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# Aromatic stability II

Video transcript

Last video, we
observed that benzene exhibits aromatic stabilization. And when chemists first
made cyclooctatetraene, which is this
molecule right here, they assumed it would
react like benzene. Because it looks like
it has alternating single, double bonds,
and it has a ring. And so they just assumed that
it would behave like benzene. It turns out cyclooctatetraene
does not react like benzene. So benzene did not react the
way cyclohexene did up here. So cyclohexene
will give us a mix of enantiomers when the bromine
adds across a double bond. Benzene does not do that. Cyclooctatetraene does,
as a matter of fact. So it will give you a
mixture of enantionmers and the bromine will add across
one of those double bonds. It turns out cyclooctatetraene
isn't even conjugated. So it looks like
it has alternating single and double bonds
in this dot structure, but it behaves like
isolated double bonds, like four different
isolated double bonds. And it's not aromatic. So let's see if we can analyze
the reasons for this reaction here. And so we're going to
just real quickly review the criteria to determine if
a compound is aromatic or not. So a compound or ion
is aromatic If it contains a ring of continuously
overlapping p orbitals, and also has 4n plus 2, where
n is an integer, pi electrons in the ring, which
is Huckel's rule. So here is my
cyclooctatetraene molecule. And if I count my
pi electrons, I can see there are
a total of eight pi electrons in this molecules. So eight pi electrons. And each carbon in
cyclooctatetraene is sp2 hybridized. So each carbon has
a free p orbital. Now the fact that
cyclooctatetraene has eight atoms
in the ring, means that there's a little bit of
angle strain in this molecule. So one possible confirmation
for the molecule to adopt is what's called a
tub confirmation. So you can see these two carbons
can swing up on either side to adopt a tub confirmation. Now if each carbon in the tub
confirmation is sp2 hybridized, I could draw a p orbital on each
carbon in the tub confirmation here. So when I do that, I'll
get this as a picture. Now the problem is, it's a
little difficult for the p orbitals to overlap. So it might be easy for these
guys to overlap right here, but because of the
tub confirmation, it'd be hard to get
overlap of these orbitals. And so it turns out that
this is the reason why the molecule acts like
it's not conjugated. It's because it does
have p orbitals, but they can't really overlap
in the tub confirmation. And so this violates
the first criteria for a compound to be aromatic. And therefore we say
that cyclooctatetraene is non-aromatic. So it's not an aromatic because
it violates the first criteria. It does not have a range of
continuously overlapping p orbitals. The p orbitals don't
really overlap very well in the tub confirmation. What if cyclooctatetraene
adopted a planar confirmation. So we're just going to pretend
like cyclooctatetraene adopts a planar confirmation down here. And once again, each
carbon is sp2 hybridized. So each carbon is going to
get a p orbital like that. And so we have a total
of eight carbons. So eight p orbitals. So eight atomic orbitals. So let me go ahead
and write that here. So we have a total of
eight atomic orbitals for cyclooctatetraene. And according to MO theory,
those eight atomic orbitals are going to give us
eight molecular orbitals. So the atomic orbitals
are going to cease to exist, and give us
eight molecular orbitals. Drawing those eight
molecular orbitals would be way too
much for this video. So we're not going to
actually draw them, but we are going to show where
they are in terms of energy using our frost circle. And so this is what we
saw in the last video. So how to draw a frost circle. I'm going to go ahead and
put a line through the center of my frost circle to help
me draw the polygon in here. So what kind of a
polygon do I draw? Well, I have an
eight-membered ring, so I'm going to draw an
eight-sided polygon here. I'm going to start
from the bottom here. So I'm going to attempt to draw
an eight-sided figure in here. So it would be
something like this. Now remember, when you are
drawing your frost circle and you're inscribing
your polygon, the important thing
is where the polygon intersects with your circle. And every point where
the polygon intersects with your circle represents
a molecular orbital. And so I can see
that I would have a total of eight
molecular orbitals, because I have eight points of
intersection between my polygon and between my circle. And again, the nice thing
about a frost circle is it shows you the relative
energies of your eight molecular orbitals. So I would have three
bonding molecular orbitals, which would be the
ones down here. So these three points
of intersection will give you the
relative energies. And so I have three
bonding molecular orbitals. And then up here
at the top, I also have three molecular orbitals. But these are my antibonding
molecular orbitals. Those are higher energy. And my two points
of intersection that are right on
the center line here, represent two non-bonding
molecular orbitals like that. So when I fill my
molecular orbitals, again it's analogous to
electron configurations. I have a total of
eight pi electrons that I need to worry about
for a planar cyclooctatetraene molecule. And so I can go ahead and start
to fill in my pi electrons like that. So that takes care of six
of them and I have two more. And since this is analogous
to electron configurations, I'm going to follow
Hund's rule and not pair up my electrons
in an orbital here. So that represents my
eight pi electrons. And since I have
unpaired electrons, I have two unpaired
electrons, that predicts a very
unstable molecule if it were to adopt a
planar confirmation. And if I think about in
terms of Huckel's rule, I know it doesn't
follow Huckel's rule. Huckel's rule is 4n plus
2, where n is an integer. And the two comes
from the fact that I have this orbital down here. And I do have 4n right here. But I don't have 4n, where n
is an integer for these two electrons up here. And so this is where
it breaks down. And so I have a total of
eight pi electrons, which does not follow Huckel's rule. And so because the number
of electrons is incorrect, this molecule is
definitely not going to adopt a planar confirmation. And so cyclooctatetraene
has a tub confirmation, and not planar. It is not aromatic. It is considered
to be non-aromatic because of the violation
of the first criteria. But it is possible to
react cyclooctatetraene. It's possible to oxidize it. And so let's see what
happens when we do that. So if we take cyclooctatetraene
and we oxidize it. So it's going to
lose some electrons. So I'm going to say that these
pi electrons are going to stay. And we're going to lose the
pi electrons on the left. So if I take away a bond
from these two carbons that used to have that
double bond there, they're going to be
positively charged. And so I could draw a
resonance structure for this. I could move these
electrons over here. And so if I go ahead and show
that resonance structure, then this carbon still
has a plus 1 charge. And the other positive charge
moves over here to this carbon, like that. And you could continue
drawing resonance structures for this molecule. I am not going to do that. I just want to show you that the
positive charges are spread out throughout the entire ion here. And so one way to represent
that would be to just show the electrons are spread out
throughout this entire ion. And the whole thing has a
2 plus charge like that. And so when I
analyze this dication that I got from
cyclooctatetraene, I realize that all the
carbons are sp2 hybridized. So if I look at these, my
carbocations those are sp2 hybridized. Everything with a double
bond on it is sp2 hybridized. And so I have eight
sp2 hybridized carbons. And so each one of those
sp2 hybridized carbons has a p orbital. And I also have
six pi electrons. I have two, four,
and six pi electrons. So six pi electrons
follows Huckel's rule. And so this ion turns
out to be planar. So it actually does
look like this up here. And so there's opportunity
for those p orbitals to overlap side by side. And so this ion actually
fulfills the first criteria. It contains a ring of
continuously overlapping p orbitals. And when you analyze
the second criteria, it has six pi electrons. And so it would fill the
bonding molecular orbitals. It no longer has these
two electrons up here. So it has six pi electrons. It fulfills Huckel's rules. So there's a total of 6 pi
electrons for this dication. And so this ion is
said to be aromatic. So this is an aromatic ion. It fulfills both of the
criteria for it to be aromatic. So it's extra stable. And that's the reason
why it turns out to be planar,
because it has eight carbons like cyclooctatetraene. And so because of
that number, it has some strain that
it has to overcome. So it would like
to turn into a tub to overcome some of that strain. But the fact that it's
a planar dication, means that there must be
some sort of extra stability in that ion that's
forcing it to be planar. And that extra stability
is due to the fact that it is aromatic. And so this ion has been
proved to be planar. So this video is just an
overview of cyclooctatetraene, why it itself is non-aromatic. But if you turn
it into a cation, it can be aromatic,
which is further proof for the concept of
aromatic stabilization.