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