of 5-membered and 6-membered rings

in this mode of reactivity

in which the π system serves as the nucleophile.

The electrophilic aromatic substitution reactions

like we were con- considering in the previous webcast,

but now with the 5-membered heteroaromatics.

We'll take as our 5-membered ring our prototypical reaction,

the reaction of thiophene, that system there,

with acetyl nitrate.

Acetyl nitrate is just a good source of an electrophilic

nitronium cation, but it's going to react

in a slightly different mode of reactivity.

Two things to note; first of all,

the reaction takes place at a very low temperature,

-10ºC.

Obviously, this is much accelerated

compared to the previous case of the 6-membered ring.

And the second thing to note is that the mode, the, of

substitution takes place in this position,

the so called 2-position.

In the 2-position, that hydrogen atom

is going to be displaced so this is carbon-2 of thiophene,

and we call that carbon-3 of thiophene,

it's the carbon-2, its hydrogen,

that undergoes substitution with the nitryl group.

There's a couple of ways we could understand why

the substitution takes place at C2 rather than C3.

One thing that we might do is to do a SHMO calculation

and look at the highest occupied molecular orbital

and we'd find that the grayest- greatest coefficient

of the highest occupied molecular orbital

is going to be on C2.

That's one thing.

Another way we might anticipate this reactivity

is to compare the two different types of intermediates

that form by attack at C2 and by attack at C3.

Let me outline that for you and what I would encourage you to do

while you're listening to this

is to make a reaction coordinate diagram

and compare the reaction pathway for C2 attack

versus C3 attack.

C2 attack involves three resonance contributors,

whereas C3 attack only involves two.

This is a π to σ* type interaction.

We're going to break the σ bond

between nitrogen and oxygen in acetyl nitrate

and we'll end up making a new carbon nitrogen bond,

and this intermediate is a carbocation intermediate

that is delocalized onto those positions in the ring

as well as on the sulfur atom.

And so a N to A type interation,

resonance interation gives us three resonance contributors

one of which, the best one, is the last one

because it has an octet of electrons on every atom.

Attack at C3 ult- provides us with a less stable intermediate.

That less stable intermediate

only has 2 resonance contributors

one of which is again, having a positive charge on sulfur,

an octet of electrons on every atom,

but you can see that we have less d- delocalization

of that positive charge.

The intermediate that results from attack at C2

is more stable, the reaction pathway proceeds

through a transition state to the C2 attack

that’s lower in energy than the C3 attack.

That pathway, the C2 pathway, is faster than C3.

Why the enhanced reactivity?

We just explained the regioselectivity,

but why the enhanced reactivity?

And that has to do with the highest occupied

molecular orbital energy level being elevated.

The higher the energy of the HOMO,

the more nucleophilic that pair of electrons.

And so that’s one- one reason.

So there’s really two reasons why benzene,

of these four, is the least reactive

towards electrophilic aromatic substitution.

So the first thing that I’ve already mentioned

is this idea down here

where the π electrons are raised up in energy,

they’re more nucleophilic than the benzene ring

pair of electrons.

And if you did a SHMO calculation

you would in fact see those electrons in the HOMO

of the thiophene ring

or the other 5-membered heteroaromatics

are higher in energy than is benzene.

But the other thing to note is that the transition state,

and the intermediate that results is more stable

and the main reason for that is that

intermediate can be drawn

with a complete octet on every single atom.

Here’s one resonance contributor and that’s the carbocation,

but we have- don’t forget

we have this nonbonding pair of electrons

that can do an end to A or an end to π*-type donation

and provide us with an intermediate-

a resonance contributor of that intermediate

that has an octet of electrons on every atom.

You cannot find that in the case of benzene,

all’s you have is open shelve carbocations.

So the intermediate is more stable

and because the transition state

is going to resemble the intermediate

the pathway is going to be lower.

So, we have two things going for us;

the energy is raised- the energy of that

higher occupied molecular orbital is raised,

and the transition state is lowered

and that’s what makes these heteroaromatics

more reactive than benzene.

Alright, let’s take a look at the imidazole

very quickly and say that, if we have

an N2 nitrogen in the ring,

we’re going to undergo deactivation

for the reasons that we encountered with pyridine.

We’re going to first do acid-base chemistry

and that’s going to facilitate lowering that HOMO,

making that π system much less reactive.

Those π electrons now being part of a positively charged ring,

are no longer very reactive and so the end-

the presence of an N2 nitrogen in a molecule like imidazole,

make it very unfavorable toward nucleophilic-

toward electrophilic aromatic substitution

in which the π-system acts as a nucleophile. 47 making that π system much less reactive.

Those π electrons now being part of a positively charged ring,

are no longer very reactive and so the end-

the presence of an N2 nitrogen in a molecule like imidazole,

make it very unfavorable toward nucleophilic-

toward electrophilic aromat