Table Of ContentAromatic and Heterocyclic Chemistry 1
Aromatic and Heterocyclic Chemistry
8 Lectures, Trinity 2013
[email protected]
handout at: http://burton.chem.ox.ac.uk/teaching.html
◾
Advanced
Organic
Chemistry:
Parts
A
and
B,
Francis
A.
Carey,
Richard
J.
Sundberg
◾
Organic
Chemistry,
Jonathan
Clayden,
Nick
Greeves,
Stuart
Warren,
Peter
Wothers
◾
Advanced
Organic
Chemistry:
Reac7ons,
Mechanisms
and
Structures;
J.
March
◾
Fron7er
Orbitals
and
Organic
Chemical
Reac7ons;
I.
Fleming
◾
Heterocyclic
Chemistry;
J.
Joule,
K.
Mills,
G.
Smith
◾
Aroma7c
Heterocyclic
Chemistry;
D.
Davies
◾
Reac7ve
Intermediates;
C.
Moody
and
G.
Whitham
◾
Aroma7c
Chemistry;
M.
Sainsbury
Aromatic and Heterocyclic Chemistry 2
Synopsis
◾
The
Origin
of
AromaGcity
and
General
CharacterisGcs
of
AromaGc
Compounds
◾
Examples
of
AromaGcity
◾
Electrophilic
AromaGc
SubsGtuGon
◾
Nucleophilic
AromaGc
SubsGtuGon
◾
Arynes
◾
ReacGons
with
Metals:
Ortho
LithiaGon,
Palladium
Catalysed
Couplings,
Chromium
Complexes
and
Birch
ReducGons
◾
IntroducGon
of
FuncGonal
Groups
◾
Pyridine:
Synthesis
and
ReacGons
◾
Pyrrole,
Thiophene
and
Furan:
Synthesis
and
ReacGons
◾
Indole:
Synthesis
and
ReacGons
◾
ReducGon
of
AromaGcs
◾
AromaGc
Chemistry
in
AcGon
Aromatic and Heterocyclic Chemistry 3
Origin
of
Aroma0city
◾
Typical
reacGons
of
alkenes
◾
Typical
reacGons
of
benzene
◾
retains
aromaGc
sextet
of
electrons
in
subsGtuGon
reacGons
◾
does
not
behave
like
a
“normal”
polyene
or
alkene
◾
benzene
is
both
kine7cally
and
thermodynamically
very
stable
◾
Heats
of
hydrogenaGon
ΔHo =
-‐120
kJmol-‐1
ΔHo =
3
x
-‐120
=
-‐360
kJmol-‐1
ΔHo =
-‐210
kJmol-‐1
hydrog hydrog hydrog
(hypotheGcal,
1,3,5-‐cyclohexatriene)
◾
benzene
≈150
kJmol-‐1
more
stable
than
expected
–
(represents
stability
over
hypotheGcal
1,3,5-‐
cyclohextriene)
–
termed
the
empirical
resonance
energy
(values
vary
enormously)
◾
empirical
reso
nance
energy
is
not
the
true
resonance
energy
as
this
would
be
the
difference
between
benzene
and
a
symmetrical
non-‐delocalised
cyclohexatriene
unit
which
does
not
exist
◾
we
know
that
delocalisaGon
is
stabilising,
but
how
much
more
stabilising
is
the
delocalisaGon
in
benzene
–
should
compare
benzene
with
a
real
molecule
–
we
will
use
1,3,5-‐hexatriene
◾
require
a
theory
which
explains
the
stability
of
benzene
Aromatic and Heterocyclic Chemistry 4
Understanding
Aroma0city
◾
Hückel’s
Rule:
planar,
monocyclic,
completely
conjugated
hydrocarbons
will
be
aroma%c
when
the
ring
contains
(4n
+2)
π-‐electrons
(n
=
0,
1,
2….posi7ve
integers)
Corollary
◾
planar,
monocyclic,
completely
conjugated
hydrocarbons
will
be
an%-‐aroma%c
when
the
ring
contains
(4n)
π-‐
electrons
(n
=
0,
1,
2….posi7ve
integers)
Hückel
Molecular
Orbital
Theory
(HMOT)
◾
applicable
to
conjugated
planar
cyclic
and
acyclic
systems
◾
only
the
π-‐system
is
included;
the
σ-‐framework
is
ignored
(in
reality
σ-‐framework
affects
π-‐system)
◾
used
to
calculate
the
wave
funcGons
(ψ)
and
hence
rela7ve
energies
by
the
LCAO
method
i
i.e.
ψ
=
c φ
+
c φ
+
c φ
+
c φ
+
c φ
…..
i 1 1 2 2 3 3 4 4 5 5
◾
HMOT
solves
energy
(E)
and
coefficients
c
i i
◾
there
are
now
many
more
sophisGcated
methods
for
calculaGng
the
stabilisaGon
energy
in
conjugated
systems;
however,
HMOT
is
adequate
for
our
purposes.
Aromatic and Heterocyclic Chemistry 5
Understanding
Aroma0city
HMOT
in
Ac0on
◾
For
cyclic
and
acyclic
systems:
Molecular
Orbital
Energies
=
E
=
α
+
mβ
i j
◾
α
=
coulomb
integral
-‐
energy
associated
with
electron
in
an
isolated
2p
orbital
(albeit
in
the
molecular
environment)
–
α
is
negaGve
(stabilising)
and
is
the
same
for
any
p-‐orbital
in
π-‐system
◾
β
=
resonance
integral
–
energy
associated
with
having
electrons
shared
by
atoms
in
the
form
of
a
covalent
bond
–
β
is
negaGve
(stabilising)
and
is
set
to
zero
for
non-‐adjacent
atoms.
◾
(all
overlap
integrals
S
assumed
to
be
zero,
electron
correlaGon
ignored)
◾
linear
polyenes
m
=
2cos[jπ/(n+1)]
j
=
1,
2……n
(n
=
number
of
carbon
atoms
in
j
conjugated
system)
◾ cyclic
polyenes
m
=
2cos(2jπ/n)
j
=
0,
±1,
±2……±[(n-‐1)/2]
for
odd
n,
±n/2
for
even
n
j
◾
1,3,5-‐Hexatriene
vs
Benzene
–
linear
polyene
◾
six
2p
atomic
orbitals
give
6π
molecular
orbitals;
n
=
6,
j
=
1,
2,
3,
4,
5,
6
◾
m
=
2cos(π/7)
=
1.80
2cos(2π/7)
=
1.25
2cos(3π/7)
=
0.45......
j
and
the
corresponding
negaGve
values
◾
energy
E
=
α
+
1.80β
α
+
1.25β
α
+
0.45β
α
–
0.45β…..etc
Aromatic and Heterocyclic Chemistry 6
◾
1,3,5-‐Hexatriene
vs
Benzene
◾
six
2p
atomic
orbitals
give
6π
molecular
orbitals;
n
=
6,
j
=
1,
2,
3,
4,
5,
6
◾
m
=
2cos(π/7)
=
1.80
2cos(2π/7)
=
1.25
2cos(3π/7)
=
0.45......
j
and
the
corresponding
negaGve
values
◾
energy
E
=
α
+
1.80β
α
+
1.25β
α
+
0.45β
α
–
0.45β…..etc
MO
no.
nodes
energy
HMOT
MO
(calculated)
ψ
5
α
-‐
1.80β
6
ψ
4
α
-‐
1.25β
5
ψ
3
α
-‐
0.45β
4
α
ψ
2
α
+
0.45β
3
ψ
1
α
+
1.25β
2
ψ
0
α
+
1.80β
1
◾
stabilisaGon
energy
=
E =
2(3α
+
3.5β)
=
6α
+
7β
stab
Aromatic and Heterocyclic Chemistry 7
◾
Benzene
m
=
2cos(2jπ/n)
j
=
0
±1,
±[(n-‐1)/2]
for
odd
n;
±n/2
for
even
n
j
◾
six
2p
atomic
orbitals
give
6π
molecular
orbitals;
n
=
6,
j
=
0
±1,
±2,
±3
◾
m
=
2cos(0)
=
2,
2cos(±2π/6)
=
1,
2cos(±4π/6)
=
-‐1,
2cos(±6π/6)=
-‐2
j
◾
energy
E
=
α
+
2β,
α
+
β,
α
-‐
β,
α
–
2β
MO
no.
nodes
energy
HMOT
MO
(calculated)
ψ
6
α
-‐
2β
6
ψ
4
α
-‐
β
4,5
α
ψ
2
α
+
β
2,3
ψ
0
α
+
2β
1
◾
stabilisaGon
energy
=
E =
2(α
+
2β)
+
4(α
+
β)
=
6α
+
8β;
stabilisaGon
energy
w.r.t.
1,3,5-‐hexatriene
=
β
stab
◾
HMOT
predicts
benzene
is
more
stable
than
1,3,5-‐hexatriene
◾
aroma7c
compounds
are
those
with
a
π-‐system
lower
in
energy
than
that
of
acyclic
counterpart
Aromatic and Heterocyclic Chemistry 8
◾
Cyclobutadiene
vs
1,3-‐butadiene
◾
four
2p
atomic
orbitals
give
4π
molecular
orbitals;
n
=
4,
j
=
0
±1,
±2
◾
m
=
2cos(0)
=
2
2cos(±2π/4)
=
0
2cos(±4π/4)
=
-‐2
j
◾
energy
E
=
α
+
2β
α
α
–
2β
cyclobutadiene
1,3-‐butadiene
no.
HMOT
MO
energy
MO
energy
nodes
(calculated)
ψ
4
α
-‐
2β
α
-‐
1.62β
4
α
-‐
0.62β
ψ
2
α
α
2,3
α
+
0.62β
ψ
0
α
+
2β
α
+
1.62β
1
◾
stabilisaGon
energy
=
E =
4α
+
4β
■ stabilisaGon
energy
stab
E =
4α
+
4.4β
stab
◾HMOT
predicts
cyclobutadiene
is
less
stable
than
1,3-‐butadiene
◾
an7-‐aroma7c
compounds
are
those
with
a
π-‐system
higher
in
energy
than
that
of
acyclic
counterpart
Aromatic and Heterocyclic Chemistry 9
Frost-‐Musulin
Diagram
–
Frost
Circle
◾
simple
method
to
find
the
energies
of
the
molecular
orbitals
for
an
aromaGc
compound
◾
inscribe
the
regular
polygon,
with
one
vertex
poinGng
down,
inside
a
circle
of
radius
2β,
centred
at
energy
α
◾
each
intersecGon
of
the
polygon
with
the
circumference
of
the
circle
corresponds
to
the
energy
of
a
molecular
orbital
α0202β α0202β
E
α020β
α α
2β α0+0β 2β
α0+02β α0+02β
cyclobutadiene benzene
General
CharacterisGcs
of
AromaGc
Compounds
◾
planar
fully
conjugates
cyclic
polyenes
◾
more
stable
than
acyclic
analogues
◾
bonds
of
nearly
equal
length
i.e.
not
alternaGng
single
and
double
bonds
◾
undergo
subsGtuGon
reacGons
(rather
than
addiGon
reacGons)
◾
support
a
diamagneGc
ring
current
-‐
good
test
for
aromaGc
character
of
a
compound
H
H
δ "="56"ppm δ "="7.26"ppm
H H
aroma.cs"δ "="78"ppm
H
Aromatic and Heterocyclic Chemistry 10
Examples
of
AromaGc
and
AnG-‐AromaGc
Compounds
◾Hückel’s
rule
[(4n
+2)
π-‐electrons
for
aromaGc
compounds;
4n
π-‐electrons
for
anG-‐aromaGc
compounds]
holds
for
anions,
caGons
and
neutrals
Cyclopropenium
caGon
◾
(4n
+2),
n
=
0,
2π
electrons;
stabilisaGon
energy
=
2α
+
4β
–
stabilisaGon
energy
of
allyl
caGon
=
2α
+
2.8β
α"%"β
E
α"+"2β
◾
insoluble
in
non-‐polar
solvents;
1
signal
in
1H
NMR
δ
=
11.1
ppm
-‐
aromaGc
and
a
caGon
H
◾
compare
with
cyclopropyl
caGon
which
is
subject
to
rearrangement
to
the
allyl
caGon
Cyclopropenones
Description:handout at: http://burton.chem.ox.ac.uk/teaching.html. ◾ Advanced Advanced Organic Chemistry: Reacfions, Mechanisms and Structures; J. March. ◾ Fronfier Orbitals and Organic Chemical Reacfions; I. Fleming. ◾ Heterocyclic
