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  • What I want to do in this video is to try to figure out

  • what type of reaction or reactions might occur if we

  • have-- what is this?

  • One, two, three, four, five It's in a cycle.

  • This is bromocyclopentane.

  • If we have some bromocyclopentane dissolved

  • and our solvent is dimethylformamide.

  • Sometimes you'll see that just written as DMF.

  • And I've actually drawn the formula for it here, so we can

  • think about what type of a solvent it is.

  • And also, in our solution, we have the methoxide ion.

  • So we also have the methoxide ion right here.

  • So let's think about what type of reaction might occur.

  • And just to narrow things down, we'll think about it in

  • the context of the last four types of reactions

  • we've looked at.

  • So this might be an Sn2 reaction, an Sn1 reaction, an

  • E2 reaction, or an E1 reaction.

  • We're going to look at all the clues and figure out what's

  • likely to occur, and then actually draw the mechanism

  • for it occurring.

  • Now, the first thing since they gave us the solvent and

  • other things that are in the solvent, let's think about how

  • those might affect the reaction.

  • So if we look at this solvent right here, whenever you look

  • at any of these reactions, when you look at the solvent,

  • you just want to think about it.

  • Is it protic or not?

  • And protic means that it has hydrogens that can kind of be

  • released or that their electrons could be nabbed off

  • and these protons could just float around.

  • And if we look over here, we do have hydrogens, but all of

  • the hydrogens are bonded to carbon.

  • And carbon is unlikely to just steal a hydrogen's electrons

  • and let the hydrogen float around.

  • Carbon is not that electronegative.

  • If you had hydrogens bonded to an oxygen, that'd be a

  • different question.

  • Then you would have a protic solvent.

  • But in this case, all the hydrogens bonded to carbons,

  • not likely to get their electrons nabbed off and float

  • around as free protons.

  • So this is an aprotic solvent.

  • Now, we've gone over this a little bit with Sn2 and Sn1,

  • but the same idea applies.

  • In order to have an Sn2 or an E2 reaction, you have to have

  • either a strong nucleophile or a strong base, and the same

  • thing could actually be both, although they're not always

  • correlated.

  • We've seen that before.

  • Now, if you had a protic solvent, it would stabilize

  • the strong base or the strong nucleophile.

  • The protons would react with them.

  • They would take the electrons from that strong base or that

  • strong nucleophile.

  • So in order to have an Sn2 or an E2, you have to have no

  • protons flying around, so you need an aprotic solvent.

  • So this aprotic solvent will favor Sn2 or an E2 reaction.

  • Now, so our mind is already thinking in Sn2 or E2, let's

  • think about the reactants themselves.

  • So over here, we have the methoxide ion.

  • And let's think about whether it's a strong or weak.

  • Let's think about it first as a strong or weak nucleophile.

  • It's actually a pretty strong nucleophile.

  • It is a strong nucleophile.

  • So that would put us in the direction of an Sn1.

  • So we have two data points.

  • I'm sorry, for an Sn2.

  • We have two data points for Sn2 because remember, it has

  • to just kind of go in there and be active.

  • It's not too big of a molecule, so it's not going to

  • be hindered.

  • But it's also an extremely strong base, even stronger

  • than hydroxide.

  • So it's also an extremely strong base, which might lead

  • us or that does imply that we're going

  • to have an E2 reaction.

  • Now, the last thing we need to think about is the carbon

  • where the leaving group might leave from.

  • And immediately, when you look at the bromocyclopentane,

  • there's only one functional group attached to the chain,

  • and that is the bromo group right here, right there.

  • It is attached to this carbon.

  • We could call that the alpha carbon, and it

  • is a secondary carbon.

  • This carbon right here is bonded to

  • one, two other carbons.

  • This alpha carbon-- let me write it this way.

  • This alpha carbon is a secondary carbon, and that

  • kind of makes it neutral in this mix.

  • If it was a methyl or primary carbon, it

  • would favor Sn2, actually.

  • I mean methyl, the only thing you could have is an Sn2.

  • And if it was a tertiary carbon, it would favor Sn1 or

  • E1 because it would favor a stable carbocation.

  • The leaving group could just leave. And if this guy was

  • bonded to another carbon, it would be very stable.

  • But in this situation, it's a secondary carbon

  • bonded to two carbons.

  • It's a little bit neutral.

  • Any of these reactions might occur.

  • When we look at all of the other data points, they're

  • pointing at both Sn2 or E2.

  • We have a strong nucleophile/base.

  • We have an aprotic solvent.

  • It's going to be Sn2 or an E2 reaction.

  • So let's actually draw the reactions.

  • Let me do the Sn2 first. So let me do it in orange.

  • So if we were to have an Sn2 reaction, let

  • me redraw the molecule.

  • Let me draw the cyclopentane part.

  • I want to make sure-- let me draw it the same way I had it

  • drawn up there.

  • So the pentagon is facing upwards.

  • And then we have our bromo group right there.

  • So we have our methoxide ion right over here.

  • So CH3O minus.

  • Or another way we could view it is that this oxygen has

  • one, two, three, four, five, six, seven valence electrons

  • with a negative charge.

  • One of these electrons right over here, this can attack the

  • substrate right over there, that carbon.

  • Right when that happens, simultaneously this bromine is

  • going to be able to nab an electron

  • from that same carbon.

  • And then we are going to be left with-- the bromine now

  • becomes the bromide anion.

  • It had one, two, three, four, five, six,

  • seven valence electrons.

  • One, two, three, four, five, six, seven.

  • Now it nabbed one more electron, making it bromide.

  • Now it has a negative charge.

  • And if we were to draw the chain, it

  • would look like this.

  • Well, we could draw it on this, and I might as well draw

  • it on this side, just so it's attacking from the other side.

  • This is the chiral substrate, so we don't have to be too

  • particular about how we draw the connections to the carbon.

  • We're not actually even showing anything

  • popping in or out.

  • But we would have the methoxide ion, where now it's

  • bonded, so it's no longer an ion, so it's

  • OCH3, just like that.

  • It has bonded to this carbon.

  • Obviously, implicitly this carbon had another hydrogen

  • that we are not showing.

  • Just that quickly, that was the Sn2 reaction.

  • That is the mechanism.

  • Now let's think about what the E2 reaction is.

  • To do the E2 properly, to give it justice, we're going to

  • have to draw some of the hydrogens.

  • So on the E2 reaction, let me draw that in blue.

  • Let me draw the cyclopentane part.

  • Let me draw it big.

  • Actually, over here, it's less important to draw it too big.

  • So let me draw the pentagon.

  • The pentagon just like that.

  • That is the bromine, three, four, five, six, and then it

  • has a seventh valence electron right over here.

  • This is the alpha carbon.

  • That right there is the alpha carbon.

  • And then there are two beta carbons.

  • There are two beta carbons right over there and there.

  • They each have two hydrogens on them.

  • They each have two hydrogens.

  • I know it's becoming a little hard to read.

  • They each have two hydrogens on them.

  • And in an E2 reaction, the strong base will react-- let

  • me make it a little cleaner than that.

  • Let me get rid of the beta.

  • The beta makes it's a little dirty.

  • OK, so they each have two hydrogens on them.

  • Now in an E2 reaction, the strong base-- over here, the

  • methoxide ion was acting as a strong nucleophile.

  • In E2, it's going to act as a strong base.

  • It's going to nab off a hydrogen off of one of the

  • beta carbons.

  • And you might want to say, OK, which one?

  • Let's look at Zaitsev's rule.

  • It doesn't matter.

  • These are symmetric.

  • They are both bonded to two other carbons.

  • They both are bonded to the same number of hydrogens.

  • It doesn't matter.

  • It's actually going to be random which one, and you

  • actually won't be able tell the difference

  • because it's symmetric.

  • So let's just draw it like this.

  • Let me draw the methoxide ion.

  • One, two, three, four-- or anion, maybe I

  • should say-- five, six.

  • And then it has one bond to the CH3.

  • It has a negative charge, very, very, very strong base.

  • It can go over here and nab the hydrogen and leave

  • hydrogen's electron behind.

  • Maybe I'll take a color.

  • This electron can be given to the hydrogen so that it forms

  • a bond with it.

  • Hydrogen's electron-- let me do this in a suitably

  • different color.

  • Hydrogen's electron that is sitting right over there can

  • now be given to the alpha carbon.

  • It can now be given to the alpha carbon to

  • form a double bond.

  • And now that the alpha carbon is getting that electron, now

  • the bromo group can leave. It's a decent leaving group.

  • And that was another thing that we should think about in

  • our equation.

  • But a good leaving group actually favors all of the

  • reactions: Sn2, E2, Sn1, E1.

  • And so the carbon's getting the electron, and then the

  • bromine can then take this carbon's electron.

  • And just in one step that's what's distinctive about the

  • E2 and the Sn2 reactions.

  • All of the reactions are involved in the

  • rate-determining step and there really is only one step.

  • Just like that, after that happens, what we're left with

  • is the methoxide anion takes the hydrogen,

  • so it becomes methanol.

  • Let me draw that.

  • So it becomes methanol.

  • So it had one, two, three, four and then five, that's

  • this one right there, but then this guy goes and bonds with

  • the hydrogen.

  • This guy goes and bonds with the hydrogen, just like that,

  • and hydrogen leaves its electron behind.

  • Let me draw the cyclopentane part now.

  • And so the cyclopentane looked like this before, if I just

  • focus on the ring.

  • Now, this guy was bonded to a hydrogen.

  • He was bonded to this hydrogen over here, but now that

  • electron is going to be used to form a bond with this alpha

  • carbon right over here.

  • Let me draw the alpha carbon.

  • And the alpha carbon is right over there.

  • Obviously, implicitly at every one of these

  • edges, we have a carbon.

  • But now, a double bond is going to form

  • with that alpha carbon.

  • We could just draw it like that, a double bond.

  • Obviously, there's another carbon here.

  • I could write up another carbon over there.

  • And now this double bond will form.

  • And now the bromide has left.

  • It's taken an electron with it from that carbon now that the

  • carbon doesn't need it.

  • It was already starting to hog it because it's so

  • electronegative.

  • So that's bromine.

  • It takes that orange electron.

  • Now, it is bromide.

  • And we're done.

  • And so just to go back to the original question here, which

  • reaction is likely to occur or which mechanism?