation
- bonded to at least two carbons, but ideally, it would
- be bonded to three carbons.
- It would be a tertiary carbon.
- Now, the carbon that the leaving group is bonded to is
- a tertiary carbon.
- It's bonded to one, two, three carbons.
- so it is a tertiary carbon.
- It can actually be a stable carbocation.
- It's a tertiary carbon, which it once again
- favors Sn1 and E1.
- So all of the clues here tell us that Sn1 and E1
- are going to happen.
- And actually, they'll both happen.
- So let's think about the mechanism.
- In the very first step, the leaving group leaves in either
- one of these reactions.
- So if we look at the iodine, it already has
- seven valence electrons.
- One, two, three, four, five, six, seven.
- And in the first step, the rate-determining step of the
- Sn1 or E1 reaction, the iodine's going to nab an
- electron off the carbon.
- I'll do that electron in green right over there.
- That's going to get nabbed onto iodine to make iodide,
- and then that carbon is going to lose an electron and become
- a carbocation.
- So after that very first step, we have something
- that looks like this.
- So that's our molecule.
- Just so it's clear what we did, this arrow I'll do all
- the way over here, so it's clear that
- this is our next step.
- And what's happened here?
- Our tertiary carbon has lost its electron.
- So now this carbon right here has a positive charge.
- It's a tertiary carbocation.
- And now the iodine has become iodide.
- It has left the molecule.
- So it had its original seven valence electrons: one, two,
- three, four, five, six, seven.
- It nabbed one more electron from the carbon and now it
- is-- I wanted to do that in green.
- It nabbed one more electron from the carbon, now it is the
- iodide anion.
- So this step right here is common to
- both Sn1 and E1 reaction.
- The leaving group has to leave.
- Now, after this, they start to diverge.
- In an Sn1, the leaving group essentially gets substituted
- with a weak nucleophile.
- In an E1, a weak base strips off one of the beta hydrogens
- and forms an alkene.
- So let's do them separately.
- So over here, I'm going to do the Sn1.
- And on the right-hand side, I will do the E1 reaction.
- So let me start over here.
- So the Sn1 is starting over here at this step.
- I'll just redo this step over here.
- So this has a positive charge.
- That has a positive charge here.
- The iodide has left.
- I don't have to draw all its valence electrons anymore.
- And what's going to happen next?
- We're going to get substituted with the weak base, and the
- weak base here is actually the methanol.
- The weak base here is the methanol.
- So let me draw some methanol here.
- It's got two unbonded pairs of electrons and one of them,
- it's a weak base.
- It was willing to give an electron.
- It has a partial negative charge over here because
- oxygen is electronegative, but it doesn't have a full
- negative charge, so it's not a strong nucleophile.
- But it can donate an electron to this carbocation, and
- that's what is going to happen.
- It will donate an electron to this carbocation.
- And then after that happens, it will look like this.
- That's our original molecule.
- Now this magenta electron has been donated to the
- carbocation.
- The other end of it is this blue electron right here on
- the oxygen.
- It is now bonded.
- That is our oxygen.
- Here's that other pair of electrons on that oxygen, and
- it is bonded to a hydrogen and a methyl group.
- And then the last step of this is another weak base might be
- able to come and nab off the hydrogen proton right there.
- Oh, I want to be very clear here.
- The oxygen was neutral.
- The methanol here is neutral.
- It is giving away an electron to the carbocation.
- The carbocation had a positive charge because it had lost it
- originally.
- Now it gets an electron back.
- It becomes neutral.
- The methanol, on the other hand, was neutral, gives away
- an electron, so now it becomes-- it now is positive.
- So now you might have another methanol.
- You might have another methanol molecule sitting out
- here someplace that might also nab the proton off of this
- positive ion.
- So this one right here, it would nab it or it
- would bond with it.
- It would give the electron to the hydrogen proton, really.
- The hydrogen's electron gets nabbed by the oxygen, and so
- then that becomes neutral.
- So in the final step, it'll all look like this.
- We have that over here.
- The methanol that had originally bonded has lost its
- hydrogen, so it looks like this.
- We just have the oxygen and the CH3 there.
- It is now neutral because it gained an electron when that
- hydrogen proton was nabbed.
- So if you wanted to draw it, it has actually those two
- extra electrons, just like that.
- And if you want to draw this last methanol, it's now a
- positive cation, so it looks like this.
- So it's OH, CH3, H, and then it has unbonded pair right
- there, and now this has a positive charge.
- So that was the Sn1 reaction.
- Now, the other reaction that's going to occur is the E1.
- Once again, our first step-- nope, I
- didn't want to do that.
- Our first step looks like that.
- So that is our first step.
- Let me get everything straight.
- So the leaving group had left.
- So in each situation, the leaving group had left.
- Our iodide is up here.
- And in an E1 reaction, you don't get substituted.
- What happens is one of the beta carbons gets a hydrogen
- swiped off of it by a weak base.
- Now, let's think about what a beta carbon is.
- The alpha carbon is the carbocation carbon.
- That's right over there.
- That's the alpha carbon.
- Beta carbons are one carbon away.
- So this is a beta carbon, this is a beta carbon, and then
- that is a beta carbon.
- Now, this carbon over here is not bonded to any hydrogens.
- It's only bonded to other carbons.
- So that one cannot lose any hydrogens.
- And then we have to pick between this carbon that's
- bonded to three hydrogens and this carbon that's actually
- bonded to two hydrogens.
- I didn't draw it before, but it's implicitly there.
- It's bonded to two hydrogens.
- Now Zaitsev's rule tells us that the dominating product is
- going to be produced when the carbon that has less hydrogens
- loses a hydrogen.
- So out of this, this carbon has two hydrogens,
- this one has three.
- So the one that's going to lose the hydrogen to the base,
- or more likely to lose the hydrogen to the base, is the
- one that has two hydrogens, not
- three, the fewer hydrogens.
- So our base in this case is once again the methanol, acted
- as a nucleophile in the Sn1, acted as a weak
- nucleophile in Sn1.
- Now it'll act as a weak base.
- So we have methanol right over here.
- That's a hydrogen.
- CH3 has some electrons right over here.
- It has a partial negative charge.
- It will give one of the electrons to the hydrogen,
- just to the hydrogen proton.
- The hydrogen's not going to take its electron with it.
- That electron is then going to be given to the carbocation to
- make it neutral, and it will form a double bond.
- So after that happens, we are left with something
- that looks like this.
- This was our original molecule.
- Now, this carbon right here in yellow, it
- has lost its hydrogen.
- This hydrogen back here is still there.
- I could draw it if I like.
- I don't need to.
- It just makes the thing messy.
- But it has now formed a double bond with the primary carbon.
- It has now formed a double bond.
- That electron that had bonded with the hydrogen was now
- given to the carbocation.
- It has a double bond with that.
- And then you have your methanol has now turned into a
- positive ion.
- It has now turned into a positive ion: CH3.
- And now it has this bond.
- It has this bond with the hydrogen.
- I'll even make that electron that it gave away in magenta.
- And then it has that extra lone pair of electrons.
- So, in this circumstance, we looked at all of the clues.
- All of the clues were against Sn2 and E2.
- They favored Sn1 and E1.
- So if you were to actually make this, if you were to
- actually try to see what happens is this reaction, you
- would get products of both Sn1 and E1 reactions, these
- products and these products over here.