Main content

## MCAT

### Unit 9: Lesson 14

Alcohols and phenols- Alcohols and phenols questions
- Alcohol nomenclature
- Properties of alcohols
- Biological oxidation of alcohols
- Oxidation of alcohols
- Oxidation of alcohols (examples)
- Protection of alcohols
- Preparation of mesylates and tosylates
- SN1 and SN2 reactions of alcohols
- Biological redox reactions of alcohols and phenols
- Aromatic stability of benzene
- Aromatic heterocycles

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# Aromatic stability of benzene

Created by Jay.

## Want to join the conversation?

- So, can someone please explain to me what exactly "n" is and what determines its value? When ever my professor does these problems and apparently the maker of this video there seems to be a bit of mistery when it comes to figuring out what "n" is. In the video he says "If n is 1 then use Huckel's rule." Well, where did 1 come from? What is he counting? To me its almost like you figure out how many pi electrons there are and then solve for "n"! But that can't be right!! Thank you...(6 votes)
- n is any positive integer. You are right in saying you figure out how many pi electrons there are and then solve for n. You just want the number of pi electrons you have to follow the rule. For example, if you had 7 pi electrons and plugged that into Huckel's rule you would get n = 1.25. Since 1.25 isn't an integer the compound isn't aromatic. If n is a positive integer then that satisfies one requirement of being aromatic.(13 votes)

- I believe the molecule in the beginning is cyclohexene because of the double bond, correct?(4 votes)
- Does anybody know if molecular orbital theory is on the MCAT?(1 vote)
- I realize its too late but for other people, yes it is remember guys its the bonding/antibonding thing(2 votes)

- The ones above the line, that is the anti-bonding molecular orbitals are of higher energies why?(1 vote)
- Why makes the middle bonding pi electrons have extremely low energies and the middle antibonding elections have such a higher potential energy?(1 vote)

## Video transcript

In this series of
videos, we're going to look at aromaticity or
aromatic stabilization. We've already seen
that bromine will add a cross double bond of a
simple alkene like cyclohexane to give us a mixture of
enantiomers for our products. If we try the same
reaction with benzene, we're not going to get
anything for our product. So there's no reaction. And so benzene is more
stable than cyclohexane. At first you might think
that the stability is due to the fact that
benzene is conjugated, but numerous other
experiments have shown that it is even more
stable than we would expect. And that extra stability is
called aromaticity or aromatic stabilization. And so benzene is an
aromatic molecule. Let's look at the
criteria to determine if a compound is aromatic. All right. So a compound is
aromatic if it contains a ring of continuously
overlapping p orbitals. And so if the
molecule is planar, that's what allows the
p orbitals to overlap. It also has to have 4n plus
2 pi electrons in the ring where n is equal to 0, 1, 2,
or any other positive integer. And this is called
Huckel's rule. So let's go ahead
and analyze benzene in a little bit more detail. So if I look at
the dot structure, I can see that benzene has two
pi electrons there, two here, and two more here for a
total of six pi electrons. If I look at the
carbons of benzene, I can see that each carbon
has a double bond to it. So each carbon is
sp2 hybridized. And if each carbon
is sp2 hybridized, that means that each carbon
has a free p orbital. So let me go ahead and sketch
in the unhybridized free p orbital on each of the
six carbons of benzene. Now since benzene is
a planar molecule, that's going to allow those
p orbitals to overlap side by side. So you get some overlap side
by side of those p orbitals. And so benzene contains a ring
of continuously overlapping p orbitals. So p orbitals are considered
to be atomic orbitals. And so there are a total of
six atomic orbitals in benzene. According to MO theory,
those six atomic orbitals are going to cease to exist. And we will get six
molecular orbitals instead. And so benzene has six
molecular orbitals. Drawing out these
molecular orbitals would be a little bit too
complicated for this video. So checkout your textbook
for some nice diagrams of the six molecular
orbitals of benzene. However, it is important
to understand those six molecular orbitals in terms of
their relative energy levels. And the simplest way to do
that is to draw a frost circle. And so here I have a
circle already drawn. And inside the circle we're
going to inscribe a polygon. And since benzene is
a six-membered ring, we're going to inscribe a
hexagon in our frost circle. I'm going to go ahead and draw
a center line through the circle just to help out with
the drawing here. And when you're inscribing your
polygon in your frost circle, you always start at the bottom. So we're going to
start down here. So we're going to
inscribe a hexagon. So let's see if we can
put a hexagon in here. So we have a six sided
figure in our frost circle. The key point about
a frost circle is everywhere your
polygon intersects with your circle, that
represents the energy level of a molecular orbital. And so this intersection right
here, this intersection here, and then all the way around. And so we have our
six molecular orbitals and we have the
relative energy levels of those six molecular orbitals. So let me go ahead and
draw them over here. So we have three
molecular orbitals which are above the center line. And those are higher in energy. And we know that
those are called antibonding molecular orbitals. So these are antibonding
molecular orbitals which are the highest in energy. But look down here, there
are three molecular orbitals which are below the
center line and those are our bonding
molecular orbitals. So those are lower in energy. And if we had some
molecular orbitals that were on the center
line, those would be non-bonding
molecular orbitals. We're going to go ahead and
fill our molecular orbitals with our pi electrons. Let me go back over here. And remember that benzene
has six pi electrons. And so filling
molecular orbitals is analogous to
electron configurations. You're going to fill the
lowest molecular orbital first. And each orbital can
hold two electrons like electron configurations. And so we're going to go
ahead and put two electrons into the lowest bonding
molecular orbital. So I have four more pi
electrons to worry about. So four more pi electrons and
I go ahead and put those in. And I have filled the bonding
molecular orbitals of benzene. So I have represented
all six pi electrons. If I think about
Huckel's rule, 4n plus 2, I have six pi electrons. So if n is equal
to 1, Huckel's rule is satisfied because I
would do 4 times 1 plus 2. And so I would get a
total of six pi electrons. And so six pi electrons
follows Huckel's rule. If we look at the
frost circle, and we look at the molecular
orbitals, we can understand Huckel's rule
a little bit better visually. So if I think about these
two electrons down here, you could think
about that's where the two comes from
in Huckel's rule. And if I think about these
four electrons up here, that would be four electrons
times our positive integer of one. So 4 times 1 plus 2
gives us 6 pi electrons. And we have filled the bonding
molecular orbitals of benzene which confers the
extra stability that we call aromaticity or
aromatic stabilization. And so benzene is aromatic. It follows our
different criteria. In the next few
videos, we're going to look at several other
examples of aromatic compounds and ions.