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Studying for a test? Prepare with these 7 lessons on Aromatic compounds.

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# Aromatic stability III

Video transcript

We have seen that a
compound or ion is aromatic if it contains a ring of
continuously overlapping p orbitals, and also if
it has 4n plus 2 pi electrons in the ring,
where n is an integer. So for example, n could be
0, or 1, or 2, or so on. We can use these
criteria to analyze the cyclopropenyl cations. So here is my cyclopropenyl
cation right there. And if I see how many
pi electrons it has, that would be 2 pi
electrons in the ion. So let me go ahead and
write 2 pi electrons here. When I look at the
carbons in the ion, this carbon has a double bond
to it, so it's sp2 hybridized. Same with this carbon. And then this top carbon
here, this carbocation is also sp2 hybridized. So we know that sp2 hybridized
carbons have a free p orbital. So each carbon in this
ion has a p orbital on it. So I can go ahead and
show that on this diagram. So I'm sketching
in the p orbital on each carbon in my ring. And so that allows me to
see that my p orbitals could overlap side by side. And so this ion fulfills
the first criteria. It has a ring of continuously
overlapping p orbitals. I know that p orbitals
are atomic orbitals. And so I have a total of
three atomic orbitals. And according to MO theory,
those three atomic orbitals are going to combine to give
me three molecular orbitals. And I can analyze the
relative energy levels of those molecular
orbitals using what's called a frost circle. And so we've seen these
frost circles before. And I'm just going
to go ahead and start by drawing a line here
to divide my circle in half, which
divides my bonding from my antibonding
molecular orbitals. And we always start at the
bottom of our frost circle. And we inscribe a
polygon to match the number of atoms in my ring. So in this case, I have
a three-membered ring. And so I'm going to inscribe a
triangle into my frost circle. So I'm going to attempt to draw
a triangle in my frost circle here. And not the best triangle,
but the important thing is where your polygon intersects
with your frost circle represents the energy level
of your molecular orbitals. So I can see I have a total
of three molecular orbitals, looking at my frost circle. And when I go and
represent them over here, I know that I have one
molecular orbital right here. And I have two molecular
orbitals up here. Bonding molecular orbitals
are always lower in energy. So I know that this
orbital down here is my bonding molecular orbital. And therefore, I have two
antibonding molecular orbitals up here. I know I have two pi
electrons to worry about. So it's completely analogous
to electron configurations. I'm going to fill my lowest
energy orbital first. And so that would be my
bonding molecular orbital. And I only have
two pi electrons. And so those 2 pi
electrons are going to completely fill my
bonding molecular orbital. Let's analyze this in
terms of Huckel's rule. So I can see that this
would be a 2 here, to represent my
two pi electrons. And then I have 0
electrons in my antibonding molecular orbitals. So it would be like 4 times 0. So 4 times 0 is, of course, 0. Plus 2, gives me a total of
2, which is Huckel's rule. So I have two pi
electrons in the ring. And this ion satisfies the
second criteria as well. So the cyclopropenyl
cation is aromatic. So if I go ahead and write
this ion is aromatic over here. And so it's extra stable. And this observation
allows us to explain some of the properties
associated with this molecule down here, which
is cyclopropenone. Cyclopropenone has a huge
amount of angle strain with a three-membered ring here. And it also has
increased dipole moment from what we might expect. And we can explain both of those
by looking at the resonance structure for the
cyclopropenone molecule. So if I think about drawing
a resonance structure, I could take these pi
electrons and move them off onto my oxygen here. And so the resonance
structure would have my three-membered
ring like that. And then it would
have this oxygen with now three lone pairs
of electrons around it, giving it a negative
1 formal charge. Took a bond away from my
carbonyl carbon right here. So that carbon is going to
get a plus 1 formal charge. And so you can see
that we've just formed the cyclopropenyl cation here,
which we know is extra stable. This is aromatic. And so while this is an
extremely minor resonance structure for most carbonyl
compounds, for this one, this contributes
a little bit more to the resonance hybrid
because of the extra stability associated with the
cyclopropenyl cation, the fact that it is aromatic. And so that affords some extra
stability to this molecule. And we could also draw it where
we show that positive charge is being spread out throughout our
three-membered ring like that. And then we have our oxygen over
here with a negative charge. And this picture allows us
to see the increased dipole moment a little bit more. So because the
part on the left is extra stable as a positive
charge, it's aromatic. That means we have an
increased dipole moment and also increasing
the stability despite the significant
angle strain associated with the cyclopropenone
molecule. So you can use the concept
of aromatic stability to analyze the structure
of other molecules as well. Let's do one more example. Let's look at this molecule,
which is cyclobutadiene. So over here on the left,
we have cyclobutadiene. If I look at the pi electrons,
here is 2 pi electrons. And here's another 2, for
a total of 4 pi electrons. So there are 4 pi
electrons in my molecule. When I look at the carbons,
each carbon has a double bond. So each carbon is
sp2 hybridized. So each carbon has
a free p orbital. And so I can sketch
in my p orbitals over here on my
diagram like that. Makes it a little bit easier to
see that these p orbitals could overlap side by side. So the first criteria
has been fulfilled. I have a ring of continuously
overlapping p orbitals. But when I look at
my second criteria, our second criteria was
4n plus 2 pi electrons. And so I don't have that. I have a total of
4 pi electrons. And so that's really 4n,
where n is equal to 1. So 4 times 1 is
equal to 4 or so. I have 4n pi electrons. I don't have 4n plus 2. So I already know that
this compound is not aromatic just by
looking at that. But let's go ahead and draw
in our molecular orbitals and see where we put
those 4 pi electrons in our molecular orbitals. So my four p orbitals
here on this diagram. It means I have a total
of four atomic orbitals for cyclobutadiene. Once again, MO theory tells
me four atomic orbitals are going to give me
four molecular orbitals. And I can draw in my
frost circle right here. So once again, we're going to
draw a line through the center to separate my bonding from my
antibonding molecular orbitals. You always start at the
bottom of your frost circle. Four-membered ring. So we're going try to
draw a four-sided figure in our frost circle. So I'm going to attempt to
draw this square in here. And once again,
where our polygon intersects with our circle
represents the energy level of our molecular orbitals. So I have a total of
four molecular orbitals. And if I go over here, I
have a molecular orbital below the line. So that's my bonding
molecular orbital. I have a molecular orbital
above the center line there. So that's my antibonding
molecular orbital. And this time I have
two molecular orbitals that are right on
the line, which represents non-bonding
molecular orbitals. When I go ahead and put
in my 4 pi electrons, once again I fill lowest energy
molecular orbitals first. So that takes care
of 2 pi electrons. And now I have two more. And so here I have the
non-bonding molecular orbitals are on the same energy level. And so if you remember
Hund's rule from electron configurations, you can't pair
up these last two pi electrons because the orbitals
are of equal energy. And so this is the
picture that we get. And you can see that I have
two unpaired electrons. And two unpaired
electrons implies that this molecule is
extremely reactive. And experimentally, it is. So cyclobutadiene will
actually react with itself. So it's experimentally
extremely reactive which tells you that
it's not extra stable. It's not aromatic. So you don't have 4n
plus 2 pi electrons. And so the second
criteria is not fulfilled, but this compound does
satisfy the first criteria. And so the term
for this compound is anti-aromatic, which means
it fulfills the first criteria, a ring of continuously
overlapping p orbitals, but does not satisfy
the second criteria. It does not have 4n
plus 2 pi electrons. It has 4n pi electrons. And so we say that
is anti-aromatic. And there are actually
very few examples of anti-aromatic compounds. But cyclobutadiene is
considered to be anti-aromatic. In the next video, we're going
to look at a few more examples of aromatic stability.