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.