Leah here from leah4sci.com and in this video we're going to discuss aromaticity for aromatic
compounds.
No this has nothing to do with aromatic or nice smelling molecules although there is
some relation in the names history, when you think of aromaticity or aromatic, which molecule
comes to mind?
Was it benzene?
You're going to see benzene a lot in organic chemistry and it's a great model for aromaticity.
What do we know about benzene, we know that it's a six carbon ring, we know that all of
the carbons are sp2 hybridized so that we have three pi bonds and resonance around the
entire ring.
Let's take a look carefully noting what happens to each different color of electrons.
The purple electrons can move clockwise or counterclockwise but let's show it moving
this way.
That would force the green electrons to collapse down in this direction, causing the blue electrons
to move in this direction.
In my resonance videos, I showed this step by step with the intermediate charged structures
but for the purpose of aromaticity, lets see what happens when they all move.
With purple pi bond here, green one at the bottom, blue on the left.
So which is correct for benzene?
Is it the structure on the right or the left?
The answer is that they are both equally contributing resonance structures and this is how we would
show them on paper but in reality because the electrons are moving around so much it
would look something like this.
And this is why when working with benzene and mechanisms or synthesis, you'll often
see it drawn with a ring in the middle because it's faster than drawing the pi bonds and
it's implying that the pi bonds are constantly resonating around the system.
What does this tell us about benzene.
Benzene is a very stable molecule and not easy to react when compared to something like
alkenes which also have pi bonds, in fact when you learn electrophilic aromatic substitution
or EAS link below, you'll see that you require a super electrophile to make benzene react.
Why is that?
What about all of this makes benzene so stable, that has to do with aromaticity and it's not
just benzene, it can be other molecules that kind of sort of look like benzene and your
challenge is going to be to figure out if a molecule has similar conditions, is it aromatic
or not.
For a molecule to be aromatic, it has to meet these four conditions, learn it, understand
it, memorize it.
The molecule has to be cyclic, it has to be planar, the ring portion of the molecule has
to be fully conjugated and finally the molecule has to follow Huckel's rule.
Let's break it down.
For a molecule to be cyclic, it has to be a cycle or a ring.
This means if you have a carbon chain, the first carbon has to connected to the last
carbon, for example, benzene.
We have six carbons in benzene where carbon number one is attached to three, four, five,
six and six is attached to one.
The end is attached to the beginning.
Let's compare benzene to this molecule, everything about it looks the same, it has six carbon
atoms, it's got three pi bonds in the same location, the only issue is that if we number
it one through six, you'll notice that carbon six is not attached to carbon one, this is
a linear or aliphatic molecule which is actually one three five hexatriene pretending to be
benzene.
This is what it looks like if we stretch it out.
The second condition is planar, this means that all of the atoms in the ring have to
be flat so that they can participate in resonance.
Think of typical carbon atom hybridization, a carbon with four bonds is SP3 hybridized
making it tetrahedral. for a review on hybridization, see the videos linked below, but tetrahedral
is a three dimensional molecule.
It is not flat and therefore not able to participate in aromaticity.
You'll commonly see SP2 atoms, carbon or other because these are trigonal planar or in simpler
terms, flat.
The key is less about the SP2 hybridization and more about the fact that we have an available
p orbital which we're going to look at in the next rule and the reason I point this
out is because if your professor tries to trick you with a very large molecule, that
is SP hybridized, an SP hybrid has two free p orbitals and that can also participate in
resonance.
Less common in aromaticity but something to keep in mind.
Looking at benzene again, notice that every carbon atom has a pi bond which is our clue
to recognize that they are all SP2 hybridized.
If I compare this to a cyclohexadiene, you'll notice that we have four SP2 carbons but we
also have two SP3 carbons.
These two carbons make this molecule non planar and that means cyclohexadiene cannot be aromatic.
Next we look for conjugation.
Conjugation should remind you of delocalized electrons or resonance as we've already seen,
benzene is capable of resonating it's pi bond throughout that entire cyclic structure but
the cyclohexadiene, even though it conjugated and it does have resonance, it can't resonate
it over the entire ring and that's the key difference, remember that you cannot resonate
onto an SP3 carbon atom because the carbon atom already has four bonds and has no way
to accept the resonating electrons.
Sometimes you will see an unconjugated system with the similar issue, in this case we have
a cyclohexadiene with two pi bonds but there is an SP3 carbon between the pi bonds so they
can't even resonate with each other, it doesn't matter where the SP3 atoms are located, if
you cannot resonate around the entire ring, that molecule cannot be aromatic.When we looked
at planar, we said it was less about SP2 and more about the availability of that P orbital
for resonance.
For example, you might see your molecule like this, it has two pi bonds and another carbon
with a lone electron pair.
That electron pair sits in a P orbital, that electron pair is able to resonate into the
molecule and even though we don't have three pi bonds, this molecule is aromatic.
Another common tricky example is a three member ring with only one pi bond.
The positive charge on the upper carbon is not a thing because a carbocation means a
lack of something or an empty P orbital however the presence of the empty P orbital means
that these electrons can move over, they can resonate up to the top allowing this molecule
to be conjugated, making this molecule aromatic.
And last but not least and certainly not easiest is Huckel's Rule.
In order for a molecule to be aromatic, it has to be cyclic, planar conjugated, and have
a low energy high stability.
There are some molecules that pretend to be aromatic but have a very high energy and instead
of going into a lab and figuring this out, Huckel gave a very simple criteria, allowing
us to quickly look at a molecule, count the pi electrons and determine if it's gonna be
aromatic or not.
Which brings us to one of the few equations you have to know in organic chemistry and
that is 4N + 2 = Pi electrons . You'll see different variations of this but the goal
is not to make it complicated.
This is an algebraic equation where you want to solve for N and N must equal to a whole
number or an integer.
If N equals to a whole number and the rest of the criteria are met, the molecule is aromatic.
If N is not equal to a whole number, the molecule is antiaromatic.
Let's go back to our benzene example, benzene has three pi bonds but six pi electrons, so
we set up the equation 4N + 2 is equal to six pi electrons and solve for N, remember
from algebra first you subtract the isolated number minus two on both sides which gives
us 4N + 0 or simply 4N is equal to six minus two or four.
Then we divide both sides by four to cancel it out on the left, divide both sides by four,
that cancels out four over four is one, telling us that N is equal to one, we don't actually
care what N is equal to because one is a whole number, that's all we care about.
N is a whole number, benzene obeys Huckel's rule, benzene is aromatic.
While this tells us how to calculate aromaticity, it's still annoying to have to do this equation
everytime you see a molecule on your exam especially because professors like to hit
you with ten different crazy molecules and ask you to determine their aromaticity.
So let's break it down and see how to quickly solve for Huckel's rule including if on a
shortcut that will leave you confident rather than questioning if you're getting it right
and that's exactly what we'll cover in the next video.
First, make sure to give this video a thumbs up and subscribe to my channel so you don't
miss any new videos and then for the complete aromaticity series including the common aromatic
compound cheat sheet and aromaticity practice quiz, visit my website leah4sci.com/aromaticity