Leah here from leah4sci.com and in this video we're going to look at charge and heterocyclic
aromatic compounds.
In order for a molecule to be aromatic it has to be cyclic, planar, conjugated and follow
huckel's rule.
That doesn't mean that the entire ring has to be composed of carbon atoms.
We're used to seeing molecules like benzene which is very simple.
6 carbon atoms in a ring, 3 pi bonds, nothing else, good to go.
When you see something else different such as a negative or positive charge, or even
another atom in the ring, you wanna evaluate their criteria regardless of what else is
going on.
You still wanna evaluate by this checklist recognizing what to expect which charges and
heterocycles.
Before we continue make sure you review the previous videos where we discussed the criteria
for aromaticity.
You can find this on my website leah4sci.com/aromaticity.
Let's start with the aromatic cation, or simply the cation.
A cation tells us that there's a positive charge for example, if we have a carbocation,
we have a carbon with bonds to three other atoms which we'll show as green lines and
then an empty P-orbital that sits above and below the carbon atom.
A cation is sp2 hybridized.
There's nothing in there, there are no electrons, instead that p-orbital is open and willing
to accept another electron pair, for example to form a pi-bond.
When you studied resonance, you saw that pi bonds or electron pairs are very happy to
move over towards a positive carbon atom which moves the carbocation to the carbon that wants
the pi bond but creates a resonating system.We've already seen this molecule with a pi bond
and a carbocation, so let's break it down.
How do we know if it's aromatic?
The molecule is cyclic, I can trace a circle around the entire compound, it's planar because
every carbon is sp2 hybridized.
The carbon's holding the pi bond as well as the carbocation.
We know that the pi bond can resonate into that empty p-orbital, either towards the left,
or towards the right making this a fully conjugated system and then we'll use the shortcut for
huckel's rule.
We have a total of 1 pi bond which is odd, an odd number of pi bonds follows huckel's
rule making this molecule aromatic.
Yes, it's tiny, yes it's under a lot of string but it's still aromatic.
We've seen that cyclopentadiene is nonaromatic because of an sp3 carbon atom.
But what if we change it to the cyclopentadiene enol cation, meaning a positive charge on
the carbon that doesn't have a pi bond?
The molecule is cyclic, I can trace a circle from start to finish.
The molecule is planar because every carbon atom in this ring is sp2 hybridized.
The carbon's holding a pi bond and the carbon holding the carbocation.
The molecule is conjugated because the pi bond can resonate in the direction of the
carbocation clockwise or counter-clockwise.
But does it follow huckel's rule?
Let's see.
We have a total of 4 pi electrons or 2 pi bonds, that means we have an even number of
pi bonds and even does not follow huckel's rule, it's anti team huckel making it anti
aromatic.
Aromatic anions are evaluated the same way with one difference, "An" means without.
But in this case it means without or negative because we're dealing with a negative charge.
If we have a system with a pi bond next to a carbon ion, the negative electrons can resonate
towards the pibond taking those electrons which gives me a resonance structure that
now has a pi bond in the middle and a lone electron pair on the left definitely capable
of resonance.
In order for these electrons to resonate, the carbon has to hybridized to sp2 putting
the lone electrons into a p-orbital so that it can participate in the resonance and this
is something to keep in mind when evaluating the aromatic anions.
We'll take our same cyclopentadiene but this time we'll give it a lone pair rather than
a carbocation on this carbon atom which means it also has a negative charge.
Let's evaluate, we already know that it's cyclic because these electrons can resonate
into the pi system, they have to sit on an sp2 hybridized carbon atom which means all
of the atom are sp2 and the molecule is planar.
Because the anion can resonate into the ring, that makes it a conjugated system, but does
it follow the huckel's rule?
We have two pi bonds which means an even number.
But remember that for huckel's rule we're not looking at the number of pi bonds as much
as the electrons are resonating, or of the shortcut the electrons pairs that are capable
of resonance.
We have a total of 6 electrons or 3 electron pairs which means we have an odd number and
the odd number means it follows huckel rule.
That makes the cyclopentadiene anion and aromatic anion.
Let's compare it to this 6-carbon anion.
The molecule is cyclic because I can trace a path around the ring.
I do notice the anion which means if my mind jumps to huckel's rule I'm thinking 6 electrons,
3 electron pairs, odd number, should be aromatic, but what else do we have going on?
Yes, the negative electrons can resonate into the ring making part of the system conjugated
but don't forget we also have this carbon over here.
This carbon is not able to participate in resonance because it doesn't have a charge
and it doesn't have a pi bond.
It's an sp3 hybridized carbon, meaning it's tetrahedral or 3-dimensional, that means the
molecule is not going to be a planar and the entire ring is not conjugated because we can't
resonate on to sp3.
This sp3 carbon is uninterested in aromaticity, it's a non baseball fan, making this a non-aromatic
compound.
Let's move on to heteroclyclic aromatic compounds remembering that hetero means different, and
an heterocyclic compound is a molecule that has something different in the ring, meaning
other than carbon atoms.
For example this molecule right here, often comes out on exams, is pyrrole an aromatic
compound or not?
More importantly, how do we evaluate when we have something like a nitrogen atom in
the ring?
Here's the trick to remember with heterocyclic compounds.
Heterocyles just like any molecule wants to be aromatic.
This is something you'll hear me say over and over and over.
It applies to organic chemistry just like it applies to your entire life.
Happy, is going to give you something that is stable, is going to be unreactive.
If something is unhappy, it is unstable and tha makes it reactive.
We said that aromatic compounds are very stable, so when possible, everything wants to be aromatic
if it can.
A heterocyclic atom, if it has the ability will try to contribute to aromaticity.
But in order to be aromatic, it has to resonate.
So here's the question to ask, does the atom participate in resonance already?
All you need is one participation.
An atom can't join resonance with two separate electron pairs at the same time.
It's one pi bond or one lone pair, that's it.
If the atom is already participating with a pi bond, you don't add the electrons in.
If the atom is not participating, you use the lone pair.
Again, IF the atom has the pibond you don't use its electron pair, If it has no pi bond,
you use its electron pair.
If it has two electron pairs, you only need one because one is enough to participate,
Let's see how this applies to pyrrole.
Is the molecules cyclic?
why yes, it is!
I can draw a circle around the ring including the nitrogen atom.
Is the molecule planar?
Let's see.
The carbon has pi bonds they call as p2 hybridized.
The Nitrogen atom appears to be sp3 hybriized.
It has 3 bonds, 1 lone pair which would sit in a hybridized orbital except for the rule
you should remember in resonance and i cover this in the series linked below.
If the atom has the ability to resonate, then it will hybridized so that it can resonate,
that means even though nitrogen appears to be sp3, the nitrogen is sp2 hybridized so
that it can participate in resonance which means the molecule is planar, everything is
resonating, and that means the molecule is conjugated.
But do you notice how we had to ask the question.
Nitrogen was not participating in resonance because there's no visible pi bond and so
we had to take its lone electron pair to contribute to the aromatic system so we have to use its
electrons for resonance.
Let's attempt the side view of this molecule and you can build this with your model kit.
We'll make these two as the rear carbons in black, we'll make these two the forward carbons
in purple.
We know that for resonance we have the p-orbital sitting above and below so we'll draw above
and below two different colors and then we have a nitrogen atom right over here.
If Nitrogen is sp2 hybridized that means it's going to have 120 degree bond angles between
nitrogen and the things that's bound to.
that makes hydrogen perpendicular to this system facing completely out of the way.
But then we have that p-orbital where the electrons would sit.
We've seen that we have a pi bond here and here like this.
Now Nitrogen can take those electrons resonate towards the pibond giving me that conjugated
system.
Let's verify with huckel's rule how many electron pairs do we have, 2 pi bonds and 1 lone pair
gives me a total of 3 electron pairs an odd number of electron pairs which means it supports
team huckel and it's aromatic.
Another common molecule you'll see is pyridine.
Pyridine like pyrrole is a heterocyclic compound that appears to be aromatic.
Let's take a look.
Is it cyclic?
yes it is! is it planar?
I see pibonds everywhere, all sp2 hybridized, yes it is.
Is it conjugated? it looks very much like benzene in fact if I pretend the nitrogen
isn't there, it's exactly benzene which means it is conjugated.
But what about huckel's rule?
I have 3 pi bonds and 1 lone pair giving me a total of 4 pi bonds or an even number which
wants to tell me that the molecule is anti aromatic.
We have to go back to the rule.
Nitrogen wants to participate in aromaticity.
In pyrrole, the only way to play was to bring its own electrons, BYOE bring your own electrons.
But in pyridine, we don't need that because nitrogen already has a pi bond, it's already
part of the resonating system, that means these electrons don't get to play along, therefore
they're not counted and so we have 1,2,3, an odd number of electron pairs, supports
team huckel, making pyridine aromatic.
Let's take a long at a side view so you'll understand why these electrons can't even
try to participate.
We'll show these 3 carbons in black, and the forward carbons in purple, and nitrogen on
this side.
We'll show the other part of the p-orbital in light blue including our nitrogen and the
lower part in grey and look what happens, we have a pi bond here, we have a pi bond
here, and we have a pi bond here.
That is a clearly resonating system, but what about the lone pair?
If Nitrogen is sp2 hybridized, then imagine that this lone pair is really a bond to itself
and an orbital that sits at 120 degrees perpendicular to the molecule.
Even if these electrons wanted to participate, they're sitting at 90 degrees, they have no
way of interacting with pi system, that is why we don't count them and that is why they're
not used.
For even more practice on heterocyclic aromatic compounds, aromatic ions and more, first,
give this video a thumbs up, then visit my website leah4sci.com/aromaticity where you'll
find the aromatic practice quiz as well as the common aromatic compounds cheat sheet.
The link again is leah4sci.com/aromaticity.