Acids, Bases and Aromaticity
Dr. G. Krishnaswamy
Faculty
DOS & R in Organic Chemistry
Tumkur University
Tumakuru

Acids and Bases
Produce H+ (as H3O+) ions in water
Produce a negative ion (-) too
Taste sour
Corrode metals
React with bases to form salts and water
Neutral Substance : Water!
General properties of Acids

General properties of Bases
Produce OH- ions in water
Taste bitter, chalky
Are electrolytes
Feel soapy, slippery
React with acids to form salts and water
Neutral Substance : Water!

Definitions of acids and bases:
• Acids – produce H+
• Bases - produce OH-
• Acids – donate H+
• Bases – accept H+
• Acids – accept e- pair
• Bases – donate e- pair
Arrehenius Concept
Only in water
Bronsted-Lowry
Any solvent
Lewis
used in organic chemistry,
wider range of substances

Examples of Brønsted–Lowry
acids and bases

6
Reactions of Brønsted-Lowry Acids and Bases
• A Brønsted-Lowry acid base reaction results in the transfer of a
proton from an acid to a base.
• In an acid-base reaction, one bond is broken, and another one is
formed.
• The electron pair of the base B: forms a new bond to the proton of
the acid.
• The acid H—A loses a proton, leaving the electron pair in the H—A
bond on A.

Each acid has a conjugate base and each base has a
conjugate acid. These conjugate pairs only differ by a
proton.
O H
H
H Br+
H O H
H
+ Br
Base Acid Conjugate Acid C. Base

There are two ways to predict when a proton transfer reaction
will occur:
(1)Quantitative approach (comparing pKa values)
(2)Qualitative approach (analyzing the structures of the acids)
Using pKa Values to Compare Acidity
• Acid strength is the tendency of an acid to donate a proton.
• Acidity is measured by an equilibrium constant.
• When a Brønsted-Lowry acid H—A is dissolved in water, an acid-
base reaction occurs, and an equilibrium constant can be written
for the reaction.

9
Because the concentration of the solvent H2O is essentially
constant, the equation can be rearranged and a new equilibrium
constant, called the acidity constant, Ka, can be defined.
It is generally more convenient when describing acid strength to
use “pKa” values than Ka values.

Selected pKa Values

The principle is this:
The stronger the acid, the weaker will be its conjugate base.
We can, therefore, relate the strength of a base to the pKa of its
conjugate acid.
The larger the pKa of the conjugate acid, the stronger is the
base.
Predicting the Strength of Bases

Using the chart of pKa values, we can also predict the position of
equilibrium for any acid-base reaction. The equilibrium will
always favor formation of the weaker acid (higher pKa value).
For example, consider the following acid-base reaction:
Using pKa Values to Predict the Position of Equilibrium
The equilibrium for this reaction will lean to the right side,
favoring formation of the weaker acid.

Qualitative approach (analyzing the structures of the
acids)
comparisons by analyzing and comparing their structures and
without the use of pKa values.
In order to compare acids without the use of pKa values, we must
look at the conjugate base of each acid:
Conjugate Base Stability
If A- is very stable (weak base), then HA must be a strong acid. If,
on the other hand, A- is very unstable (strong base), then HA must
be a weak acid.

Consider the deprotonation of HCl:
Chlorine is an electronegative atom,
and it can therefore stabilize a negative
charge. The chloride ion (Cl-) is in fact
very stable, and therefore, HCl is a
strong acid. HCl can serve as a proton
donor because the conjugate base left
behind is stabilized.
The following discussion will develop a methodical approach for
comparing negative charge stability. Specifically, there are four
factors to consider:
(1)Bond strength and which atom is the charge on
(2)resonance
(3)Induction and
(4)orbitals
Factors Affecting the Stability of Negative Charges/Acidity

Consider Halogen Acid Series
Acid: H—F H—Cl H—Br H—I
pKa: 3.2 -7 -9 -10
Stronger Bonds Weaker Bonds
H—X Bond Strength
Weaker Bonds  Stronger Acids

Which atom is the charge on?
The first factor involves comparing the atoms bearing the negative
charge in each conjugate base.
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Across a row of the periodic table, the acidity of H—A
increases as the electronegativity of A increases.
Positive or negative charge is stabilized when it is spread
over a larger volume.

For example, C- and O-appear in the same row of the periodic table. When two
atoms are in the same row, electronegativity is the dominant effect.
The story is different when comparing two atoms in the same column of the
periodic table.
For example, the acidity of water and hydrogen sulfide:
In this example, we are comparing O- and S-, which appear in the same column
of the periodic table. In such a case, electronegativity is not the dominant effect.
Instead, the dominant effect is size. Sulfur is larger than oxygen and can
therefore better stabilize a negative charge by spreading the charge over a larger
volume of space. The HS- is more stable than HO-, and therefore, H2S is a
stronger acid than H2O. We can verify this prediction by looking at pKa values
(the pKa of H2S is 7.0, while the pKa of H2O is 15.7).

Resonance Effects
• Resonance is a second factor that influences acidity.
• In the example below, when we compare the acidities of ethanol and
acetic acid, we note that the latter is more acidic than the former.
• When the conjugate bases of the two species are compared, it is evident
that the conjugate base of acetic acid enjoys resonance stabilization,
whereas that of ethanol does not.

• Resonance delocalization makes CH3COO¯ more stable than CH3CH2O¯, so
CH3COOH is a stronger acid than CH3CH2OH.
• The acidity of H—A increases when the conjugate base A:¯ is resonance
stabilized.

Inductive Effects
• An inductive effect is the pull of electron density through 
bonds caused by electronegativity differences between atoms.
• Also Called Electron Withdrawing Effect
• In the example below, when we compare the acidities of ethanol
and 2,2,2-trifluoroethanol, we note that the latter is more acidic
than the former.
• The reason for the increased acidity of 2,2,2-trifluoroethanol is
that the three electronegative fluorine atoms stabilize the
negatively charged conjugate base.

• When electron density is pulled away from the negative charge
through  bonds by very electronegative atoms, it is referred to
as an electron withdrawing inductive effect.
• More electronegative atoms stabilize regions of high electron
density by an electron withdrawing inductive effect.
• The more electronegative the atom and the closer it is to the site
of the negative charge, the greater the effect.
• The acidity of H—A increases with the presence of electron
withdrawing groups in A.

Hybridization or orbital effect
More ‘s’ character in the orbital  more stable anion
• The higher the percent of s-character of the hybrid orbital, the
closer the lone pair is held to the nucleus, and the more stable the
conjugate base.

Summary of Factors Affecting the Stability of Negative
Charges/acidity

A convenient way to look at basicity is based on electron pair
availability.... the more available the electrons, the more readily
they can be donated to form a new bond to the proton and, and
therefore the stronger base.
Key factors that affect electron pair availability in a base
Electronegativity. When comparing atoms within the same row
of the periodic table, the more electronegative the atom donating
the electrons is, the less willing it is to share those electrons with a
proton, so the weaker the base.
Increasing basicity

Size. When comparing atoms within the same group of the
periodic table, the larger the atom the weaker the H-X bond and
the lower the electron density making it a weaker base.
Increasing basicity
Resonance. In the carboxylate ion, RCO2
- the negative charge is
delocalised across 2 electronegative atoms which makes it the
electrons less available than when they localised on a specific
atom as in the alkoxide, RO-.
Strong base Weak base

Lewis Acids and Bases
The Lewis definition of acids and bases is more general than the
BrØnsted-Lowry definition.
•A Lewis acid is an electron pair acceptor.
•All BrØnsted-Lowry acids are also Lewis acids, but the reverse is not
necessarily true.
•Any species that is electron deficient and capable of accepting an electron
pair is also a Lewis acid.
•Common examples of Lewis acids (which are not BrØnsted-Lowry acids)
include BF3 and AlCl3.

•Lewis bases are structurally the same as BrØnsted-Lowry bases.
Both have an available electron pair—a lone pair or an electron pair
in a  bond.
•A BrØnsted-Lowry base always donates this electron pair to a
proton, but a Lewis base donates this electron pair to anything that
is electron deficient.
•A Lewis base is an electron pair donor.

•Lewis acid-base reactions illustrate a general pattern in organic
chemistry. Electron-rich species react with electron-poor species.
•In the simple Lewis acid-base reaction one bond is formed and no
bonds are broken.
•This is illustrated in the reaction of BF3 with H2O. H2O donates an
electron pair to BF3 to form a new bond.
• A Lewis acid is also called an electrophile.
• When a Lewis base reacts with an electrophile other than a proton,
the Lewis base is also called a nucleophile.
electrophile nucleophile

The word steric is derived from ‘stereos’ meaning space. So this effect is
manifested when two or more groups or atoms come in close proximity to each
other (precisely within each other’s van der Waals radii (definition of van der
Waals radii can be found in any standard textbook)) and result in a mutual
repulsion. This makes the molecule unstable.
Steric effects

The usual physical clash between groups, almost always is accompanied by
an electronic component as well. This is called stereoelectronic effect, which
is not the same as the electronic effects discussed above and does not carry
have an effect on some other part of the molecule like inductive and
resonance effects.
When the two atoms get to close, into each other’s van der Waal’s radii, the
electron cloud surrounding each atom repel each other leading to a lot of
destabilization.
Steric effect affects different properties of molecules, like acidity, basicity and
general reactivity.

• Resonance involves delocalization of π electrons, leaving the σ
bond untouched.
• However in some cases, a σ bond and an adjacent π bond may
get involved in resonance.
• Such a delocalization is called as Hyper conjugation.
• The electrons of the sigma bond between C and H are involved in
delocalization.
• In structure to the right: No bond between C and H due to
migration of the sigma bond. Hence Hyper conjugation is also
called as ‘NO BOND RESONANCE’.
Hyper conjugation

 When carbon possessing atleast one Hydrogen is attached to another carbon
bearing an unshared orbital or to an unsaturated carbon, there are more than
one structure (canonical forms) possible for the molecule.
 These result from the overlap of the bonding electrons of C – H σ bond with
2p or π orbital of the adjacent carbon atom.
 As a result, the bond between carbon and hydrogen does not exist anymore.

• The shared pair of electrons is now borne by carbon alone and hydrogen is in
its close proximity as proton.
• The negative charge developed on the carbon gets delocalized by overlap
with adjacent p orbital.

This does not indicate that hydrogen is completely detached from the structure,
but some degree of ionic character in the C – H bond and some single bond
character between carbon – carbon double bond.
Such an interaction is also referred to as ‘Heterovalent’ or ‘Sacrificial
hyperconjugation’.
This is so named because the contributing structure contains one two-electron
bond less than the normal Lewis formula of the compound.

Effect of Hyperconjugation on the chemical properties
Alkyl cations and their relative stability
• Carbocations have an electron deficient (positively charged) carbon.
• The empty p orbital of this sp2 carbon can overlap with σ orbital of C – H
bond of adjacent alkyl group (α C – H bond).
• This overlap permits individual electrons to help bind together three nucleus: 2
carbon and 1 hydrogen.
• The positive charge thus gets dispersed over large volume of space and is
stabilized.

The more the number of alkyl groups on the carbocation, more is the number of α
C – H σ bonds and hence more are the possibilities for hyperconjugation which
makes the carbocation more stable.
The order of stability of the Carbocations is:
9 Hyperconjugable
H
6 Hyperconjugable
H
3 Hyperconjugable
H
No Hyperconjugable
H
30alkyl carbocation > 20alkyl carbocation > 10alkyl carbocation > methyl carbocation

Alkyl radicals and their relative stability:
• Alkyl radicals have p-orbital of carbon occupied by an odd electron which gets
delocalized over three nuclei (2 carbon and 1 hydrogen) by the overlap with
the σ orbital of C – H bond of adjacent alkyl groups.
• The bond is formed between 2 carbons and odd electron is held by the
hydrogen atom.
• The order of stability depends on the extent of delocalization. The greater the
delocalization, the more is the stability which similar to alkyl cations.

• Overlap of σ orbital of C – H bond with π orbital of adjacent C – C double
bond gives rise to canonical structures.
• Delocalization of electrons occurs over three nuclei and thus stabilizes the
alkene.
Alkenes and their stability
More the substituent, more is the opportunity for hyperconjugation and more
stable is the alkene.

Alkene Number of Hyperconjugable Hydrogen (α to unsaturated
function)
CH2=CH2 0
(CH3)CH=CH2 3
(CH3)2C=CH2
(CH3)CH=CH(CH3)
6
(CH3)2C=CH(CH3) 9
(CH3)2C=C(CH3)2 12
Stability of alkenes will increase with increase in number of Hydrogen α to unsaturated
system.
Hyper conjugation leads to shortening of sigma (σ) bond.
Eg. C – C bond in 1,3-butadiene and methylacetylene is 1.46 A0 in length,
much less as compared to 1.54 A0 found in saturated hydrocarbons.
Bond length

Bredt's Rule stated that bridged ring systems like camphane and pinane cannot have a
double bond at the bridgehead position.
This rule came from observations on dehydration
of alcohols in these ring systems.
Bredt's Rule

For example, the bicyclooctyl 3º-chloride shown below appears to be similar to tert-butyl
chloride, but it does not undergo elimination, even when treated with a strong base (e.g.
KOH or KOC4H9). There are six equivalent beta hydrogens that might be attacked by
base (two of these are colored blue as a reference), so an E2 reaction seems plausible.
The problem with this elimination is that the resulting double bond would be constrained
in a severely twisted (nonplanar) configuration by the bridged structure of the carbon
skeleton. Because a pi-bond cannot be formed, the hypothetical alkene does not exist.
Bredt's Rule should not be applied blindly to all bridged ring systems. If large rings are
present their conformational flexibility may permit good overlap of the p-orbitals of a
double bond at a bridgehead.

Benzene, first isolated by Michael Faraday in 1825 is the simplest and the ideal molecule
to illustrate electron delocalization, resonance and aromaticity.
Important milestones during structure elucidation of benzene include:
• Friedrich Kekule’s (1866) proposal of cyclic equilibrating structures I and II which
partially explained the existence of three isomers (instead of four) for disubstituted
benzene
• Hydrogenation of benzene to cyclohexane by Paul Sabatier (1901) which confirmed
its cyclic structure.
Debate over the structure of benzene came to an end in 1930s when X-ray and electron
diffraction studies confirmed that it is a planar, regular hexagon in which all the
carbon-carbon bond lengths are 1.39 Å, which is shorter than C-C single bond (1.54
Å), but slightly longer than C-C double bond (1.33 Å). Such a structure is possible
only if all the carbon atoms have the same electron density, with π electrons
delocalized over the entire skeleton of ring carbons.
AROMATICITY

Delocalization means possibility of new orbital overlap and additional stabilization of
the system. The extra stability (in terms of energy) gained through delocalization is
called delocalization energy or resonance energy.
The heat of hydrogenation of cyclohexene has been experimentally determined to be
28.6 kcal/mol. If we consider C6H6 as just a cyclohexatriene, the heat of hydrogenation
should be 3 x 28.6 kcal/mol = 85.8 kcal/mol. However, when the heat of hydrogenation
was experimentally determined for benzene, it was found to be 49.8 kcal/mol. Since
hydrogenation of cyclohexatriene and benzene both lead to cyclohexane, reason for the
difference in their heat of hydrogenation should be due to the difference in their
stabilities. From this, it is clear that benzene is 36 kcal/mol (ie. 85.8-49.8 kcal/mol) more
stable than ‘cyclohexatriene’. i.e. benzene with six delocalized π electrons is 36 kcal/mol
more stable than ‘cyclohexatriene’ with six localized π electrons. Here, 36 kcal/mol is
the resonance energy of benzene (Heat of hydrogenation is the quantity of heat released
when one mole of an unsaturated compound is hydrogenated).

In 1931 the German physicist Erich Hückel carrie dout a series of mathematical
calculations.
Hückel’s rule is concerned with compounds containing one planar ring in which each
atom has a p orbital as in benzene.
Based on the analysis of a number of compounds with unusual resonance stabilization
energies, the following characteristics have been accepted as criteria for aromaticity.
 The molecule must be cyclic, planar with uninterrupted cloud of π electrons above
and below the plane of the ring.
 It should have 4n+2 π electrons. Where n = 0, 1, 2, 3 and so on
Hückel’s Rule: The 4n+2 π Electron Rule

Non aromatic compounds, as the name implies, are not aromatic due to reasons such as
lack of planarity or disruption of delocalization. They may contain 4n or 4n+2 π
electrons.
Antiaromatic compounds are planar, cyclic, conjugated systems with an even number
of pairs of electrons.
Such compounds satisfy the first three criteria for aromaticity. i.e. they are planar, cyclic
with an uninterrupted ring of p orbital bearing atoms.
But they have an even number of pairs of π electrons (4n, n = 1, 2, 3 etc).
It should be noted that an aromatic compound is more stable compared to an analogous
cyclic compound with localized electrons, where as an antiaromatic compound is less
stable compared to an analogous cyclic compound with localized electrons (in 4n+2
systems delocalization increases the stability where as in 4n systems, delocalization
decreases stability)

The bond angles of 120°in benzene suggests that C atoms are sp2 hybridized. An
alternative representation therefore starts with a planar framework and considers overlap
of the p orbitals (π electrons).
MOLECULAR ORBITAL REPRESENTATION OF BENZENE
(MO THEORY)

Each MO can accommodate 2 electrons, so for benzene we see all electrons are paired and
occupy low energy MO’s (bonding MO’s). All bonding MO’s are filled. Benzene is
therefore said to have a closed bonding shell of delocalized π electrons and this accounts
in part for the stability of benzene.
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.

Frost diagrams

Points to remember while making predictions on aromaticity using Frost’s circle
Aromatic compounds will have all occupied molecular orbitals completely filled where
as antiaromatic compounds would have incompletely filled orbitals.
The word annulene is incorporated into the class name for monocyclic compounds
that can be represented by structures having alternating single and double bonds. The
ring size of an annulene is indicated by a number in brackets. Thus, benzene is
[6]annulene and cyclooctatetraene is [8]annulene.
Hückel’s rule predicts that annulenes will be aromatic if their molecules have 4n +2 π
electrons and have a planar carbon skeleton.
The Annulenes

During the 1960s, and largely as a result of research by F. Sondheimer, a number of
large-ring annulenes were synthesized, and the predictions of Hückel’s rule were
verified.
Consider the [14], [16], [18], [20], [22], and [24]annulenes as examples. Of these, as
Hückel’s rule predicts, the [14], [18], and [22]annulenes (4n+2, when n 3, 4, 5,
respectively) have been found to be aromatic. The [16]annulene and the [24]annulene are
not aromatic; they are antiaromatic. They are 4n compounds, not 4n +2 compounds:

In [10]-annulene, there is considerable steric interaction between hydrogens at 1 and 6
positions. Further, a planar form (regular decagon) requires an angle of 144o between
carbon atoms which is too large to accommodate in a sp2 framework. The system
prefers a nonplanar conformation and is not aromatic (the fact that angle strain need
NOT always be a problem in achieving planarity is evident from examples such as
cyclooctatetraenyl dianion, which is stable and aromatic). Bridging C1 and C6 in [10]-
annulene leads to the compound VII (Figure 9) which is reasonably planar with all the
bond distances in the range of 1.37-1.42Ao and show aromaticity

If a stabilized cyclic conjugated system (4n+2 e s) can be formed by bypassing one
saturated atom, that lead to homoaromaticity. Compared to true aromatic systems, the
net stabilization here may be low due to poorer overlap of orbitals. Cyclooctatrienyl
cation (homotropylium ion) formed when cyclooctatetraene is dissolved in
concentrated sulfuric acid is the best example to demonstrate homoaromaticity. Here,
six electrons are spread over seven carbon atoms as in Tropylium cation.
HOMOAROMATICITY