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Organic chemistry
Aromatic stability III
Aromaticity of the cyclopropenyl cation and antiaromaticity of cyclobutadiene. Created by Jay.
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If a molecule is antiaromatic, would it still be considered aromatic, or non-aromatic?(4 votes)
An antiaromatic compound has a planar ring of overlapping p orbitals, the same as an aromatic
compound. The difference is that delocalization of the electrons over the ring raises the energy,
rather than lowering it, usually because it is a system of 4n electrons (Huckel's rule for antiaromatic).
Cyclobutadene is antiaromatic, as would be cyclooctatetraene if it were planar, which it is not.
A non-aromatic compound lacks the planar ring of overlapping p orbitals, so Huckel's rule does not apply. Examples are cyclohexadiene, cycloheptatriene, and even the actual cyclooctatetraene, which folds into a bathtub shape to avoid overlap of the p orbitals, which would make it antiaromatic.(11 votes)
How that carbo cation is sp2 hybridized ?
Please help.
Thank you.(6 votes)
Because it is bonded to two adjacent C atoms and one H atom ( which is invisible in the structure) .
Since, The steric Number = No. of sigma bonds + No. of Lone pairs
For that Carbocation , it will be 3 thus it is sp2 hybridised ( remember it has no lone pair !)
Thus, that carbocation is sp2 hybridised !!
Remember= Steric number is always use to determine the Hybridisation !!(6 votes)
- what s the difference betwen antibonding and non bonding? i thought the orbitles above the line are non-bonding.. but according to this video they are antibonding ?(2 votes)
- Orbitals above the line are antibonding.
Only orbitals *on the line* are nonbonding. Examples are the two degenerate orbitals of cyclobutadiene at8:15.(6 votes)
- what is the significance of non-bonding molecular orbitals(4 votes)
A non-bonding orbital, also known as non-bonding molecular orbital (NBMO), is a molecular orbital whose occupation by electrons neither increases nor decreases the bond order between the involved atoms. Non-bonding orbitals are often designated by the letter n in molecular orbital diagrams and electron transition notations. Non-bonding orbitals are the equivalent in molecular orbital theory of the lone pairs in Lewis structures. The energy level of a non-bonding orbital is typically in between the lower energy of a valence shell bonding orbital and the higher energy of a corresponding antibonding orbital(1 vote)
- Can someone please answer this, thanks a a lot!
How can we tell the difference between anitaromatic and non aromatic? In this video, it is mentioned that the cyclobutadiene is antiaromatic because it meets the first criteria for aromaticity but not the second criteria, and for cyclooctatetraene it is said to be non aromatic because it doesn't meet either the first or second criteria. My question is how can we determine if a compound is antiaromatic or non aromatic if the first criteria that requires a compound to have continuously overlapping p orbitals is not clear to find/see? Because in the cyclooctatetraene example there is no way for us to determine on our own that it is a tub shape.(2 votes)- The answer is that any molecule that has more than 6 carbon atoms in a regular polygon with alternating double and single bonds will not be planar, because the internal bond angles deviate from the preferred bond angle of 120°.
The ring will pucker to reduce the bond angle strain.
You may not be able to predict the shape, but you know that the molecule will not be planar.
However, the ions C₇H₇⁺ and C₈H₈²⁻ can be planar because their resonance energy is large enough to compensate for the bond angle strain energy.
And larger rings can be planar and aromatic by having trans double bonds in the ring.
For example, search on-line for [18]annulene.(3 votes)
- at9:00, you said the cyclobutadiene is antiaromatic, but you didn't really mention this molecule is planar.
only at the start when you talk about this molecule, you mentioned the p orbitals may overlap each other.
but may doesn't mean definitely(1 vote)- If a compound doesn't follow Huckel's rule it can't be aromatic. In fact cyclobutadiene has 4n pi electrons which would make it antiaromatic. And yes it is planar. In fact what happens to avoid this issue: cyclobutadiene will distort into a rectangle with 2 long sides and 2 short sides, this will make more sense if you draw the frost circle for the rectangle. Hopefully that makes sense.(4 votes)
at7:01you are talking about anti aromatic, when 4n=4 is that the immediate indicator that it is anti aromatic? like if on an exam i see that it has 4 pi electrons I will know its anti, or is it specific to this example?(2 votes)
Are MO that are on the "center line" considered bonding or non bonding?(1 vote)
They are non-bonding – the orbitals with even higher energies are called anti-bonding.(2 votes)
- in case of carbocations even if they are sp2 hybridised how come a free p orbital arises over here . because that one does not possess any eletron . i mean we count any orbital's existence only when the electrons are present in it . right ?(1 vote)
- A p orbital exists whether or not there are any electrons in it. A p orbital without any electrons is called a vacant p orbital.(2 votes)
- why is it that the orbitals beneath the frost circle line are bonding orbitals and those above are anti bonding orbitals and those on the line are non bonding orbitals? what do these names mean and why are they like that(1 vote)
Bonding, antibonding and nonbonding orbitals come from molecular orbital theory.
The basic concept is that two atomic orbitals can combine in phase (think constructive interference of waves) which gives a bonding orbital and out of phase (destructive interference) which gives an antibonding orbital.
Electrons in bonding orbitals strengthen the bond between atoms, electrons in antibonding orbitals weakens the bond between atoms.
Nonbonding does neither of these.
A frost circle is a quick way to create the pi molecular orbitals of rings. There’s no real way to answer why this works, it’s a quick trick that saves having to model it on a computer.(2 votes)
8:15Video 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.