What makes a compound “aromatic”?
(i.e. What is required for complete
delocalization?)
1. Planar System (for perfect p orbital overlap)
2. Sp2 hybridization (every atom has a p orbital)
3. Cyclic system (“monocyclic”)
4. 4n + 2 π electrons
Huckel’s Rule for pi electrons:
The perfect number of π electrons for an aromatic system will fit the equation 4n
+2, where n is any whole number (0, 1, 2, etc).
Perfect values would then be 2, 6, 10, etc.
This perfect number of π electrons is perfect because they completely fill the bonding
molecular orbitals for varying types of aromatic rings.
• More than this number and electrons must go into higher energy antibonding
orbitals
• Less than this number and radical systems may exist and unpaired electrons will
be in the orbitals.
If the number of π e- fits the equation 4n, the compound is called “antiaromatic” and is
HIGHLY reactive.
Consider each of the following: Aromatic?
Ex.
Ex.
Ex.
Larger rings with conjugated cis double bonds cannot be planar due to angle strain.
Annulenes are a class of cyclic compounds that contain both cis and trans double bonds
in the ring system.
Ex. [18] Annulene
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Consider these:
a.
Only four of the carbons are sp2 hybridized. The fifth carbon is sp3 with four covalent
bonds. Not aromatic.
b.
H
=
Cyclic. All carbons are sp2 hybridized (carbocations are sp2 hybridized). The total
number of π electrons is 4. For 4n+2 = 4, n = ½ (not a whole number). Not aromatic.
c.
H
=
The cyclopentadienyl anion (seen in the compound Ferrocene) is cyclic. Four of the
carbons are sp2 hybridized and the fifth carbon (three covalent bonds and a lone pair)
appears as though it should be sp3, according to common assignment of hybridization of
atoms. This would make it NOT aromatic.
Consider what would happen if that fifth carbon were sp2 hybridized:
=
H
H
When the fifth carbon is sp2 hybridized, all carbons are sp2. The fifth carbon has
three covalent bonds (two to carbons in the ring, one to the hydrogen atom still
attached), and the lone pair of electrons is in a perpendicular p orbital, which would
place the lone pair in the delocalized pi system. There are five p orbitals, aligned and
2
overlapping, and six π electrons (four from the pi bonds and a pair of electrons in the p
orbital of the fifth carbon, as shown above). For 4n+2 = 6, n = 1.
If a compound can be low energy and super stable by containing an aromatic ring system,
it will. Thus, this cyclopentadienyl anion IS an aromatic compound (and that fifth
carbon is definitely sp2 hybridized!).
Supporting fact: The pKa value of the sp3 C-H of cyclopentadiene is 16, not 60.
Why?
The anion of cyclopentadiene is aromatic and extremely stable.
Now, how about:
d.
Cyclic and has 6 π electrons but one carbon is sp3 hybridized (with four covalent bonds)
so overall, the system is not aromatic.
e.
H
=
The anion of cycloheptatriene is a cyclic compound. If the anionic carbon is traditionally
considered sp3 hybridized stays sp3, then complete delocalization isn’t possible (need
sp2) and not aromatic. If the anionic carbon were to be hybridized sp2, then there will
be 8 π electrons and 4n+2 = 8 and n = 1½ (not a whole number). Still not aromatic. Can’t
make this one aromatic.
f.
H
=
3
The cation of cycloheptatriene is cyclic, sp2 hybridized all around and the cation carbon
has a p orbital that is empty. Total number of pi electrons? 6 pi electrons. If 4n+2 = 6
then n = 1. Aromatic!
Consider the following three: Which would be aromatic?
Heterocyclic compounds are cyclic compounds that contain atoms other than C in their
ring structure. Consider whether these are aromatic and if so, how that is possible.
Pyridine:
N
There are six atoms (five carbons, one nitrogen) in this ring and they are already all sp2
hybridized. Inside the ring, there are shown three pi bonds with six pi electrons (4n+2
= 6; n = 1). Pyridine is therefore “aromatic”. Note that the lone pair on nitrogen is NOT
included in the pi system. In what type of orbital is that lone pair existing?
The lone pair on nitrogen is in an sp2 hybrid orbital and is not part of the π system so its
not being delocalized and it is available so that nitrogen can act as a base.
Pyrrole:
H
N
There are five atoms in this ring and the four of the carbons are visibly in double bonds
and thus recognizable as sp2 hybridized. The nitrogen atom, with three covalent bonds
and a lone pair, is classically considered to be sp3 hybridized, which would exclude this
compound from being “aromatic”.
Consider the result if the nitrogen atom was sp2 hybridized. As we saw for the
cyclopentadienyl anion, the atom has three covalent bonds (two to carbons and one to
hydrogen) and a lone pair. The lone pair would exist in the perpendicular p orbital, giving
a total of 6 pi electrons. Cyclic, sp2, Huckel’s Rule. Aromatic.
4
=
N
N
H
H
Note: Unlike pyridine, Pyrrole cannot act as a base. The lone pair shown drawn on
nitrogen is busy participating in the pi system of the aromatic ring.
Imidazole:
H
N
N
An interesting ring, imidazole has two nitrogen atoms to consider.
H
N1
N2
The nitrogen labeled “N2” is in the ring, appearing as part of a double bond so
the sp2 hybridization is obvious and the lone pair on N2 will not involved in the
pi system. That lone pair will be found in an sp2 hybrid orbital, much like the
lone pair found on the nitrogen in pyridine.
The nitrogen labeled “N1” must be considered though. If it remains sp3
hybridized (due to the three covalent bonds and a lone pair), it would defeat
N1
the requirement that all atoms must be sp2 hybridized and imidazole would not
be aromatic. If the nitrogen of N1 is sp2 hybridized, then the lone pair will be
N2 in a perpendicular p orbital and there will be six pi electrons (4n+2 = 6; n = 1).
Since this will make the compound aromatic, N1 IS sp2 hybridized. N1 is not
basic though since its lone pair is involved in the delocalized pi system.
H
Furan:
O
Consider that oxygen atom (with its two lone pairs). If it remains sp3 hybridized, then
the ring cannot be aromatic. If the oxygen is sp2 hybridized, what happens to its two
lone pairs of electrons?
O
=
5
This may look very odd, but one lone pair is going to wind up in an sp2 hybrid orbital, like
the lone pair on nitrogen in pyridine. The other lone pair will be in the perpendicular p
orbital and part of the pi system. The total number of pi electrons will then be six and
the system is aromatic. Oxygen still has basic character, due to the remaining lone pair
on oxygen.
A comment about “Monocyclic” – All aromatic rings must be “monocyclic” yet we have
many rings that appear to have multiple rings in their structures:
Ex. Naphthalene (mothballs) and Anthracene
In both of these examples, one can see what appears to be a benzene ring on the left of
the resonance forms that are drawn. Obviously this is aromatic. But how extensive is
the delocalization?
Look at naphthalene, for instance. All ten π electrons of naphthalene are able to
delocalize themselves through the entire ring system of both rings (but if you look at
the p orbitals, you’ll see that it really is one large ring, not two separate rings).
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