Aromaticity

Part B
Dr. P. Vasu Govardhana Reddy
Assistant Professor
Department of Chemistry
Yogi Vemana University, Kadapa
Books Referred

Advanced organic Chemistry F A Carey and R J Sundberg

Pathasala
UNIT – I: Electronic Effects and Criteria of Aromaticity
15Hrs

(A) Electronic Effects

Electronic effects: Inductive effect, mesomeric effect (Resonance), hyperconjugation, steric effect, tautomerism;
hard and soft acids and bases, acidity and basicity of organic molecules.

(B) Criteria of Aromaticity

The energy, structural and electronic criteria for aromaticity; relationship among energetic, structural and electronic
criteria; Huckle`s rule and molecular orbital theory, aromaticity in benzenoid, non-benzenoid compounds; aromaticity
in charged ring fused-ring systems; heteroaromatic systems; annulenes: cyclobutadiene, benzene,1,3,5,7-
cyclooctatetraene, [10] annulenes-1,3,5,7,9-cyclodecapentaene isomers, and [12]-, [14]-, [16]- and [18]-annulenes;
azulenes; fulvenes; fullerenes; ferrocene; anti-aromaticity; homo-aromaticity.
 Aromaticity in Benzenoid compounds
Benzene containing compounds are known as Benzenoid compounds
Aromaticity rules
1.Compound should be cyclic and planar
2.Compound should contain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow Huckel (4n+2) ℼ rule , n=0,1,2,3, 4,
etc

1 Ex: Benzene
Benzene: (4n+2)ℼ=6 ℼ
(4n+2)=6
4n=6-2
4n=4
n=1
The benzene fulfilled above all the aromaticity rules, hence this molecule exhibits the aromaticity
 1.Compound is cyclic and planar
2 2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc
Naphthalene: (4n+2)ℼ=10 ℼ
(4n+2)=10
4n=10-2
Hence, Naphthalene exhibits the aromaticity
4n=8
3 n=2
1.Compound is cyclic and planar
2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc

Naphthalene: (4n+2)ℼ=14 ℼ
(4n+2)=14
4n=14-2=12, n=3, Anthracene exhibits the aromaticity
4 Phenanthrene
1.Compound is cyclic and planar
2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc

4n+2)ℼ=14 ℼ
(4n+2)=14
4n=14-2=12, n=3, Phenanthrene exhibits the aromaticity

Aromaticity in non-benzenoid compounds

Non-benzenoid compounds: Compounds which are not having benzene ring is called non-benzenoid compounds

Ex:
 1.Compound is cyclic and planar
2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc

4n+2)ℼ=2 ℼ
(4n+2)=2
4n=2-2=0, n=0/4=0, it exhibits the aromaticity

Anti aromaticity rules

1.Compound is cyclic and planar
2.Compoundiscontain conjugated ℼ electrons
1 3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
4nℼ=4 ℼ 5. Compound should follow (4n) ℼ rule , n=0,1,2,3, 4, etc
(4n)=4, n=1, hence the molecule exhibits
The anti-aromaticity 2
 Non-aromatic
1.Compound is cyclic and planar
2.Compound is not containing conjugated ℼ electrons
3. SP3Hybridized p-orbitals
4.Delocalization is not possible

Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc

4n+2)ℼ=2 ℼ
(4n+2)=2
4n=2-2=0, n=0
eventhough it follow (4n+2) ℼ rule , since it is having sp3 hybridized orbitals, there is no
delocazation, hence the molecule does not exhibits the aromaticity, so it is non-
aromatic
1.Compound is cyclic and planar
2.Compound is not containing conjugated ℼ electrons
3. SP3Hybridized p-orbitals
4.Delocalization is not possible

Compound should follow (4n ℼ rule , n=0,1,2,3, 4, etc

4n ℼ= 4 ℼ , n==1,
Eventhough the molecule follows Huckel 4n ℼ rule, there is no conjugation and it is
not possible for delocalization, hence it is a non-aromatic
1.Compound is cyclic and planar 4nℼ=4 ℼ
2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals (4n)=4, n=1, hence the molecule exhibits
4.Delocalization and exhibits various resonance structures The anti-aromaticity
5. Compound should follow (4n) ℼ rule , n=0,1,2,3, 4, etc

1.Compound is cyclic and planar

2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc
(4n+2)ℼ=6 ℼ
(4n+2)=6
4n=6-2
4n=4
n=1, it exhibits the aromaticity
1.Compound is cyclic and planar
2.Compound is not containing conjugated ℼ electrons
3. SP3Hybridized p-orbitals
4.Delocalization is not possible

Compound should follow (4n+2)ℼ=6 ℼ

(4n+2)=6
4n=6-2
4n=4
n=1
Eventhough the molecule follows Huckel 4n+2) rule, there is no
conjugation and it is not possible for delocalization, hence it is a
non-aromatic
 1.Compound is cyclic and planar (4n+2)ℼ=6 ℼ
2.Compoundiscontain conjugated ℼ electrons (4n+2)=6
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures 4n=6-2
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc 4n=4
n=1, it exhibits the aromaticity
Tropilium cation

1.Compound is cyclic and planar 4nℼ=8 ℼ

2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals (4n)=8, n=2, hence the molecule exhibits
4.Delocalization and exhibits various resonance structures the anti-aromaticity
5. Compound should follow (4n) ℼ rule , n=0,1,2,3, 4, etc
(4n+2)ℼ=6 ℼ
1.Compound is cyclic and planar (4n+2)=6
2.Compoundiscontain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals 4n=6-2
4.Delocalization and exhibits various resonance structures 4n=4
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, etc
Tropone n=1, it exhibits the aromaticity
 Annulenes
Annulenes are conjugated monocyclic polyenes
Annulenes molecular formula CnHn
Annulenes are represented as [n] annulene

Cyclobutadiene C4H4 [4] annulene

1.Compound is cyclic and planar
2.Compoundiscontain conjugated ℼ electrons 4nℼ=4 ℼ
3. SP2 Hybridized p-orbitals (4n)=4, n=1, hence the molecule exhibits
4.Delocalization and exhibits various resonance
The anti-aromaticity
structures
5. Compound should follow (4n) ℼ rule , n=0,1,2,3, 4, etc
 2 Ex: Benzene C6H6 [6]-Annulene Benzene: (4n+2)ℼ=6 ℼ
(4n+2)=6
4n=6-2
4n=4
n=1

1.Compound should be cyclic and planar

2.Compound should contain conjugated ℼ electrons
3. SP2 Hybridized p-orbitals
4.Delocalization and exhibits various resonance structures
5. Compound should follow Huckel (4n+2) ℼ rule , n=0,1,2,3, 4, etc

The benzene fulfilled above all the aromaticity rules, hence this molecule exhibits the aromaticity
3. Cyclooctatetraene C8H8 [8]-Annulene

Cyclooctatetraene can be assumed to have a planar cyclic conjugated system which has 4n ℼ e where n = 2 as
shown in the figure

This regular planar octagon has bond angles of 135° with large bond angle strain due to large deviation from
sp2 bond angles of 120°. To overcome this strain molecule assumes a non-planar, tub-shaped geometry with
angles C=C−C = 126.1° and C=C−H = 117.6°.

Tub shaped structure is not planar, not aromatic and neither antiaromatic (as nonplanairty hinders delocalization).
Hence it is non-aromatic.
 [10] Annulene C10H10
The prospect of observing aromatic character in conjugated polyenes having 10, 14, 18, and so on _ electrons spurred
efforts toward the synthesis of higher annulenes
(4n+2)ℼ=10 ℼ
(4n+2)=10
4n=10-2
4n=8
n=2
Although 10pelectrons give an indication of aromaticity as per the (4n+2) ℼelectrons requirement for Huckel’s rule, but
the fact is that it is non-aromatic. This is because, [10] annulene is unable to adopt the necessary planar configuration

A problem immediately arises in the case of the all-cis isomer of [10]-annulene, the structure of which is shown in
the preceding problem. Geometry requires a ten-sided regular polygon to have 144° bond angles; sp2
hybridization at carbon requires 120° bond angles.

Therefore, aromatic stabilization due to conjugation in all-cis-[10]-annulene is opposed by the destabilizing effect of 24° of
angle strain at each of its carbon atoms. All-cis-[10]-annulene has been prepared. It is not very stable and is highly reactive. A
second isomer of [10]-annulene (the cis, trans, cis, cis, trans stereoisomer) can have bond angles close to 120° but is
destabilized by a close contact between two hydrogens directed toward the interior of the ring. In order to minimize the van
der Waals strain between these hydrogens, the ring adopts a nonplanar geometry, which limits its ability to be stabilized by ℼ
electron delocalization
 [12] Annulene-C12H12
The structure of [12] Annulene is planar and shown in the figure below

4nℼ=12 ℼ
(4n)=12, n=3, hence the molecule exhibits
The anti-aromaticity

The three H in-between the ring are far enough and do not create any strain for the planar arrangement. So this is a
cyclic, planar system having continuous delocalisation of pi electrons and fulfilling the first condition. But the number
of pi electrons continuously delocalised are 12 i.e 4n pi electrons, where n = 3. Since it is a system, it is anti-aromatic
in nature 4n ℼ e

(4n+2)ℼ=14 ℼ
[14] Annulene
(4n+2)=14
4n=14-2
4n=12
14-Annulene is non-aromatic Dehydro 14-Annulene is aromatic n=3

This is a 14 esystem i.e., a (4n + 2) ℼ system and can be presumed to be aromatic. However, it was found not to undergo
substitutive nitration or sulphonation reactions indicative of its nonaromatic behaviour.
 As can be seen from the figure that ‘H’ present at the interior of the ring interfere with each other, and X ray
analysis shows that the molecule is not planar. Also it was observed that Dehydro – [14] annulene formed by
removal of two interfering hydrogen leads to formation of a triple bond, and a planar molecule. The two e- s
from one of the ℼ bond of are delocalized into aromatic system and the molecule becomes aromatic. The
other pair of ℼ e-s does not interact with the delocalized system as it is at right angles to the conjugated system
of ℼ e-s.

[16] Annulene-C16H16

The first condition is fully satisfied with the molecule being cyclic, planar and continuous delocalization of pi
electrons. In the second condition, it is a 16 pi electrons system and thus a (4n) pi electron system where, n= 4.
Hence, we can conclude that it is anti-aromatic

4nℼ=16 ℼ
(4n)=16, n=4, hence the molecule exhibits
The anti-aromaticity
 [18] Annulene C18H18

(4n+2)ℼ=18 ℼ
(4n+2)=18
4n=18-2
4n=16
n=4

When the ring contains 18 carbon atoms, it is large enough to be planar while still allowing its interior hydrogens
to be far enough apart that they do not interfere with one another. The [18]-annulene shown is planar or nearly
so and has all its carbon–carbon bond distances in the range 137–143 pm—very much like those of benzene. Its
resonance energy is estimated to be about 418 kJ/mol (100 kcal/mol). Although its structure and resonance
energy attest to the validity of Hückel’s rule, which predicts “special stability” for [18]-annulene

[18] annulene is a planar, Hence the molecule exhibits the aromaticity

 Azulenes

The name ‘Azulene’ is derived from the Spanish word ‘azul’, meaning ‘blue’. Azulene (1) is a dark blue colour organic
compound. It is an isomer of naphthalene (2) having molecular formula, C10H8. Azulene is a completely conjugated non-
benzenoid fused ring hydrocarbon, which consists of a seven membered ring fused with a five membered ring. While
naphthalene is a benzenoid hydrocarbon consisting of two benzene rings fused together at ortho-position.

Azulene (bicycle[5.3.0]decapentaene, 1) is a non-benzenoid aromatic compound, while naphthalene is a benzenoid

aromatic compound. The higher stability of naphthalene over azulene is due to resonance stabilization, since only two
resonance structures can be written for azulene compared to three for naphthalene
Synthesis of Azulenes
Aromaticity of Azulenes

1.Compound is cyclic and planar

2.Compoundiscontain conjugated ℼ electrons (4n+2)ℼ=10 ℼ
3. SP2 Hybridized p-orbitals (4n+2)=10
4.Delocalization and exhibits various resonance structures
4n=10-2
5. Compound should follow (4n+2) ℼ rule , n=0,1,2,3, 4, et
4n=8
n=2

Azulene is an aromatic compounds and hence easily undergoes electrophilic substitution reactions like halogenations,
nitration, Friedel-Craft alkylation reaction etc.
 Fulvenes

Fulvalenes represent another interesting class of compounds to look for potential aromaticity. Among possible symmetrical
structures, pentafulvalene and heptafulvalene have been prepared, but were found to exhibit polyene character. However,
when a combination of rings, such as cyclopentadiene and cyclopropene were examined, results were in support of the
existence of dipolar resonance structures. The large measured dipole moments of phenyl substituted analog x and reduced
barrier of rotation (as revealed by NMR) of the dialkyl substituted analog x’ are manifestations of such effects.
Aromaticity in fulvenes
Ferrocene is an organometallic compound with the formula Fe(C5H5)2. The molecule consists of
two cyclopentadienyl rings bound on opposite sides of a central iron atom
(4n+2)ℼ=18 ℼ

Synthesis of ferrocene (4n+2)=18

4n=18-2
4n=16
n=4
 Aromaticity in Ferrocene

The number of π-electrons on each ring is then six, which makes

it aromatic according to Hückel's rule. These twelve π-electrons are then
shared with the metal via covalent bonding. Since Fe2+ has six d-electrons, the
complex attains an 18-electron configuration, which accounts for its stability.

Ferrocene is an aromatic compounds and hence easily

undergoes electrophilic substitution reactions like
halogenations, nitration, acylation reaction etc.
(4n+2)ℼ=18 ℼ
(4n+2)=18
4n=18-2
4n=16 Ferrocene exhibits the
aromaticity
n=4
 FULLERENES
Fullerenes are allotropes of carbon

Allotropes of an element are different forms of the same element with different arrangement of the bonded atoms,
in same physical state.
 Carbon has many allotropes namely – diamond , graphite, graphene , carbon nanotubes and fullerenes.
 In diamond, all carbon atoms are sp3 hybridized. Thus, in diamond the carbon atoms are in a tetrahedral arrangement.
 In graphite , the carbons are sp2 hybridized and they occur in layer form. Graphene is a single layered sheet of carbon
atoms and carbon nanotubes are one layer tubes of carbon atoms.
 Fullerenes have carbon atoms in hollow sphere form
Fullerenes discovered by Harry Kroto and Robert Curl and Richard Smalley

In 1996, Nobel prize in chemistry was awarded to these three scientists for their work on carbon clusters.

Fullerenes are sphere-like and cage- like molecules of carbon atoms which look like a geodesic dome.
Fullerenes consist of fused pentagons and hexagons. No two pentagons lie side by side. They have
soccer ball like structure.
The smallest , most stable fullerene obtained by usual preparation methods is the buckyball or C60

C70 has an ellipsoidal cage-like structure. The red atoms indicate five hexagons additional to the
C60 molecule
C60 ⇒ 20 hexagonal faces + 12 pentagons.
C70 ⇒ 25 hexagonal faces + 12 pentagons.

C60
Aromaticity of Fullerenes – Spherical aromaticity.

 Each carbon atom in a fullerene is sp2 hybridized and forms σ bonds with three other carbon atoms.
 The fourth electron is delocalized into the system. The five membered rings are antiaromatic and six
membered rings aromatic.
 Thus, fullerenes have both stabilizing and destabilizing elements in its structure. So, it is very difficult to
explain aromaticity of fullerenes.
 Aromaticity of fullerenes is not the same as that for benzene or other planar compounds. Fullerenes
have a conjugated π system of electrons.
 They are resistant to high temperatures. They have stable structures. However , some
characteristic reactions for planar aromatic compounds are not observed for fullerenes. For example,
the typical aromatic substitution reactions are not possible for fullerenes.
Molecular orbital description of aromaticity and antiaromaticity

Benzene

According to molecular orbital theory, these six p orbitals combine to form six molecular orbitals, three of which are
bonding and three, anti-bonding. Six π electrons occupy the bonding orbitals, which are lower in energy compared
to the un-hybridized p orbitals (atomic orbitals).

The relative energies of p molecular orbitals in planar cyclic conjugated systems can be determined by a simplified
approach developed by A. A. frost in 1953.
This involves the following steps: 1) Draw a circle 2) Place the ring (polygon representing the compound of interest) in
the circle with one of its vertices pointing down.
Each point where the polygon touches the circle represents an energy level. 3) Place the correct number of electrons
in the orbitals, starting with the lowest energy orbital first, in accordance with Hund’s rule.