Ask A Scientist Chemistry Archive

(Created prior to 1993)

Question: In order for a molecule to be aromatic, it must be flat for p-
orbitals to overlap; it must be conjugated, and it must have 4n+2 pi 
electrons.  This is understandable for monocyclic compounds since the energies 
for an n-site cyclic system is given by:  Ek = -2*t*cos(2*pi*k/n)  where 
k=0,=+-1,=+-2,...,+-(n-1)/22)/2 if n is even and, and k=0,+11-1,+12+-2,...,+1-
(n-1)/2 if n is odd.  Thus, when electrons fill MO's, it takes 2 electrons to 
fill the lowest lying orbital and 4 electrons to fill each succeeding 
degenerate energy level.  However, for polycyclic aromatic compounds such as 
naphthalene, the 4n+2 rule still applies for determining aromaticity, but in 
computing the energies via Huckel, there are no degeneracies. There seems to 
be something fishy here.  How are we still be allowed to use the 4n+2 rule 
when there are no degeneracies in energy?  Also, the buckyball is considered 
to be aromatic, but it is not planar.  Why is it aromatic?
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I am thinking about this one carefully.  I am acquainted with a 
world-famous physical organic chemist at an Ivy League university.  In these, 
the latter days of his career, he has embarked on a major research program to 
more fully define the meaning of "aromaticity" in chemistry.  He claims that 
it is a complex phenomenon not completely understood, and I am inclined to 
agree with him.  That said, here goes:  There are many working definitions of 
aromaticity.  One book I have says that an aromatic compound is "a cyclic 
compound containing some number of conjugated double bonds, characterized by 
an unusually large resonance energy."  (L.G.Wade Jr., "Organic Chemistry" 
(Prentice-Hall, New York, 1987.)  Note that this definition says nothing about 
the number of electrons, or whether the molecule is flat; it just has to have 
conjugated double bonds, be arranged in some sort of cyclic fashion, and have 
something called a "resonance energy" which is unusually large. The resonance 
energy is obtained by comparing the heats of hydrogenation of alkene fragments 
to the heat of hydrogenation of the aromatic molecule.  Essentially, this 
resonance energy represents a breakdown of the "local bond" picture of 
chemistry.  Molecular orbital theory is required to take resonance, or 
delocalization of a molecular wave function over an entire molecule, into 
account.  That said, let us look at sfung's question/comment more directly.  
First of all, the expression given for the energies of an n-site cyclic 
system, i.e., Ek = -2*t*cos(2*pi*k/n), is only strictly correct for a 
monocyclic system (i.e. with one ring).  For such systems, the type of pi 
molecular orbitals depends on whether the ring has an odd or an even number of 
carbons.  If N (N=number of carbon atoms) is odd, there will be (N-2)/2 
bonding, (N-2)/2 antibonding, and 2 nonbonding MOs, and the nonbonding MOs 
will be occupied by one electron each.  If N is even,there will be N/2 bonding 
and N/2 antibonding MOs, and all electrons will be paired up in bonding MOs.  
This turns out to be equivalent to the 4n+2 rule, where n is the number of pi 
electrons.  If you look at the bicyclic compounds in the same way, (see O. 
Sinanoglu, Tetrahedron Letters 29, pp.889-892 (1988)), the situation is much 
more complicated and depends on whether each ring has an even or odd number of 
carbons...for example, note that cyclobutadiene is antiaromatic! So, in the 
end, question whether one is "allowed" to use the 4n+2 rule on bicyclic 
compounds.  Whether degeneracy is relevant or not is another story . . .  But 
I will not get into that.  Next time I log in I will comment on "buckyball" 
and aromaticity.  PS: Good question!!!!!
Topper
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I made a slight mistake in the previous discussion.  The trends I 
mentioned for the planar rings are not quite right.  If one has an even number 
of carbons, then as one increases the number of carbons (4,6,8,etc...) one 
observes the alternation I mentioned between having the HOMO (highest occupied 
molecular orbital) be non-bonding and having the HOMO be bonding.  If the 
number of carbons is odd, (3,5,7, etc) then the system is never aromatic.  
Sorry again for my mistake!  Just goes to show that no one is infallible, not 
even a "scientist."
Topper
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