Aromaticity and Benzene.ppt

BENZENE
AND
AROMATIC
COMPOUNDS
Bedanta Kr Borah
Assc. Prof.
D. R. College, Golaghat
Assam
bedanta69@gmail.com
2/14/2024 1

Aliphatic hydrocarbons are open-chain and
ring compounds that react like open chain
compounds:
saturated: alkanes and cycloalkanes
(typical reaction = substitution)
unsaturated: alkenes, alkynes, dienes,
cycloalkenes (typical reaction = addition).
Aromatic hydrocarbons are unsaturated
ring compounds that resist the typical
addition reactions of aliphatic unsaturated
compounds, instead undergoing
substitution reactions. They are also much
more stable than they should be.
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4
• Benzene (C6H6) is the simplest aromatic hydrocarbon (or
arene).
• Benzene has three degrees of unsaturation, making it a
highly unsaturated hydrocarbon.
• Whereas unsaturated hydrocarbons such as alkenes,
alkynes and dienes readily undergo addition reactions,
benzene does not.
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KMnO4 oxidation no reaction
Br2/CCl4 addition no reaction
HI addition no reaction
H2/Ni reduction no reaction
Reagent Cyclohexene Benzene
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7
• Benzene does react with bromine, but only in the presence of
FeBr3 (a Lewis acid), and the reaction is a substitution, not an
addition.
• Proposed structures of benzene must account for its high
degree of unsaturation and its lack of reactivity towards
electrophilic addition.
• August Kekulé proposed that benzene was a rapidly
equilibrating mixture of two compounds, each containing a six-
membered ring with three alternating  bonds.
• In the Kekulé description, the bond between any two carbon
atoms is sometimes a single bond and sometimes a double
bond.
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• These structures are known as Kekulé structures.
• Although benzene is still drawn as a six-membered ring with
alternating  bonds, in reality there is no equilibrium between the
two different kinds of benzene molecules.
• Current descriptions of benzene are based on resonance and
electron delocalization due to orbital overlap.
• In the nineteenth century, many other compounds having
properties similar to those of benzene were isolated from natural
sources. Since these compounds possessed strong and
characteristic odors, they were called aromatic compounds. It
should be noted, however, that it is their chemical properties,
and not their odor, that make them special.
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Any structure for benzene must account for the following facts:
1. It contains a six-membered ring and three additional
degrees of unsaturation.
2. It is planar.
3. All C—C bond lengths are equal.
The Kekulé structures satisfy the first two criteria but not the
third, because having three alternating  bonds means that
benzene should have three short double bonds alternating with
three longer single bonds.
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• The resonance description of benzene consists of two equivalent
Lewis structures, each with three double bonds that alternate
with three single bonds.
• The true structure of benzene is a resonance hybrid of the two
Lewis structures, with the dashed lines of the hybrid indicating
the position of the  bonds.
• We will use one of the two Lewis structures and not the hybrid in
drawing benzene. This will make it easier to keep track of the
electron pairs in the  bonds (the  electrons).
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• Because each  bond has two electrons, benzene has six
 electrons.
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• In benzene, the actual bond length (1.39 Å) is
intermediate between the carbon—carbon single bond
(1.53 Å) and the carbon—carbon double bond (1.34 Å).
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17
Draw all possible resonance structures for biphenyl?
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What orbitals are used to form the indicated bonds, and of
those which is the shortest?
H
Csp2-Hs
Csp2-Csp3 Csp2-Csp2
Cp-Cp
Csp2-Csp2
Csp2-Csp2
Cp-Cp
shortest
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• To name a benzene ring with one substituent, name the
substituent and add the word benzene.
Nomenclature of Benzene Derivatives
• Many monosubstituted benzenes have common names
which you must also learn.
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• There are three different ways that two groups can be
attached to a benzene ring, so a prefix—ortho, meta, or
para—can be used to designate the relative position of
the two substituents.
ortho-dibromobenzene
or
o-dibromobenzene
or 1,2-dibromobenzene
meta-dibromobenzene
or
m-dibromobenzene
para-dibromobenzene
or
p-dibromobenzene
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• If the two groups on the benzene ring are different,
alphabetize the names of the substituents preceding the
word benzene.
• If one substituent is part of a common root, name the
molecule as a derivative of that monosubstituted
benzene.
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For three or more substituents on a benzene ring:
1. Number to give the lowest possible numbers around the ring.
2. Alphabetize the substituent names.
3. When substituents are part of common roots, name the
molecule as a derivative of that monosubstituted benzene. The
substituent that comprises the common root is located at C1.
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• A benzene substituent is called a phenyl group, and it can be
abbreviated in a structure as “Ph-”.
• Therefore, benzene can be represented as PhH, and phenol
would be PhOH.
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• The benzyl group, another common substituent that contains a
benzene ring, differs from a phenyl group.
• Substituents derived from other substituted aromatic rings are
collectively known as aryl groups.
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27
Give the IUPAC name for each compound.
PhCH(CH3)2 isopropylbenzene
OH
m-butylphenol
Cl
Br
2-bromo-5-chlorotoluene
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28
Which structure matches the given name?
o-dichlorobenzene
Cl Cl
A
Cl
Br
B
Cl
Cl
C
Cl
Cl
D
Cl
Cl
C
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29
4-chloro-1,2-diethylbenzene
Cl
A
Cl
B
Cl
C
Cl
D
Cl
A
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30
Spectroscopic Properties of Benzene
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31
Figure 17.2
13C NMR absorptions
of the three isomeric
dibromobenzenes
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32
Give the structure of the compound, C10H14O2, with an IR absorption
at 3150-2850 cm-1. Also has +1H NMR peaks:
1.4 ppm (triplet 6H)
4.0 ppm (quartet 4H)
6.8 ppm (singlet 4H)
2(10)+2-14=8/2=4
IR absorption tells that there are sp2 and sp3 C-H bonds
6.8 ppm tells us its aromatic but with only one type of proton,
what does this tell us?
So aromatic plus 4 DOUS equals benzene ring.
So where would the substituents have to be?
We know we get a doublet of doublets when benzene is para
substituted but with two different groups.
When the substituents are the same what do you get?
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One type of proton due to symmetry. So we know that we have a
benzen ring with para subsituents that are the same.
X X
So this leaves us 4 Cs and 2 Os. And we have a quartet and a
triplet up field for nonaromatic protons, so that tells us that we
could have an ethyl group present without any effect from the
ring so we must have an ether.
O O
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• Consider the heats of hydrogenation of cyclohexene, 1,3-
cyclohexadiene and benzene, all of which give cyclohexane
when treated with excess hydrogen in the presence of a metal
catalyst.
Stability of Benzene
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Figure 17.6 compares the hypothetical and observed heats of
hydrogenation for benzene.
Resonance Energy: The energy difference between most stable
canonical structure and resonance energy.
High resonance energy more will be stability.
Figure 17.6
A comparison between the
observed and hypothetical
heats of hydrogenation
for benzene
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• The low heat of hydrogenation of benzene means that benzene is
especially stable—even more so than conjugated polyenes. This
unusual stability is characteristic of aromatic compounds.
• Benzene’s unusual behavior is not limited to hydrogenation.
Benzene does not undergo addition reactions typical of other
highly unsaturated compounds, including conjugated dienes.
• Benzene does not react with Br2 to yield an addition product.
Instead, in the presence of a Lewis acid, bromine substitutes for
a hydrogen atom, yielding a product that retains the benzene
ring.
2/14/2024

Eric Hückel
1896
"Eric Hückel received his PhD in experimental
Physics in 1921 at the University of Göttingen.
After spending a year with David Hilbert
in mathematics and Max Born, he left Göttingen
for a position at the ETH Federal Institute of
Technology in Zurich with Peter Debye.
While at Zurich, Hückel and Debye
developed the Debye-Hückel theory of strong
electrolytes. In 1930 Hückel received
an appointment in chemical physics
at the Technical Institute in Stuttgart.
In 1937 he was appointed a professor of theoretic
physics at the University of Marburg,
where he remained until his retirement in 1962."
(Source: DA McQuarrie "Quantum Chemistry",
University Science Books, 1983)
In the quantum chemistry community,
Hückel is best known for introducing in 1930
a simple theory for the treatment of
conjugated molecules and aromatic molecules.
This theory came to be known as"Hückel molecul
orbital theory" or simply"Hückel Theory".
This was later extended by Roald Hoffmann
and has been widely used in organic
and inorganic chemistry.

40
Four structural criteria must be satisfied for a compound
to be aromatic.
The Criteria for Aromaticity—Hückel’s Rule
[1] A molecule must be cyclic.
To be aromatic, each p orbital must overlap with p orbitals
on adjacent atoms.
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[2] A molecule must be planar.
All adjacent p orbitals must be aligned so that the 
electron density can be delocalized.
Since cyclooctatetraene is non-planar, it is not aromatic,
and it undergoes addition reactions just like those of other
alkenes.
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[3] A molecule must be completely conjugated.
Aromatic compounds must have a p orbital on every atom.
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[4] A molecule must satisfy Hückel’s rule, and contain
a particular number of  electrons.
Benzene is aromatic and especially stable because it
contains 6  electrons. Cyclobutadiene is antiaromatic and
especially unstable because it contains 4  electrons.
Hückel's rule:
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Note that Hückel’s rule refers to the number of  electrons,
not the number of atoms in a particular ring.
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1. Aromatic - A cyclic, planar, completely conjugated
compound with 4n + 2  electrons.
2. Antiaromatic - A cyclic, planar, completely conjugated
compound with 4n  electrons.
3. Not aromatic (nonaromatic) - A compound that lacks
one (or more) of the following requirements for
aromaticity: being cyclic, planar, and completely
conjugated.
4. Homoaromatic – Aromatic ring intercepted by –CH2
group
Considering aromaticity, a compound can be classified in
one of four ways:
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Note the relationship between each compound type and a similar
open-chained molecule having the same number of  electrons.
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• [10]-Annulene has 10  electrons, which satisfies
Hückel's rule, but a planar molecule would place the two
H atoms inside the ring too close to each other. Thus,
the ring puckers to relieve this strain.
• Since [10]-annulene is not planar, the 10  electrons
can’t delocalize over the entire ring and it is not
aromatic.
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• 1H NMR spectroscopy readily indicates whether a
compound is aromatic.
• The protons on sp2 hybridized carbons in aromatic
hydrocarbons are highly deshielded and absorb at 6.5-8
ppm, whereas hydrocarbons that are not aromatic
absorb at 4.5-6 ppm.
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Examples of Aromatic Rings
• Completely conjugated rings larger than benzene are
also aromatic if they are planar and have 4n + 2 
electrons.
• Hydrocarbons containing a single ring with alternating
double and single bonds are called annulenes.
• To name an annulene, indicate the number of atoms in
the ring in brackets and add the word annulene.
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CH2
A B
Which of these is aromatic?
A) Is aromatic. Count the number of pi
bonds in the outer ring. A has 5 which
means 10 pi electrons, 4(2)+2=10. While B
has 6 pi bonds and 12 pi electrons,
4(3)=12. Doesn’t meet the Huckel rule
requirements for aromaticity.
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53
• Two or more six-membered rings with alternating double and
single bonds can be fused together to form polycyclic aromatic
hydrocarbons (PAHs).
• There are two different ways to join three rings together, forming
anthracene and phenanthrene.
• As the number of fused rings increases, the number of
resonance structures increases. Naphthalene is a hybrid of
three resonance structures whereas benzene is a hybrid of two.
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54
What is the correct name for this compound?
A) 3-Nitrotoluene
B) 4-Nitromethylbenzene
C) p-Nitrotoluene
D) (4,1)-Methylnitrobenzene
O2N
C) p-Nitrotoluene
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55
A) 3,5-difluoroanisole
B) Difluoromethoxybenzene
C) 1,5-difluoro-3-methoxybenzene
D) 1,3-difluoro-5-methyl-O-benzene
O
F
F
3,5-difluoroanisole
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56
A) 4-bromo-m-xylene.
B) 1-bromo-2,4-dimethylbenzene.
C) p-bromo-m-methyltoluene.
D) o-bromo-m-methyltoluene.
Br
B) 1-bromo-2,4-dimethylbenzene.
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Br
1
2
3
4 5
6
1,2,4
Br
4
3
2
1 6
5
1,3,4
Br
2
1
6
5 4
3
1,2,5
Br
6
1
2
3 4
5
1,3,6
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58
What is the correct name?
A) 1-fluoro-3-isopropyl-5-ethylbenzene
B) 1-ethyl-3-isopropyl-5-fluorobenzene
C) 1-ethyl-3-fluoro-5-isopropylbenzene
D) 1-isopropyl-3-fluoro-5-ethylbenzene
F
C) 1-ethyl-3-fluoro-5-isopropylbenzene
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59
Is this compound aromatic or antiaromatic?
Antiaromatic – cyclic, planar, conjugated , but does
not meet Huckel’s rule.
4 doulbe bonds and 2 triple bonds so 4(2) + 2(4)=16
pi electons. 4n+2 or 4n? 4(4)=16
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60
Indicate which of the following are aromatic and
antiaromatic?
A
B
C
D
C is aromatic 4(3)+2=14
A is antiaromatic 4(2)=8
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• Heterocycles containing oxygen, nitrogen or sulfur, can also be
aromatic.
• With heteroatoms, we must determine whether the lone pair is
localized on the heteroatom or part of the delocalized  system.
• An example of an aromatic heterocycle is pyridine.
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• Pyrrole is another example of an aromatic heterocycle. It
contains a five-membered ring with two  bonds and one
nitrogen atom.
• Pyrrole has a p orbital on every adjacent atom, so it is
completely conjugated.
• Pyrrole has six  electrons—four from the  bonds and two from
the lone pair.
• Pyrrole is cyclic, planar, completely conjugated, and has 4n + 2 
electrons, so it is aromatic.
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• Histamine is a biologically active amine formed in many tissues.
It is an aromatic heterocycle with two N atoms—one which is
similar to the N atom of pyridine, and the other which is similar
to the N atom of pyrrole.
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67
Both negatively and positively charged ions can be aromatic if
they possess all the necessary elements.
We can draw five equivalent resonance structures for the
cyclopentadienyl anion.
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68
• Having the “right” number of electrons is necessary for a
species to be unusually stable by virtue of aromaticity.
• Thus, although five resonance structures can also be
drawn for the cyclopentadienyl cation and radical, only
the cyclopentadienyl anion has 6  electrons, a number
that satisfies Hückel’s rule.
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• The tropylium cation is a planar carbocation with three
double bonds and a positive charge contained in a
seven-membered ring.
• Because the tropylium cation has three  bonds and no
other nonbonded electron pairs, it contains six 
electrons, thereby satisfying Hückel’s rule.
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70
Which of the following is aromatic?
A B
C D
C is aromatic 10 pi electrons, 4(2)+2=10 and
completely conjugated b/c lone pair is in a p
orbital.
Which are antiaromatic?
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71
Which of these is antiaromatic?
A B
C
D
B 8 pi electrons 4(2)=8
C and D as well, 8 and 4 respectively
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72
Carcinogens
• Fused ring of benzene cause cancer
• Contain four or more fused rings
• Present in tobacco smoke, auto exhaust, and
burned food
• The more a person smokes, the greater the risk of
developing cancer
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The Basis of Hückel’s Rule
• Why does the number of  electrons determine whether
a compound is aromatic?
• The basis of aromaticity can be better understood by
considering orbitals and bonding.
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74
• Thus far, we have used “valence bond theory” to
explain how bonds between atoms are formed.
• Valence bond theory is inadequate for describing
systems with many adjacent p orbitals that overlap, as
is the case in aromatic compounds.
• Molecular orbital (MO) theory permits a better
explanation of bonding in aromatic systems.
• MO theory describes bonds as the mathematical
combination of atomic orbitals that form a new set of
orbitals called molecular orbitals (MOs).
• A molecular orbital occupies a region of space in a
molecule where electrons are likely to be found.
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75
• When forming molecular orbitals from atomic orbitals,
keep in mind that a set of n atomic orbitals forms n
molecular orbitals.
• If two atomic orbitals combine, two molecular orbitals
are formed.
• Recall that aromaticity is based on p orbital overlap.
• Also note that the two lobes of each p orbital are
opposite in phase, with a node of electron density at the
nucleus.
• When two p orbitals combine, two molecular orbitals
should form.
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• The combination of two p orbitals can be constructive—that is,
with like phases interacting—or destructive, that is, with
opposite phases interacting.
• When two p orbitals of similar phase overlap side-by-side, a 
bonding molecular orbital results.
• When two p orbitals of opposite phase overlap side-by-side, a
* antibonding orbital results.
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• The * antibonding MO is higher in energy because a
destabilizing node results, which pushes nuclei apart when
orbitals of opposite phase combine.
Figure 17.8
Combination of two p orbitals
to form π and π* molecular
orbitals
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• The molecular orbital description of benzene is much
more complex than the two MOs formed in Figure 17.8.
• Since each of the six carbon atoms of benzene has a p
orbital, six atomic p orbitals combine to form six 
molecular orbitals as shown in Figure 17.9.
The six MOs are labeled 1- 6, with 1 being the lowest
energy and 6 being the highest.
• The most important features of the six benzene MOs are
as follows:
The larger the number of bonding interactions, the
lower in energy the MO.
The larger the number of nodes, the higher in energy
the MO.
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• The most important features of the six benzene MOs
(continued):
• The larger the number of bonding interactions, the lower in
energy the MO.
• The larger the number of nodes, the higher in energy the
MO.
• Three MOs are lower in energy than the starting p orbitals,
making them bonding MOs, whereas three MOs are higher
in energy than the starting p orbitals, making them
antibonding MOs.
• Two pairs of MOs with the same energy are called
degenerate orbitals.
• The highest energy orbital that contains electrons is called
the highest occupied molecular orbital (HOMO).
• The lowest energy orbital that does not contain electrons
is called the lowest unoccupied molecular orbital (LUMO).
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Consider benzene. Since each of the six carbon atoms in benzene
has a p orbital, six atomic p orbitals combine to form six  MOs.
To fill the MOs, the
six electrons are
added, two to an
orbital. The six
electrons
completely fill the
bonding MOs,
leaving the anti-
bonding MOs
empty. All bonding
MOs (and HOMOs)
are completely
filled in aromatic
compounds. No 
electrons occupy
antibonding MOs.
Figure 17.9
The six molecular orbitals
of benzene
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The Inscribed Polygon Method of Predicting Aromaticity
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• This method works for all monocyclic completely conjugated
systems regardless of ring size.
• The total number of MOs always equals the number of vertices of
the polygon.
• The inscribed polygon method is consistent with Hückel's 4n + 2
rule—there is always one lowest energy bonding MO that can
hold two  electrons and the other bonding MOs come in
degenerate pairs that can hold a total of four  electrons.
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Figure 17.10
Using the inscribed polygon
method for five- and seven-
membered rings
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84
• For the compound to be aromatic, these MOs must be
completely filled with electrons, so the “magic numbers” for
aromaticity fit Hückel’s 4n + 2 rule.
Figure 17.11
MO patterns for cyclic,
completely conjugated systems
2/14/2024

Aromaticity and Benzene.ppt

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