Aromatic Compounds

Early in the history of organic chemistry (late 18th, early 19th century) chemists discovered a class of compounds which were unusually stable

A number of these compounds had a distinct odor

Hence these compounds were called “aromatic”

Today the term aromatic is used regardless of the odor of the compound Some “aromatic” compounds have little to no odor

The parent aromatic compound was discovered to have a molecular formula of C6H6

(called benzene)

This 1:1 ratio of carbon to hydrogen is extremely low compared to other known compounds

It was also quickly discovered that these aromatic compounds did not react like other alkene compounds

Structure of Benzene

Before NMR and other spectroscopic tools it was hard to determine the structure of organic compounds

Ultimately the symmetry of the molecule revealed its structure

All carbon atoms, and all carbon-carbon bonds, are symmetrically equivalent

To account for these observations the proposed structure consisted of a cyclic compound stabilized by resonance

Each resonance structure is equal in energy and thus each contributes equally to the overall structure

Stability

The resonance structures imply an extra stability, but the amount of stability in benzene is much more than a typical resonance structure

Consider reactivity:

But the presence of a conjugated ring is not enough to cause this extra stability

HBrBr

HBr BrBr

HBr No reaction

HBrBr

What causes this extra stability in benzene and why is a 6-membered conjugated ring more stable than an 8-membered conjugated ring?

Reaction is faster than 1-butene due to more stable carbocation intermediate

Having conjugation in ring somehow stabilizes

compound

Stability of Aromatic Compounds

Can measure stability by hydrogenation

H2catalyst

The energy required for this hydrogenation indicates the stability of the alkene

28.6 Kcal/mol

57.4 Kcal/mol 55.4 Kcal/mol

Almost double in energy

2 Kcal/mol Conjugation stability

How much energy should be in the hydrogenation of Benzene? Have three double bonds in conjugation, so therefore should expect ~79 Kcal/mol

(~24 Kcal/mol more than 55 Kcal/mol for 1,3-cyclohexadiene)

? Kcal/mol

49.8 Kcal/mol

Benzene is ~ 30 Kcal/mol more stable than predicted!!

Aromatic Stabilization

This ~30 Kcal/mol stabilization is called “aromatic stabilization”

It is the cause of the difference in reactivity between normal alkenes

It would cost ~30 Kcal/mol to break the aromaticity and thus the normal alkene reactions do not occur with benzene

Somehow having these three double bonds in resonance in a cyclic system offers a tremendous amount of energy

Cyclic system alone, however, is not sufficient for aromatic stabilization

Consider a four membered ring

Cyclobutadiene also has a ring structure with conjugated double bonds that could resonate

This compound however is highly reactive and does not exist with equivalent single and double bonds

In solution it reacts with itself in a Diels-Alder reaction

Aromatic Stabilization

Why the Difference in Stability?

Can already see in electron density maps that cyclobutadiene is not symmetric

Benzene 6-fold symmetry

Cyclobutadiene Not symmetric

Aromatic Stabilization

Molecular Orbitals for Benzene

For benzene there are 6 atomic p orbitals in conjugation, therefore there will be 6 MO’s -As the number of nodes increase, the energy increases

For lowest energy MO there are zero nodes, therefore bonding interactions between each carbon-carbon bond

Benzene model Top view with orbitals Side view

Entire MO Picture for Benzene

E

Zero nodes

2 nodes

6 nodes

2 nodes

4 nodes 4 nodes

Nonbonded energy level

Molecular Orbitals for Benzene

Notice all electrons are in bonding MO’s

All the antibonding MO’s are unfilled With a cyclic system we obtain degenerate orbitals

(orbitals of the same energy)

Overall this electronic configuration is much more stable than the open chain analog

This is now the definition of an aromatic compound (not aroma), Flat conjugated cyclic system is MORE stable than the open chain analog

>

Molecular Orbitals for Cyclobutadiene

E

Unlike benzene, cyclobutadiene has two electrons at the nonbonding energy level (these electrons do not stabilize the electronic structure)

Nonbonded energy level

Antiaromatic

Cyclobutadiene is less stable than butadiene

If a cyclic conjugated system is less stable than the open chain analog it is called antiaromatic

Part of the reason for cycobutadiene to be antiaromatic is the presence of two MO’s at the nonbonding level

In butadiene all electrons are in bonding MO’s therefore the electrons are more stable in butadiene relative to cyclobutadiene

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Frost Circle

A simple method to determine the relative molecular orbital energy levels for a conjugated ring is called a Frost circle (or Frost Mnemonic)

First just draw a circle

Next draw a polygon with equal length of sides corresponding to the number of atoms in the ring being considered

Place the polygon inside the ring having a vertex point directly at the bottom

Wherever a vertex point of the polygon hits the ring corresponds to an energy level

The electronic configuration would be obtained by placing the correct number of electrons in the molecular orbitals

(the relative energy levels are also obtained as the ring drawn initially has a radius of 2β)

Will work for any flat, conjugated ring system to determine energy levels

Hückel’s Rule

In order to determine if a system is aromatic or antiaromatic, without needing to determine the overall electronic energy of the closed form versus the open form, Hückel’s rule was developed

First the cyclic system must have a p orbital on all atoms in a continuous cyclic chain (if there is an atom without a p orbital in the cycle then the system is nonaromatic)

In practice this means the cyclic system must be flat (to allow overlap of p orbitals)

If these criteria are met then:

If the system has 4n+2 π electrons, it is aromatic

If the system has 4n π electrons, it is antiaromatic

6 π electrons, 4n+2 Therefore aromatic

4 π electrons, 4n Therefore antiaromatic

No p orbital on one atom Therefore nonaromatic

Hückel’s Rule

What is the underlying cause for the symmetry in Hückel’s rule?

Ultimately the stabilization is due to the relative electronic configuration for a flat, conjugated ring system

The symmetry is also observed with the Frost circle

4 π electron system

Obtain 2 electrons at nonbonding level

6 π electron system

All electrons are at bonding level

8 π electron system

Obtain 2 electrons at nonbonding level

4n+2 systems allow all electrons to be in bonding molecular orbitals, therefore more stable 4n systems, however, will place 2 electrons at nonbonding level and thus be less stable

Remember that the cyclic ring must have overlap of p orbitals to be considered aromatic or antiaromatic by Hückel’s rule

Molecule, however, adopts a non-flat low energy conformation

top view side view

Hückel’s Rule

Cyclooctatetraene If flat this molecule is antiaromatic

with the 8 π electrons

This is an example of a rare case where delocalization is avoided to increase stability!

Aromatic Ions

Benzene is a neutral aromatic compound

Any compound with 4n+2 electrons in a continuous loop is considered aromatic regardless of the number of carbons in the loop

There are many aromatic compounds with a different number of electrons than atoms in the loop

Due to this difference usually these compounds are ions, hence aromatic ions

Cyclopentadienyl Anion

Cyclopentadiene is nonaromatic since there is not a p orbital on one of the carbons in the ring

Upon removal of a proton, however, there is now a p orbital on each carbon 6 electrons in system, therefore according to Hückel this is aromatic

nonaromatic pKa ~15

base

H

H

pKa ~50-60

base

pKa ~44

base

Unactivated alkanes have much higher pKa

Simple conjugation only explains small portion of

stability

Extremely low pKa is due to aromatic stabilization

Aromatic Ions

Any compound that will have 4n+2 electrons in a continuous loop for planar conjugated compound will be favored due to aromatic nature

OH

H2SO4 Tropylium ion 6 π electrons, 4n+2

K 10 π electrons, 4n+2

Need correct number of conjugated electrons, not all conjugated ions are aromatic

OHH2SO4 base

Does not form! High pKa

Benzene Derivatives

The IUPAC name of 1,3,5-cyclohexatriene is never used The common name of benzene dominates naming of these structures

In addition, another common naming tool for benzene derivatives is for disubstituted compounds (ortho, meta, para)

ortho- dimethylbenzene

meta- dimethylbenzene

para- dimethylbenzene

Number along ring to give lowest number First priority substituent is at the 1-position

Other common names

CH3 OHO

OH

toluene phenol benzoic acid

Benzene Derivatives

If benzene group is being considered as a substituent, instead of root name, then use

“phenyl” prefix, from phenol name OH

4-phenyl-2-butanol

Another common name used for a substituted toluene is called “benzyl” group

Br

Benzyl bromide

OH

Benzyl alcohol

Heterocyclic Aromatic Compounds

Compounds that contain atoms besides carbon can also be aromatic

Need to have a continuous loop of orbital overlap and follow Hückel’s rule for the number of electrons in conjugation

Common noncarbon atoms to see in aromatic compounds include oxygen, nitrogen, and sulfur

furan thiophene pyrimidine imidazole pyrrole pyridine

N HN O S N N NHN

All of these compounds have 6 electrons conjugated in ring Consider where the lone pair(s) are located for each heteroatom

Pyridine

One common aromatic compound with nitrogen is pyridine

N

One carbon atom of benzene has been replaced with nitrogen

N

Consider the placement of electrons

Lone pair is orthogonal to conjugated electrons in ring

The number of electrons in conjugation is 6 (don’t include lone pair that is orthogonal to ring)

therefore pyridine follows Hückel’s rule and is aromatic

Pyridine can be protonated in acidic conditions and it will still be aromatic, protonation occurs at lone pair

Pyrrole

A similar aromatic compound is pyrrole

With pyrrole the lone pair is included in the conjugated ring Have 6 electrons in loop and therefore this compound is aromatic

If protonated, however, pyrrole will become nonaromatic since the nitrogen would thus be sp3 hybridized without a p orbital for conjugation

HN

N H

Difference in electron placement affects properties

Excess electron density of lone pair is localized orthogonal to ring in pyridine while the electron density is conjugated in ring with pyrrole

pyridine pyrrole

Heterocyclic Aromatic Compounds

Fused Rings

Compounds with more than one fused ring can also be aromatic

The simplest two ring fused system is called naphthalene

Like benzene, naphthalene is an aromatic compound with 10 electrons in a continuous ring around the cyclic system

(one p orbital on each carbon is conjugated)

Naphthalene

Consider one electron Electron can resonate in p orbitals

Will occur with all 10 electrons

Br

The reactivity of naphthalene is similar to benzene

It is unreactive toward normal alkene reactions because any addition would lower the aromatic stabilization

If it did react, however, there would still be one benzene ring intact

Hypothetical reaction – does not occur

With larger fused ring systems normal alkene reactions start to occur

Fused Rings

HBr

Anthracene

Reactions occur at central ring due to large aromatic stabilization remaining

Two intact benzene rings

Diels-Alder reactions can also occur about this central ring

NO2

Br2

Br

Br

The dienophile approaches the central ring from top or bottom

And then Diels-Alder reaction occurs to leave two intact benzene rings

Two intact benzene rings

NO2

Fused Heterocyclics

Fused ring systems with heterocyclics can also be aromatic

Extremely important compounds biologically and medicinally

Two of the four constituents of base pairs in DNA consist of fused aromatic rings, the other two bases, cytosine (C) and thymine (T), are one ring aromatic base pairs

N

NN

NH

NH2

Adenine (A) 10 π electrons

N

NHN

NH

O

NH2

Guanine (G) 10 π electrons

HN

NH

O

O

CH3

Thymine (T)

N

NH

O

NH2

Cytosine (C)

Aromatic Base Pairing

The four bases shown in the preceding page (A, G, C, T) are the bases used in DNA The bases are attached to a sugar through the NH group on each ring

and the sugars are linked through a phosphate backbone

The bases are complementary to each other and bind through hydrogen bonding (C binds with G and A binds with T)

This complementarity allows genetic information to be passed along as the DNA is replicated

P OOO

O

Sugar

Sugar

O

OP OOO

P OOO

Base

Base

PO OO

O

Sugar

Sugar

O

OPO OO

PO OO

Base

Base

Hydrogen bonding

π stacking

Things that disrupt this complementarity can cause cell death or possibly cancer

N

N

NN

OHN

Sugar

HH

N

N

O NH

Sugar

H

G

C

HOOH

O

Polycyclic Aromatic Hydrocarbons

Polycyclic aromatic hydrocarbons (PAH’s) have been shown to disrupt this base pairing

The PAH is first oxidized by enzymes (these enzymes are essential to remove hydrophobic compounds in the body)

P450

O HOOH

P450

Need PAH to react, Benzene or naphthalene would

not undergo this reaction

N

NHN

N

O

NH2Sugar

Guanine can react with this epoxide, however, which will destroy its hydrogen-bonding complementarity in the DNA base pairing, thus causing cells to eventually die

Benzyl Group

The benzyl group behaves similar to the allyl group seen previously, the orbitals on this group are stabilized through resonance with the adjacent benzene group

CH3OHSN1

Cl OCH3

Cl

2˚ cation Intermediate: Relative

rate: 1 100,000

2˚ cation resonance

Cl

SN1

SN2 CH3ONaI OCH3

Intermediate NUC

Br

Transition State

T.S.: Relative

rate:

1˚ 1˚ in resonance 1˚ in resonance 1 33 78

III

SN2

Benzyl Group

A unique reaction of benzyl groups is that the benzyl carbon can be oxidized with either potassium permanganate (KMnO4) or dichromic acid (H2Cr2O7) to a carboxylic acid

CH31. KMnO42. H+,H2O

O

OH

If alkyl chain is longer, then carbon-carbon bonds are broken and left with benzoic acid

1. KMnO42. H+,H2O

1. KMnO42. H+,H2O

1. KMnO42. H+,H2O

Must have hydrogen on benzylic carbon, though, as a t-butyl group will not be oxidized Realize also this reaction is not selective, any alkyl chain on benzene will be oxidized

1. KMnO42. H+,H2O

CO2H

HO2C

Benzyl Group

As seen in chapter 12, halogenation reactions can occur with either chlorine or bromine under photolytic conditions

Reaction proceeds through a radical intermediate

The benzylic radical is more stable due to resonance with aromatic ring

CH2

Remember that chlorination was more reactive, bromination though occurred selectively

Cl2, h! ClCl

Br2, h!

Br

Realize reaction does not occur on aromatic ring, do not obtain radical at sp2 hybridized carbon

Birch Reduction

While typical alkene reactions do not occur on benzene, the aromatic ring can be reduced by adding electrons to the system

(in essence a nucleophilic addition)

The electrons need to be generated in situ

This electron is called a “solvated” electron

The reduction is similar to the dissolving metal reduction of alkynes to E-alkenes

NH3(l) Na NH3(l) e Na+

In the presence of an aromatic ring this electron will react

Addition of one electron thus generates a radical anion

This strongly basic anion will abstract a proton from alcohol solution

e

Birch Reduction

ROH

H H

The radical will then undergo the same operation a second time

The final product has thus been reduced from benzene to a 1,4-cyclohexadiene (always obtain a 1,4 relationship of the dienes in a Birch reduction – they are not conjugated)

The aromatic stabilization has been lost

Birch Reduction

H H

e

H H

ROH

H H

H H

Birch Reduction

What happens if there is a substituent on the aromatic ring before reduction?

Which regioisomer will be obtained?

Similar to every other reaction studied need to ask yourself, “What is the stability of the intermediate structure?”

The preferred product is a result of the more stable intermediate

XNH3(l), Na

ROH

X X

The intermediate in a Birch reduction is the radical anion formed after addition of electron

With electron withdrawing substituent:

With electron donating substituent:

O NH3(l), NaCH3OH

O

Placing negative charge adjacent to carbonyl allows resonance

O O

NH3(l), NaCH3OHOCH3

Want negative charge as far removed from donating group as possible

OCH3 OCH3

Birch Reduction

Spectroscopy of Aromatic Compounds

We have already seen how aromatic benzene compounds have a relatively large downfield NMR shift due to aromatic ring current

Therefore any of these aromatic systems, which by definition have a ring current, have a large downfield shift

Can use as a characteristic of aromaticity

S N

Mass Spectrometry

A characteristic peak in a MS for a benzenoid compound is the presence of a peak at m/z 91 (if formation is possible)

Due to resonance stabilized benzyl cation