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CHM136H1 (147)
Lecture

aromaticity.pdf

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Department
Chemistry
Course Code
CHM136H1
Professor
Kris Quinlan

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Description
Chemistry Tutorial: Aromaticity Based on a Chemistry 14C Honors Project 
 
 
 
 
 
 Six
carbons
once
formed
in
a
ring,
 with
sp 
hybridization.
 The
strain
was
relieved,
 and
all
six
achieved
 electron
delocalization
 
 
 
 
 
 ‘The
stability,
itself
is
dramatic,’
 said
a
puzzled
o­chemist

fanatic.
 ‘All
these
factors
at
work
 just
add
a
new
perk.’
 And
thus
was
proclaimed
aromatic.
 
 
 
 
 
 
 
 Contents
 
 Section
1:

Vocabulary
 Section
2:

Determining
Aromaticity
 Section
3:

Conjugation
and
Aromaticity
 Section
4:

Resonance
and
Aromaticity
 Section
5:

Frequently
Asked
Questions
 Section
6:

Common
Errors
 Section
7:

Readings
and
Sources Section
1:

Vocabulary
 
 




(a)

Define
each
term
 




(b)
Explain
how
it
relates
to
aromaticity
 ‐ Aromaticity
 ‐ Conjugation
 ‐ Cyclic
delocalization
 ‐ Resonance
 ‐ Pi
bonds
 ‐ Partial
Pi
bonds
 ‐ Ring
Strain
 
 Section
1:

Vocabulary
–
Solutions
 
 Aromaticity

 a)
Extra
stability
possessed
by
a
molecule
that
meets
specific
criteria:

pi
bonds
all
must
lie
 within
a
cyclic
structure,
loop
of
p
orbitals,
p
orbitals
must
be
planar
and
overlap,
must
 follow
Hückel’s
Rule.


 b)
n/a
 
 Conjugation
 a)
The
special
stability
provided
by
three
or
more
adjacent,
parallel,
overlapping
p
orbitals.


 b)
Aromatic
molecules,
by
default,
have
conjugation.

As
it
takes
a
minimum
of
three,
adjacent,
 overlapping
 p
 orbitals
 for
 planarity,
 aromaticity
 requires
 a
 minimum
 of
 this
 for
 conjugation.

Aromaticity
is
like
conjugation,
but
extra
stable.


 
 Cyclic
delocalization
 a)
Electron
delocalization,
or
distribution
of
electron
density,
occurs
as
a
result
of
overlapping
 p
orbitals
in
a
planar,
cyclic
structure.


 b)
A
closed
loop
of
overlapping
p
orbitals
must
be
present
for
aromaticity
to
occur;
therefore,
 cyclic
delocalization
occurs
in
aromatic
molecules
and
is
a
contributor
to
the
molecule’s
 extra
stability.
 
 Resonance
 a)
A
situation
in
which
a
molecule
can
be
represented
by
two
or
more
valid
Lewis
structures.


 b)
By
looking
at
resonance
structures
such
as
a
benzene
ring,
we
can
determine
where
p
 orbitals
and
partial
pi
bonds
occur.

Partial
pi
bonds
and
planar
p
orbitals
can
contribute
 to
aromaticity,
so
resonance
also
contributes
to
the
stability
of
aromatic
molecules.
 
 Pi
bonds
 a)
Pi
bonds
are
formed
by
the
overlap
of
p
orbitals
between
two
adjacent
atoms.


 b)
In
order
for
a
molecule
to
have
aromaticity,
it
must
first
have
pi
bonds
so
that
overlapping
 p
orbitals
and
electron
delocalization
are
present.


 
 Partial
Pi
bonds
 a)
Partial
pi
bonds
are
formed
as
a
result
of
close
pi
bonds
lying
in
the
same
plane.


 b)
Through
resonance,
aromatic
molecules
achieve
conjugation
and
a
ring
of
overlapping
p
 orbitals
through
partial
pi
bonds
within
a
structure.


 
 Ring
Strain
 a)
Ring
strain
occurs
as
a
combination
of
torsional
and
angle
strain
and
from
deviation
from
 the
ideal
or
preferred
bond
angles.


 b)
Because
aromatic
molecules
include
a
closed
ring
of
p
orbitals,
ring
strain
occurs;
however,
 the
benefits
of
aromaticity
generally
outweigh
the
ring
strain
in
terms
of
molecular
 stability.
 
 Check
the
Illustrated
Glossary
of
Organic
Chemistry
(available
at
the
course
web
site)
for
 more
definitions. Section
2:

Determining
Aromaticity
 
 Overview:


 
 Aromaticity
is
special
stability
provided
to
a
molecule
upon
possessing
four
specific
 qualities
mentioned
below.

To
determine
if
a
molecule
is
aromatic,
investigate
its
structure
 for
the
qualifiers.

If
all
are
present
within
the
molecule,
then
it
is
aromatic.


 
 Criteria
for
Aromaticity:


 ‐
Pi
bonds
must
lie
within
cyclic
structure
 ‐
Each
atom
in
the
cycle
must
have
p
orbital,
forming
p
orbital
loop
 ‐
All
p
orbitals
in
the
loop
must
overlap
(planarity)
 ‐
Hückel’s
Rule:

orbital
arrangement
must
result
in
a
lowering
of
energy.

4n
+
2
pi
 electrons
(n
is
an
integer:

0,
1,
2,
3,
etc…)
in
the
loop
 
 Example
Problem:


 Determine
if
the
following
molecule
is
aromatic.


 
 
 
 
 
 
 
 • Step
1:

Do
pi
bonds
lie
within
a
cyclic
structure?


 
 
 
 
 
 
 
 
 ‐
Three
pi
bonds
are
present,
each
lying
within

 
 
 
 
 
 
 

the
cyclic
structure
of
benzene
 
 
 
 
 
 
 
 
 
Yes
 
 
 • Step
2:

Does
each
atom
in
the
cycle
have
a
p
orbital,
forming
a
p
orbital
loop?
 
 
 
 
 
 
 
 
 ‐
Each
atom
has
a
p
orbital,
forming
a
loop
 
 
 
 
 
 
 
 
 
Yes
 
 
 • Step
3:

Do
all
p
orbitals
overlap
and
lie
within
the
same
plane?
 
 
 
 
 
 
 
 
 
 
 ‐
By
rotating
the
molecule,
we
see
that
all


 
 
 
 
 
 
 

orbitals
lie
within
the
same
plane
 
 
 
 
 
 
 
 
 
 
Yes
 • Step
4:

Does
the
molecule
follow
Hückel’s
Rule?
 
 
 
 
 
 
 
 6
total
pi
electrons
 
 
 
 
 
 
 
 
 4n
+
2
=
6
 n
=
1
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Yes
 
 • Conclusion:

Benzene
is
an
aromatic
molecule
 
 
 Practice
Problems:


 
 Determine
whether
or
not
the
following
molecules
are
aromatic.


 
 
 
 
 Practice
Problems:

Solutions
 
 1.

Yes;
meets
all
criteria
 
 2.

No;
does
not
include
loop
of
overlapping
p
orbitals
 
 3.

No;
does
not
contain
loop
of
overlapping
p
orbitals
 
 4.

Yes;
meets
all
criteria
 
 5.

No;
does
not
follow
Hückel’s
Rule
 
 6.

Yes;
meets
all
criteria
 
 Section
3:

Conjugation
and
Aromaticity
 
 Overview:


 
 Conjugation
requires
at
least
three
overlapping
p
orbitals
in
the
same
plane
so
that
 electrons
 can
 be
 delocalized
 for
 better
 stability.
 
 Aromaticity
 cannot
 exist
 without
 conjugation
 because
 aromatic
 molecules
 require
 planarity
 and
 overlapping
 p
 orbitals.

 However,
conjugation
can
exist
in
a
molecule
without
being
aromatic.


 
 Example
Problem:


 
 Determine
if
the
molecule
below
has
conjugation,
aromaticity,
both,
or
neither.
 
 
 
 
 
 
 
 • Step
1:

Does
the
molecule
have
conjugation?
 
 
 
 
 
 
 
 
 
 
Yes,
the
atoms
are
all
planar
with
more
than

 
 
 
 
 
 
 




three
overlapping
p
orbitals
 
 
 
 • Step
2:

Is
the
molecule
aromatic?
 
 
 
 
 
 
 
 
 
 
 
 
 
Pi
bonds
are
present
within
a
cyclic
structure
 
 
 
 
 
 
 
 
 
 
 
 
Each
atom
has
a
p
orbital,
forming
a
loop
 
 
 
 
 
 
 
P
orbitals
overlap
and
lie
in
the
same
plane
 
 
 
 
 
 
 
 
 
 
Violates
Hückel’s
Rule
(4n+2
=
8;
n
=
6/4)
 
 
 
 
 
 
 
 
Not
aromatic
 
 • Conclusion:

The
molecule
has
conjugation,
but
is
not
aromatic.


 
 
 Practice
Problems:
­
Conjugation
and
Aromaticity:


 
 Determine
if
the
molecules
below
have
conjugation,
aromaticity,
both,
or
neither.


 
 1.
 2.
 3.
 
 
 
 
 
 
 
 
 
 4.
 
 5.
 N 6.
 H 
 
 
 
 Practice
Problems
­
Solutions
 
 1.
 
 Not
 conjugated
 (the
 molecule
 has
 a
 tub
 shape
 and
 the
 C=C
 are
 perpendicular);
 Not
 aromatic
–
violates
Hückel’s
rule
 
 2.
 
 Not
 conjugated
 –
 does
 not
 contain
 at
 least
 three
 atoms
 with
 planar
 p
 orbitals;
 Not
 aromatic
–
does
not
contain
closed
loop
of
planar,
overlapping
p
orbitals
 
 3.

Conjugated;
Aromatic
–
meets
all
criteria
 
 4.

Conjugated,
Not
aromatic
–
does
not
contain
closed
loop
of
planar,
overlapping
p
orbitals
 
 5.

Conjugated;
Aromatic
–
meets
all
criteria
 
 6.

Conjugated;
Aromatic
–
meets
all
criteria
 Section
4:

Resonance
and
Aromaticity
 
 Overview:


 
 Resonance
 exists
 as
 a
 result
 of
 electron
 delocalization
 in
 a
 molecule.
 
 Different
 patterns
 emerge
 as
 a
 result
 of
 structure
 and
 atom
 arrangement
 within
 a
 molecule.

 Resonance
provides
an
extra
stability
due
to
electron
delocalization,
and
consequently;
 aromatic
 rings
 have
 resonance
 structures
 due
 to
 cycling
 double
 bonds.
 

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