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## Organic chemistry

# Aromatic stability IV

Aromaticity of the cyclopentadienyl anion and cycloheptatrienyl cation. Created by Jay.

## Want to join the conversation?

- At about9:10, how do you know there is 3 bonding and 4 anti bonding instead of the other way around?(5 votes)
- When constructing molecular orbitals using a Frost circle, one of the vertices is always drawn in the centre pointing down. In an aromatic compound there will only be one arrangement where all of the p orbitals are in phase. This arrangement has maximal orbital overlap so will be the lowest energy bonding orbital (in the pi system). This single lowest-energy MO is represented by the vertex at the bottom of the Frost circle (it will always stand alone without any degenerate orbitals). Having a vertex at the bottom of the Frost circle (as opposed to a side of the polygon at the bottom) means that there are 3 bonding and 4 anti-bonding MOs.(8 votes)

- Is there ever a situation where you calculate the pi electrons at the start and when you get to the frost circle, they don't create a 'full outer shell'? Or is the Frost circle just a way to demonstrate BMO etc, but in reality you don't need it to work out if a molecule is AROMATIC?(3 votes)
- The Frost circle is just an easy visual way to work out if a molecule is aromatic without having to do the mathematical calculations.(5 votes)

- What is pKa and what does it numeric value suggest?(1 vote)
- Ka is the value of the equilibrium constant for the ionization of an acid in water. The greater the Ka, the more of the acid ionizes, and the stronger the acid. Most common weak acids have Ka values ranging from 10^-2 to 10^-14. A Ka value of 10^-2 means that an acid is stronger than one with a Ka of 10^-12.

pKa is the negative logarithm (to the base 10) of Ka. Thus the pKa values of these acids range from 2 to 14. The lower the value of pKa, the stronger the acid. So an acid with pKa = 2 is stronger than an acid with pKa = 14.(8 votes)

- (1:30) A pKa value of 16 indicates a weak acid, doesn't it?(3 votes)
- in the very first example... arent those hrydrogens on the sp2 double bonded carbons acidic... i thought that becoz sp2 orbitals are more acidic than sp3 orbitals??(2 votes)
- Quite correct! Normal sp3 bonded H atoms have a pka ~ 51, while sp2 bonded H atoms have pKa ~44, and acetylenic (sp bonded) H atoms have pKa ~25. But in cyclopentadiene the sp3 H atoms have pKa ~16. They are 10^9 times more acidic than even an alkyne H.

The reason is that the cyclopentadienide ion (C5H5-) is exceptionally stable because it is aromatic. So it is much easier to lose the sp3 bonded H (the activation energy is much less), and the H is therefore more acidic than you might expect.(4 votes)

- I've read in my textbook that Hükel's rule doesn't work with polycyclic aromatic compounds does anyone know why?(1 vote)
- For the last figure, at8:16, why does the does the top carbon have a pi orbital. Isn't it only the electrons involved with the pi bond?(1 vote)
- The top carbon is positively charged, so is sp2 hybridized. Thus, it has an unhybridized p orbital. This p orbital has the necessary orientation to be involved in resonance structures with the double bonds in the ring. Thus, through resonance, the
*p_ orbital on the carbocation is part of the _pi*system.(3 votes)

- do frost circles only work with cyclic compounds?(1 vote)
- It's even more limited than that. They work only for
*planar*cyclic conjugated compounds.(4 votes)

- if having a high pKa value indicates that it is a weak acid, doesnt that mean that the molecule is not willing to give up a hydrogen very easily to reach this apparently very stable conformation? so why is it constantly referred to as "so acidic" in the video?(1 vote)
- It is very acidic
**relative to**other hydrocarbons. When Jay first talks about this1:12he says "extremely acidic**for a hydrocarbon**".

Typical pKa values:

alkanes pKa ~50

alkenes are pKa ~45.

Various sources list the pKa of cyclopentane as being ~45 to >60 and I can't find anything for cyclopentene.

So the acidity of cyclopentadiene about 10^29 times greater than for a typical alkene!(2 votes)

- how can we say that a carbon cation is sp2 hybridised. pls explain.(1 vote)
- It has 2 electrons each in s and 2 p. Try drawing the electronic configuration of Carbon. It'll look something like, 1s² 2s² 2px² 2py² 2pz0. Or use VSEPR theory formula. It's pretty straightforward.(1 vote)

## Video transcript

Here we have the
cyclopentadiene molecule. And let's analyze
it to see if it fits the criteria
to be aromatic. And so we're going to start
with the first criteria. So does it contain a ring of
continuously overlapping p orbitals. Well, if we analyze
these carbons here, this carbon is double bonded,
so it's sp2 hybridized. And therefore has
a free p orbital. Same with this carbon. And same with these
other two, here. So those four carbons
are sp2 hybridized. If I look at this carbon,
however, it's sp3 hybridized. That's a little bit easier
to see if I go ahead and draw in some
hydrogens on there. So since this carbon right here
has four single bonds to it, it is sp3 hybridized,
which means that carbon does not
have a p orbital. And so cyclopentadiene
is not aromatic. It violates the first criteria. It does not contain a ring
of continuously overlapping p orbitals. Since it violates
the first criteria we can go ahead and say
that cyclopentadiene is non-aromatic. So you don't even need to worry
about the second criteria. It doesn't fulfill the
first criteria for something to be aromatic. However, cyclopentadiene
has an interesting property. It's extremely acidic
for a hydrocarbon. It actually has a pka
of approximately 16 for one of these two protons
here that I drew in yellow. And so, there must be some
sort of stability associated with the conjugate base in
order for cyclopentadiene to be so acidic. And so, if we think about
a base coming along, so maybe a lone pair of
electrons, a negative charge, just some generic base. It's going to take this proton
right here, leaving these two electrons behind
on that top carbon. So if we go ahead and
draw the structure of the conjugate base, we
have our pi electrons here. And now we have a
lone pair of electrons on our top carbon, which
makes this negatively charged. So here's the conjugate base. And there must be some sort
of stabilization associated with this conjugate
base because of the fact that cyclopentadiene
is so acidic. Remember, the more stable
the conjugate base, the more acidic the compound. And so, in this case,
the extra stability is associated with the fact
that this ion is aromatic. And it might not look
like that, because if we look at this top carbon,
this top carbon still looks like it's sp3 hybridized
because it's a carbanion. However, that lone
pair of electrons can participate in resonance. So if we take this
lone pair of electrons and we move them
in here, and that would kick these electrons
in here off onto this carbon. If we draw one of the possible
resonance structures for this. I'm going to go ahead
and draw and show that these electrons
have now moved into here. These electrons are now off
on this carbon right here. So it's a negative
1 formal charge. And then we have these pi
electrons over here like that. Now, if we analyze
this top carbon here, now we can see that it's
actually sp2 hybridized. So in this resonance
structure, it looks like it's sp2 hybridized. And that lone pair of
electrons here in magenta is now occupying a
p orbital because it is participating in resonance. And so there are actually
many more resonance structures you could draw. We're not going to do
that here in this video. Here, I'm just
trying to point out that the electrons in magenta,
the lone pair of electrons in magenta participate
in the resonance. And therefore, that
lone pair of electrons is actually delocalized
and occupying a p orbital in the ring. And so now we have a ring of
continuously overlapping p orbitals. So over here on the
left, we'd already said that these carbons
are sp2 hybridized. And if you think about
these resonance structures, you can have all
those carbons in that ring are now sp2 hybridized. And so you fulfill
the first criteria. You have a ring of continuously
overlapping p orbitals. Let's analyze this
anion a little bit more using our frost circles. So if I go ahead and
sketch in the fact that this is our anion, with
a negative of formal charge. When I'm looking for pi
electrons in this molecule I will use this color here. So here are 2 pi electrons,
and then 4 pi electrons. And then we've seen that
this lone pair participates in resonance. So this lone pair of electrons
occupies a p orbital. And those are actually
pi electrons now. So we have a total
of 6 pi electrons. We have 6 pi electrons
for this anion. And we've now seen that all
of these carbons in here are sp2 hybridized when you
draw all of your resonance structures. And so each one of those
carbons has a p orbital. And so I can sketch in a
p orbital on this diagram. So we have five
carbons in the ring. Each carbon has a p orbital. And so there are five p
orbitals in this ring. And you can see that those
p orbitals in the ring can overlap to delocalize
those electrons. So the overlap of
those p orbitals satisfies our first criteria. It contains a ring of
continuously overlapping p orbitals. I have a total of
five p orbitals. And I know that those p
orbitals are atomic orbitals. So five atomic orbitals,
according to MO theory, are going to give me
five molecular orbitals. And I can represent
those molecular orbitals in terms of their energy
on my frost circles. I'm going to go ahead and
draw this line in here. When you're doing
frost circles, remember to always start at
the bottom here. A five-membered
ring means I'm going to try to inscribe a five-sided
polygon in my circle. So I'm going to try to
draw pentagon in here. So let's see if we can do it. So we'll put those
lines in here like that. And then you can
see that here is my pentagon inscribed
in my circle. Once again, the
important thing is where that polygon
intersects with my circle. That represents the energy
level of my molecular orbitals. And so you can see,
I'm going to have three bonding
molecular orbitals. The ones below the center line. And two antibonding
molecular orbitals like that. I need to fill my molecular
orbitals with my 6 pi electrons. And so I go ahead and put in
my 6 pi electrons like that. And you can see that I
have filled my bonding molecular orbitals analogous
to having a full outer shell. And I have Huckel's rule. I have 4n plus 2 pi electrons. And once again, we can see
that using this energy diagram here would be the two electrons. And then 4 times 1. So n is equal to 1 here. Gives me a total
of 6 pi electrons. And so this ion
satisfies both criteria. It contains a ring of
continuously overlapping p orbitals. It also has 4n plus 2 pi
electrons in the ring. And so this conjugate
base up here is stable because it's
an aromatic anion, which is the reason for such a low
pka value for cyclopentadiene. So even though cyclopentadiene
itself as non-aromatic, this ion over here
turns out to be aromatic which
explains the stability. Let's analyze one more ion here. So here we have the
cycloheptatrienol cation. And if I'm looking
for my pi electrons, I have 2, 4, and 6 pi
electrons in this ion. So a total of 6 pi electrons. When I'm looking for
sp2 hybridized carbons, everything that
has a double bond. These are all sp2 hybridized. And then my carbocation
right here at this carbon is also sp2 hybridized. So all seven carbons
are sp2 hybridized, which means that each one of
those carbons has a p orbital. And so a little bit
difficult to sketch in here. But you can see that each
one of my seven carbons has a p orbital, which makes
it relatively easy for those orbitals to overlap in my
ring to delocalize those pi electrons. And so I satisfy
my first criteria, a ring of continuously
overlapping p orbitals here. So a total of seven p orbitals. So seven atomic orbitals. And once again those
seven atomic orbitals give me seven
molecular orbitals, which I can represent
using my frost circle. So I go over here
to my frost circle and I draw my dividing
line between my bonding and my antibonding
molecular orbitals. And now since I have
a seven-membered ring, I have to attempt to sketch
in a seven-sided polygon into my frost circle. And so this is probably
one of the trickiest ones to draw here. And so let's see if we can do
something that approximates a seven-sided figure here
inside of my circle here. So something like this gives
me a seven-sided figure. And once again we care about
the points of intersection because that
represents the energy levels of my molecular orbitals. And so now it's
easy to see that we have three bonding
molecular orbitals. And four antibonding
molecular orbitals like that. And we're going to fill 6
pi electrons in our diagram. So once again, electrons fill
the lowest orbitals first. And so we can take care of
our 6 pi electrons like that. And you can see
that we've filled all of our bonding
molecular orbitals. So that's the extra
stability that goes along with an ion being aromatic here. So n is equal to 1. So 4 times 1 gives
me 4, plus the 2, gives me total of
6 pi electrons. And so both criteria
have been fulfilled. We have a ring of continuously
overlapping p orbitals, which is over here. So this, over here on the
left, is our first criteria. So this is our ring. And then we also have
4n plus 2 pi electrons. And so the cycloheptatrienol
cation is aromatic. So this is an aromatic cation. We can expect some extra
stability associated with it.