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期中考试
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Chapter 5 Structures of Polyatomic
Molecules (I)
1. Hybrid Orbital Theory
2. VSEPR Model (after-class reading)
3. Delocalized p-electrons/conjugation
and Delocalized Molecular Orbital Theory
4. Hückel Molecular Orbital Theory and
Conjugated Systems
5. Symmetry Rules for Molecular Reactions
Chapter 5 Structures of Polyatomic
Molecules (I)
Please refer to Webtext:
http://www.chem1.com/acad/webtext/chembond/cb06.html
http://www.chem1.com/acad/webtext/chembond/cb07.html
Linus Pauling (1901-1994) was the most famous American chemist
of the 20th century and the author of the classic book The Nature of
the Chemical Bond. He won the Nobel Prize in Chemistry in 1954
and the Nobel Peace Prize in 1962.
§5.1 Hybrid Orbital Theory and Atomic Orbital
Hybridization
1. Hybrid orbital theory -first proposed by Pauling in 1928
Two Modern Theories of Chemical Bonding
VALENCE BOND THEORY
• valence electrons are localized between two atoms (or as
lone pairs).
• half-filled atomic orbitals overlap to form bonds.
VB theory demands orbital hybridization if necessary!
MOLECULAR ORBITAL THEORY
• valence electrons are delocalized.
• valence electrons spread over entire molecule.
MO theory describes hybridization of atomic orbitals
more naturally and inherently !
How does MO theory deal with hybridization of AO’s?
 CMO   cm mAO
2u*
c2 > c 1
 g (2s)  (2 sA  2 sB )
 g (2 pz )  (2 p A  2 p B ) c1 > c2
z
LUMO
1pg
2p
2p
Energy
(LCAO-MO)
1st-order approximation:
HOMO
2g
1pu
z
2nd-order approximation:
 g (2 sp )  c1 g ( 2 s )  c2 g ( 2 pz )
c1 (2 sA  2 sB )  c2 (2 pz A  2 pz B ) 
( c12 sA  c22 pz A )  ( c12 sB
1u*
sp-mixing
2s
1g
2s
N
N
 c22 pz B ) N : KK(1σ )2(1σ )2(1π)4(2σ )2
2
g
u
g
Inequivalent sp-hybridization at each N atom
Similarly,  u ( 2 sp )  ( c12 sA  c22 pz A )  ( c12 sB  c22 pz B )
The -MOs of N2 consisting jointly of 2s and 2pz AOs,
 g / u (2 sp)  (c12 sA  c22 p A )  (c12 sB  c22 p B )
z
z
z
For each N atom,
(a) 2s+2pz --HO1
1  (c12 s  c22 pz )
z
(b) 2s2pz
--HO2
2  (c12 s  c22 pz )
• Hybridization of AOs is inherently, though not
explicitly, taken into account within the framework
of MO theory!
a. Central Themes of Valence Bond Theory
Basic Principle of Valence Bond Theory: a covalent bond
forms when the orbitals from two atoms overlap and a pair of
electrons occupies the region between the nuclei.
1) Opposing spins of the electron pair.
2) Maximum overlap of bonding orbitals.
3) Hybridization of atomic orbitals, if required.
Pauling proposed that the valence atomic orbitals in a molecule
are different from those in the isolated atoms. We call this
Hybridization!
e.g., H2O
b. Why do atomic orbitals need hybridization?
• Suppose each  bond arises
from the overlap of an H1s
orbital with one of the O2p
orbitals.
• This model suggests that the
H-O-H bond angle should be
90, which differs significantly
from the experimental value.
Even worse, such a bonding motif poses very
Use of O
p - orbitals
strong repulsion among the occupied 2s (O) AO
and the two O-H bonds!
b) Hybridization –sp3
a) No hybridization
104.5
90
b1
H2O
b2
Inequivalent sp3
hybridization of
O(2sp) AOs
c. Why do atomic orbitals need hybridization?
CH4
• Carbon has four valence
electrons that are typically
involved in the formation of
4 C-H bonds.
• For a carbon atom bonded
to four other atoms,
experimental evidences
suggested that all of the
bonds have similar bonding
orbitals.
Hybrid Orbital Theory is phenomenological!
d. How do atomic orbitals hybridize?
• Hybridization of n AO’s of an atom gives rise to n
hybrid orbitals (HO)!
n
{iAO }( i  1,..., n )  { jHO   c jiiAO } ( j  1,..., n )
i 1
e.g., a sp3-hybridized orbital can be expressed as:
h  c12 s  c22 px  c32 py  c42 pz
• All n hybrids are equivalent except for directionality
 same energy. (Valid for equivalent hybridization!)
2. Construction of hybrid orbitals
a. sp hybridization
b. sp2 hybridization
c. sp3 hybridization
d. dsp3 (sp3d) hybridization (trigonal bipyramidal or
square pyramidal)
e. dsp2 (sp2d) hybridization (square planar)
f. d2sp3 (sp3d2)hybridization (octahedral)
g. More … (e.g., sp3d3f, or sp3d3)
Bonding ability of AO: F  2l  1
Fs  1, Fp  3 , Fd  5 ...
a. sp hybridization (linear species)
• One s and one p AOs mix to form a set of two hybrid orbitals.
+
-
+
+
-
+
-
+
+
-
For equivalent sp-hybridization:
sp, =180, linear,
1
h1 
(2 s  2 px )
Normalized & Orthogonal
2
1
hjhi d   ij

h 2 
(2 s  2 px )
 = |c2s|2=1/2 (component of s orbital)
2
Equivalent /inequivalent sp-hybridization
Whether an atom adopts equivalent or inequivalent
hybridization depends on its chemical environment!
sp-hybridization at Be atom
Cl-Be-Cl
equivalent
Cl-Be-Br
inequivalent
b. sp2 hybridization (trigonal planar)
• One s and two p (px and
py) AOs mix to form a set
of three hybrid orbitals.
• Equivalent case:
1
h1 
s 
3
1
h 2 
s 
3
1
h 3 
s 
3
2
2 px
3
1
2 px 
6
1
2 px 
6
1
2 py
2
1
2 py
2
sp2, D3h, =120º, trigonal
 = 1/3 (component of s-orbital)
 hi  ai s  bi 2 px  ci 2 py
For equivalent hybridization: the weighting of s orbital in each
hybrid orbital is 1/3, and therefore
ai  1 / 3
hi  (1 / 3 )s  bi2 px  ci2 py
Supposing h1 is parallel to the x-axis and is perpendicular to the
y-axis, then we have
h1  1 / 3s  bi2 px
h1  1 / 3s  2 / 32 px
Normalization
 h1  1 / 3 s  2 / 3 2 px
 hi  1 / 3 s  bi2 px  ci2 py
Normalization and orthogonality
(i = 2 or 3)
a1a2  b1b2  c1c2  0
1 / 3  2 / 3b2  0  c2  0
a2  b2  c2  1
1 / 3  b2  c2  1
2
2
2
b2   1 / 6
2
2
c2   1 / 2
 h 2  1 / 3 s  1 / 62 px  1 / 2 2 py
 h3  1 / 3 s  1 / 6 2 px  1 / 22 py
c. sp3 hybridization (tetrahedral)
H
equivalent
hybridization
H
=1/4
H
C
H
h2
1
h 2  [ s  px  p y  pz ]
2
z
h1
y
x
h4
1
h1  [ s  px  p y  pz ]
2
h3
1
h 3  [ s  px  p y  pz ]
2
1
h 4  [ s  px  p y  pz ]
2
• One s and 3 p AO’s mix to form a set of four hybrid sp3 orbitals.
2
CH 4 , SiH 4 , GeH 4 , PX 4 , SO4 , PO4
3
Formation of four LMOs of CH4:
h2
b
LMO
HO of C
H 1s
 a  ( s  px  p y  pz ) / 2  1sa
z
a
h1
y
x
d
h4
c
h3
 b  ( s  px  p y  pz ) / 2  1sb
 c  ( s  px  p y  pz ) / 2  1sc
 d  ( s  px  p y  pz ) / 2  1sd
 iCMO CMO AO C AO' s of H atoms
1 a1 : s  s  (1sa  1sb  1sc  1sd ) / 2
 2 t1 : x  px  (1sa  1sb  1sc  1sd ) / 2
 2 t1 : y  p y  (1sa  1sb  1sc  1sd ) / 2
2
t1 : z  pz  (1sa  1sb  1sc  1sd ) / 2
iCMO   cim mAO
m
CMOs of CH4
In light of the superposition principle, each LMO of CH4 can be
expressed by a linear combination of its CMOs, i.e.,
LMO
CMOs
HO of C,
H 1s
 a  s  x  y  z  ( s  p x  p y  pz ) / 2  1sa
 b  s  x  y  z  ( s  px  p y  pz ) / 2  1sb
 c  s  x  y  z  ( s  px  p y  pz ) / 2  1sc
Yet to be
normalized!
 d  s  x  y  z  ( s  px  p y  pz ) / 2  1sd
It is provable that the four LMOs have the same energy!
As CMOs are eigenfunctions of one-particle eigenequations, we have
CMO
CMO
CMO
CMO
hˆ i
  i i
&  ( j
)* i d   ij
 j
LMO
4
  ci i
j
CMO
i 1
4
4
& ci j   /  1  E j
4
LMO
  ( j
  ( ci )  i /  ( ci )  (   i ) / 4  ( 1  3 2 ) / 4
j 2
i 1
j 2
i 1
i 1
LMO
LMO
LMO
LMO
) * hˆ  j d /  ( j ) * j d
(for j  1,...,4)
F
d. dsp3 (sp3d) hybridization (bipyramidal)
1
2
 h1 
s
px
3
3
1
1
1
h 2 
s
px 
py
2
3
6
1
1
1
h3 
s
px 
py
2
3
6
1
h 4 
[ pz  d z 2 ];
2
z
h3
x
S
h4
h1
h2
F
F
F
F
h5
F
P
y
1
h 5 
[ pz  d z 2 ]
2
F
F
The bond lengths will not be the same because there is more
d contribution to the axial hybrid orbitals.
In the case of dsp3, the axial bonds are shorter.
In the case of sp3d, the axial bonds are longer.
F
e. dsp2 hybridization (square planar)
d x 2  y 2 , s, p x , p y
1
h1  s 
2
1
h 2  s 
2
1
h 3  s 
2
1
h 4  s 
2
 = 1/4
1
1
 px  d x 2  y 2
2
2
1
1
 px  d x 2  y 2
2
2
1
1
 py  d x 2  y 2
2
2
1
1
 py  d x 2  y 2
2
2
Square planar: D4h
h3
y
h1
h2
x
h4
e.g., Ni(CN)42, AuCl4
f. d2sp3 (sp3d2) hybridization (Octahedral)
1
1
1
oc1 
s
px  d x 2  y 2 
z
2
6
2
1
1
1
3 52 y
oc2 
s
p y  d x2  y2 
4
2
6
2
1
6
1
1
1
x
oc3 
s
px  d x2  y2 
2
2
6
F
F
F
F
S
F
F
d z 2 , d x 2  y 2 , s,
1
d z2
12
1
d z2
12
1
d z2
12
1
1
1
1
oc4 
s
p y  d x2  y2 
d z2
2
2
6
12
1
1
oc5 
s
pz 
2
6
px , p y , pz
1
1
oc6 
s
pz 
2
6
1
d z2
3
1
d z2
3
Hybridization schemes
• spndm gives a “complete” set of hybrid orbitals for
“any” geometry.
sp
sp2
sp3
sp3d (dz2)
sp3d(dx2-y2)
linear
trigonal planar
tetrahedral
trigonal bipyramidal
square pyramidal
sp2d2
square pyramidal
sp3d2
octahedral
sp2d
square planar
3. The angle between two hybrid orbitals
1) spn hybridization
Let’s define
hi
h   2 s  1   2 p
 hi   i  2 s  1   i 2 p
i
hj   j 2 s  1   j 2 pj
where pi is
 ij
 hj
2 p  xi px  yi2 py  zi2 pz
i
Note the angle between the two vectors pi and pj is the angle ij
between the two hybrid orbitals hi and hj , i.e.,
cos ij    pi pj d
Normalization and orthogonality
0   hihj d   ( i s  1  i  pi )(  j s  1   j  pj )d
 i j   d  (1  i )(1   j )   pi pj d  0  0
2
s
 0  i j  (1  i )(1   j ) cos ij
 cos ij  
i j
(1  i )(1   j )
 0  ij  90
 hi
 ij
 i   j Equivalent hybridization. e.g., CH4
 i   j Non-equivalent hybridization. e.g., CHCl3, CH3Cl
sp:  = 0.5

cos = -1
sp2 :  = 1/3  cos = 0.5
  = 180.0
  = 120.0
 hj
Equivalent hybridization
1
sp  hybridization   
4
3
i j
1
cos ij  

3
(1  i )(1   j )
   109 28"
0
Non-equivalent hybridization
sp3  hybrides
z
H
t2
N
H
PH3, PF3, NF3,
H
t1
Other examples
y
X
s-component of the N sp3
hybridized orbitals to form
the N-H bonds.
t3
t4
for NH 3 :   HNH  107.3
  (1   ) cos 107.3  0
   0.23
  HO ( N  H )  0.23s  0.77 p  0.48s  0.88 p
 Ls  1  3 * 0.23  0.31
 Lp  3  3 * 0.77  0.69
 lone pair  0.31s  0.69 p  0.56s  0.83 p
s-component of
the lone pair
p-component of
the lone pair
For d-s-p hybridization, the angles between two hybrid orbitals can be
calculated by: (where ,  and  are the component of s, p and d
orbitals)
3
1
2
   cos    ( cos   )  0
2
2
1
1
1
d sp :   ,   ,  
3
2
6
1
1 3
1 1
2
 cos   ( cos   )  0
2
3 2
6 2
 cos 1  0, cos  2  1
2
z
3
 1  90 ,  2  180
3
y
5
1
x
Octahedral
4. The bonding ability of hybrid orbitals
Atomic orbital bonding ability : s, p, d, f :1, 3, 5 , 7
n
The bonding ability of a hybrid for the sp hybridizat ions,
orbital is given by :
1
1
fh    3   5
Hybridization enhances the
bonding ability of atomic
valence orbitals!!
1
1
1
  ,   , 
6
2
3
  , 
f h  1.932
sp
f h  1.992
sp2
2
2
1
2
  , 
3
3
1
3
  , 
4
4
  0,
f h  2.00
sp3
fh  3
p
  1,
fh  1
s
f h  2.925
d 2sp 3
5. Discussions
F
B
sp2
e.g., BF3(D3h), BH3 (D3h), NO3, CO32 
Example: BH3
• The B atom has 3 sp2 hybrid
orbitals each with one
electron.
• This one electron pairs with
the hydrogen 1s electron.
• The 2pz AO of B atom is thus
empty.
F
F
Example:
p pz+pz
H
C
H
tr3
H
C
C2H4
x
H
tr1
tr2
y
C2H4
The sp2-hybridization and chemical bonding in C2H4.
Example: nitrate (NO3-) --hybridization and delocalization
dot-electron
structure
Resonance of three equivalent dot-electron structures
–delocalization of p-electrons
Example: Benzene --hybridization and delocalization
The sp2 hybrid orbitals of C
atoms and the 1s AOs of H
atoms form the -framework
of benzene.
The unhybridized pz AOs
of C atoms form a
continuous cyclic 66
bond of benzene.
6. “Bent bond” model -- An alternative of hybrids
“Bent bond” model of ethylene
• No need to assume
hybridization in the C
atoms at all.
• Directions of the p-orbitals in C
atoms should be distorted
sufficiently to provide the
overlap for bonding. These bent
bonds are called “banana bonds”.
6. “Bent-bond” Model -- an alternative of hybrids
cyclopropane
sp3 hybridization of C
H2
C
H2C
CH2
Note the C–C bond angles are
120°— quite a departure
from the tetrahedral angle of
109.5°associated with sp3
hybridization!
Three “banana bonds” formed!
In terms of MO theory, Dewar proposed that this molecule has
-aromaticity and the C atoms adopt sp2-hybridization!
You should understand that
• Hybridization is not a physical phenomenon; it is merely a
mathematical operation that combines the atomic orbitals we are
familiar with in such a way that the new (hybrid) orbitals possess
the geometric and other properties that are reasonably consistent
with what we observe in a wide range (but certainly not in all)
molecules.
• Hybrid orbitals are abstractions that describe reality fairly well in
certain classes of molecules (and fortunately, in much of the very
large class of organic substances) and are therefore a useful
means of organizing a large body of chemical knowledge... but
they are far from infallible.
Concept
Map (I)
Concept Map II
§5.2 Valence Shell Electron-Pair Repulsion
(VSEPR) Model
• Quantum mechanical treatments have a number of
advantages. However, the VSEPR model allows a simple
qualitative prediction of molecular geometry.
• In a molecule, the valence electron pairs are stabilized by
electrostatic attraction exerted by the nucleus, but
destabilized by Pauli repulsion exerted by neighboring
electron pairs.
•
A lone pair is more diffuse than a bonding pair, exerting
larger repulsion to a neighboring pair.
A stable molecule should adopt such a geometry that
minimizes the repulsions among its various electron pairs!
VSEPR
Repulsion of electron pairs
LP-LP > LP-BP > BP-BP
Atom B in ABn lies on the surface of a sphere; electron pairs
are “localized” on a sphere of smaller radius at maximum
distances apart, so as to minimize overlap of different
electron pairs.
1. Repulsion between two lone pairs (LPs) is the greatest.
2. Repulsion between a lone electron pair and a bonding
electron pair (BP) is less.
3. Repulsion between two bonding pairs is the least.
4. p electron pair does not influence stereochemistry.
VSEPR model is used in combination with the Lewis dot-electron
structures and hybrid orbital theory.
Arrangements of Maximum distance between valence shell
electron pairs.
No. of pairs
arrangement
2
Linear
3
Equilateral triangle
4
Tetrahedron
5
Trigonal bipyramid
6
Octahedron
AX2
BeCl2, CaCl2
(linear)
(2+2)/2=2
AX3
BF3
(trigonal plane)
(3+3)/2=3
AX2E
CCl2, SnCl2
(bent)
(4+2)/2=3
AX4
CH4, AlH4-, CCl4, NH4+
(tetrahedral)
(4+4)/2=4
AX3E
NH3, NF3
(pyramidal)
(5+3)/2=4
AX2E2
H2O, H2S
(bent)
(6+2)/2=4
(5+5)/2=5
(6+4)/2=5
(7+3)/2=5
(8+2)/2=5
(6+6)/2
(7+5)/2=6
For such molecule as XeF4, isomers exist!
(8+4)/2=6
Repulsion of electron pairs
LP -LP > LP-BP > BP-BP
• N(VEP) = (7 + 4 + 1)/2 = 6; N(BP) = 4; N(LP) = 2
• Simply considering the pairs with a bond angle ~ 90º!!!!!!!
• The first structure is preferred!
Repulsion of electron pairs
BP-BP < BP-LP < LP-LP
• N(VEP) = (7+3)/2 =5; N(BP) = 3; N(LP) = 2
• The second structure is preferred!
Molecular Stereochemistry
No. of pairs of e formula
stereochemistry point group
1
A2, AB
none
Dh or Cv
2
AB2
linear
Dh
3
AB2e1
bent
C2v
3
AB3
triangular
D3h
4
AB2e2
bent
C2v
4
AB3e1
pyramidal
C3v
4
AB4
tetrahedral
Td
Molecular Stereochemistry
No. of pairs of e- formula
stereochemistry point group
5
AB2e3
linear
Dh
5
AB3e2
T-shaped
C2v
5
AB4e1
distorted tetrahedron C2v
5
AB5
trigonal bipyramid
D3h
6
AB4e2
square planar
D4h
6
AB5e1
square pyramid
C4v
6
AB6
octahedron
Oh
VSEPR Example
Lp
Lp
N
N
H
H
F
102°
107°
H
F
F
• Bond angles decrease with increasing electronegativity of
the ligand or decreasing electronegativity of the central
atom.
• Why? The BP is more distant from the central atom upon
increasing electronegativity of the ligand.
ABnem (n = coordination number; m = number of lone pairs)
An alternative view on the chemical bonding within
hypervalent compounds
3-center 4-electron bond – VB (Coulson model) vs. MO
(Rundle–Pimentel model ) description
Example 1. XeF2
a) Classic view – Pauling’s model (d-orbital included).
sp3d hybridization,
3 lone pairs + 2 axial Xe-F bonds
b) 3-center 4-electron bond (d-orbital excluded).
sp2 hybridization  3 lone pairs (equatorial)
Axial 3c4e bond: Xe 5p (2e) + 2F 2p (2x1e)
VB:
Xe-F bond order =1/2
MO:  1  A(F 1  2 Xe  F 2 )  2  A(F 1  F 2 )
(occupied MOs)
 LUMO  A(F 1  2 Xe  F 2 )
Example 2. SF4
a) Classic view (inclusion of valence d orbital).
sp3d hybridization,
1 lone pair + 4 S-F bonds
b) 3-center 4-electron bond (exclusion of valence d orbital).
sp2 hybridization  1 lone pair + 2 S-F (equatorial)
Axial 3c4e bond (MO): S 3p (2e) + 2F 2p (1e)
  1  AF 1  BS  AF 2  2  A' (F 1  F 2 ) (occupied MOs)
 LUMO  AF 1  BS  AF 2 (A  B)
• This model counts well for the fact that the axial S-F
bonds are longer than the equatorial ones.
• Such bonding model can be used for similar hypervalent
compounds, e.g., XeF4, BrF5 etc!
§5.3 Delocalized p-electrons/conjugation
and Delocalized Molecular Orbital Theory
5.3.1 Normal p bond
H
H
C
H
C
H
C2H4
tr3
x
tr1
tr2
y
C2H4
i.e., 2c-2e p-bond
Except the -framework, the remaining,
unhybridized p orbital of C atom is
perpendicular to the molecular plane,
forming a p bond with that of a
neighboring C atom.
5.3.2 Delocalized p bond

m
n
n-orbital numbers,
m-electron numbers (n>2)
Conditions for the formation of a delocalized p bond
•
The atoms are coplanar, with every atom contributing a porbital orientated in the same direction.
•
The number of p electrons is less than twice of the number of
participating p-orbitals.
A. Some Inorganic conjugated molecules
i.
Linear type (AB2 16 valence electron) C: sp-hybridization + px + py
p electrons : 16 - 4 - 4  8
3c-4e p-bond
2 34 :  34 ( x )  34 ( y )
For x or y, we have three p-MOs,
1
2
1
p 2 
[O (1)  O (2)]
2
1
 p 3  [O (1)  2C  O (2)]
2
 p 1  [O (1)  2C  O (2)]
bonding
 p1
p2
non - bonding
anti - bonding
p3
Isoelectronic analogy : CO2, NO2+, N2O, N3, COS, BeCl2, HgCl2
Note: While NO2+ has two equivalent 34 bonds, NO2 has just a
34 bond!
ii. Non linear type(bent) (17~19 valence electron)
sp2 orbital
..
e.g., O3 (18 valence electrons)
pz O
1
 p 1  [1  22  3 ]
2
1
p 2 
[1  3 ]
2
1
 p 1  [1  22  3 ]
2
O
..p
z
..
p electrons : 18 - 6 - 8  4
 34
 34
O..
pz
..
Central atom O: sp2 hybridization + pz
bonding
 p1
non - bonding(HO MO)
p2
anti - bonding(LU MO)
p3
Isoelectron analogy:
SO2
19 valence electrons: ClO2

4
3
 53

anti-bonding

non-bonding

bonding
iii. Equilateral triangular type: BCl3 D3h
24 valence electrons
24
Central atom B: sp2 hybridization
non-bonding
Cl
-6
 p  c1 B (1)  c2 Cl ( 2)  c3 Cl ( 3)  c4 Cl (4)
anti-bonding
Cl
B
18
-12
Cl
6
p electrons :
bonding

6
4
Isoelectronic analogy: AlF3, NO3-, CO32-, SO3, Cl2CO,
(H2N)2C=O, Cl2C=S
• For four-atom triangular type molecules, the pp-orbitals form one
bonding, two nonbonding and one antibonding orbitals, in general.
• For four-atom linear type molecules, the pp-orbitals form two
bonding and two antibonding orbitals!
B. Some organic conjugated molecules


6
6
10
10
NO2
  
H
C
CH2
C
H
H2C
 44
10
9
6
6
4
3
5.3.3 The conjugation effects.
i. The electrical conductivity is enhanced by the delocalization of
p bond, e.g., graphene.
ii. Color. The formation of delocalized p bonding increases the
delocalized extent of the p electrons and causes the energy of the
system to decrease.
[n]Polyacene: laterally fused benzene rings
2n+1
4n+2
2n+1

4n  2
4n  2
Benzene, naphthalene, anthracene,.., pentacene, hexacene,…
1) HOMO-LUMO gap of oligoacene decreases with increasing of n.
2) Pristine oligoacene becomes instable with n > 5.
3) When n > 6, pristine oligoacene adopts an open-shell singlet
ground state, i.e., biradical. (J. Am. Chem. Soc., 2004, 126, 7416)
SOMO()
SOMO()
Two degenerate disjoint non-bonding orbitals of [n]polyacene. (n>6)
iii. Acidity and basicity.
A - NH

3
+ H+
- H+
OH
- H+
O-

8
7

4
3

4
3
NH2
8
7

Electron delocalization results in enhanced acidity of phenol and
carboxylic acids as well as weakened basicity of amides and
aniline.
iv. Delocalization/conjugation effects on Chemical reactivity.
e.g. H2C=CH-Cl. The formation of 34 causes a contraction
of the C-Cl bond, and reduces the lability of Cl.
e.g., the p-bond of Benzene shows much lower reactivity than
that of simple alkenes.
5.3.4 Hyperconjugation.
•First introduced in 1939 by R.S. Mulliken.
•Hyperconjugation is the stabilizing interaction that results from
the interaction of the electrons in a -bond (usually C–H or C–C)
with an adjacent empty (or partially filled) non-bonding p-orbital
or antibonding p orbital to give an extended molecular orbital
that increases the stability of the system.
•Only electrons in bonds that are  to the positively charged
carbon can stabilize a carbocation by hyperconjugation.
1.46 Å < 1.55 Å Hyperconjugation due to electron delocalization
in simple alkanes. from (C-H) to p*(C≡C).
Effects of hyperconjugation on Chemical properties.
A. Bond length and bond energy:
-bond: Shortening of bond length and increasing of bond energy.
Bond type
Hybridization
C-C bond length (Å) C-C bond energy(kJ.mol-1)
sp3-sp3
1.54
346.3
sp3-sp2
1.51
357.6
sp3-sp
1.46
382.5
B. Dipole moment
The dipole moment of 1,1,1-trichloroethane is much larger than
that of chloroform due to hyperconjugation.
C. Stabilizing carbocations: (C-H)  pp(C+)
(CH3)3C+ > (CH3)2CH+ > (CH3)CH2+ > CH3+
§5.4 Hückel molecular orbital（HMO）
theory and conjugated system
p-type AO
5.4.1 HMO method
• Proposed in 1931 by Hückel to deal
with molecules with p-conjugation.
 nm
• Later Modified by R. Hoffmann
 In light of the single-particle approximation, the p-electron
Hamiltonian for a nm system is approximated by the form,
m
Hˆ p   hˆ eff
j
j 1
(i.e., sum of one-electron Halmitonians)
 The total wavefunction of the nm system can be expressed as,
m
p   j
j 1
n
ˆ   E   hˆ     p - MO：  c 
& H
p p
p p
j
j j
 ji i
j
eff
j
i 1
LCAO-MO: n-the number of C atoms; i-the p-AO of ith atom.
hˆ eff
j  j   j j
n
eff
( j   c jii ) The exact form of h is unknown.
Now use the variational method!
i 1
    * hˆ eff d
&
let E   
E
Then make E E

 ...... 
0
Variation Theorem
c1 c2
cn
Thus we have n secular equations ( in the matrix form),
 H 11  ES11 H 12  ES12 ... ... H 1n  ES1n  c1   0 
 c
 H  ES
H

ES
H

ES
...
...
0 
2n
2 n  2 
21
22
22
 21

 ...   ...
 ......
......
......
 H  ES
   
H

ES
...
...
H

ES
n1
n2
n2
nn
nn  c
 n1
 n  0 
This demands the secular determinant to be zero,
H 11  ES11
H 21  ES 21
......
H n1  ES n1
H 12  ES12 ... ... H 1n  ES1n
H 22  ES 22 ... ... H 2 n  ES 2 n
No. of integrals:
 0 H ~ n(n+1)/2
ij
......
......
Sij ~ n(n+1)/2
H n 2  ES n 2 ... ... H nn  ES nn
To solve the secular equation, Hückel further introduced the
following approximations,
H ii   i hˆ i d  
*
--Coulombic integral
H ij   i hˆ  j d   (if i  j  1) --Resonance integral
*
 0 (if i  j  1)
Sij   i  j d  1
*
(if i  j)
--Overlap integral
For a linear p-conjugation system,
0
(if i  j  1)
the secular determinant can thus be
0
(if i  j  1) simplified as,
αE
β
... 0
0
Get {Ej} (n roots)
For each E, get {Cji}
How to determine the  & 
integrals semiempirically?
β
0
...
0
αE
β
...
0
... 0
0
... 0
0 0
... ...
...
β αE
Semiempirical determination of the integrals
H ii   i hˆ i d  
*
Coulombic integral: p-type AO energy of C!
H ij   i Hˆ  j d   (if i  j  1)
*
 0 (if i  j  1)
e.g. C2H4
Resonance integral
E

0

E
x 1
 0 ( x  ( - E)/ )
1 x
x  1  E    
E2    
Ep bond  2 E1  2 E0  2 
  Ep bond / 2
The p-bond energy of ethene can be
determined by UV-vis spectroscopy.
2pA
2pB
E0  
E1    
5.4.2 The HMO treatment for the p-bonding within 1,3-butadiene
H2C=CH-CH=CH2
ψ  c1φ1  c2φ2  c3φ3  c4φ4
Variation theorem
Simplified secular equation by using Hückel approximations:

0
0  c1   0 
  E
   

E

0  c2   0 
 
 



 0
c3

E

0
   

0

  E  c4   0 
 0
Seqular determinant:
E

0
0

E

0
0
0

E

0
0

E
αE
let x 
, E  α-βx
β
HMO method
E

0
0

E

0
0
0

E

0
0

E
x 1 0
11
( 1 x 1 x 1
0 1 x
1 1 0
1 2
 ( 1)  1 0 x 1  0
0 1 x
x 1 0 0
1 x 1 0
0
0 1 x 1
0 0 1 x
1
1

11 x
1 2 1

x  ( 1 x
 ( 1 1

1
x
0
x


1
11 x
 ( 1 1
0
1 x
(

(

x x 1  x  x 1  0
2
2
2
x 4  3x 2  1  0
2
αE
let x 
, E  α-xβ
β
x 1 0 0
1 x 1 0
0
0 1 x 1
0 0 1 x
HMO method
3
H2C=CH-CH=CH2
4
2
1
x 4  3x 2  1  0
 x 2  0.38  x2,3  0.618
 2
 x  2.62  x1,4  1.618
thus,
E 4    1.618 Anti-bonding
E3    0.618
Anti-bonding
E 2    0.618
Bonding
E1    1.618
Bonding
(Note: ,  < 0)
Now we can solve the secular equations to get the coefficients {ci}:
 x 1 0 0  c1 

 
 1 x 1 0  c2 
 0 1 x 1  c   0

 3 
 0 0 1 x  c 

 4 
c1 x  c 2  0
c1  c 2 x  c3  0
c 2  c3 x  c 4  0
c3  c 4 x  0
normalizat ion condition ： c12  c22  c32  c24  1
1) By substituting x1 = -1.618, we get
c1  c4  0.372，c2  c3  0.602
i.e., the p-MO is in the form
 1  0.3721  0.6022  0.6023  0.3724
with E1   - x1    1.618
3
4
2
1
E1    1.618 
E2    0.618 
E3    0.618 
E4    1.618 
 1  0.3721  0.6022  0.6023  0.3724
 2  0.6021  0.3722  0.3723  0.6024
 3  0.6021  0.3722  0.3723  0.6024
 4  0.3721  0.6022  0.6023  0.3724
Antibonding MO
(three nodes)
E4    1.618
Antibonding MO
(two nodes)
E3    0.618
Bonding MO
(One node)
E2    0.618
Bonding MO
(No node)
E1    1.618
Delocalized energy
Total p-electron energy in butadiene
Epe (C4 H 6 )  2 E1  2 E2  2(  1.618 )  2(  0.618 )
 4  4.48
Total p-electron energy of two isolated C=C bonds
Epe ( 2C  C )  2  2(   )
 4  4 
Thus, the delocalized energy or resonance
energy is
E2
||

||
E1 =+
E deloc  Epe ( C4 H 6 )  Epe ( 2C  C )  0.48   0
Electron delocalization (p-conjugation) enhances the stability!
5.4.3 Population analysis and molecular diagrams of conjugated system
i.
Charge density --- the total p-electron density on the ith atom
i   nk ci2 (k )
k
The probability of an electron in the
kth MO to appear on the ith atom
nk: number of electrons in the kth MO.
 (k )   ci (k )i
i
The p-electron density
on the ith atom is:
• For 1,3-butadiene,
the two doubly
occupied MOs are
1    2 (k )d   ci2 (k )
i
i   nk ci2 (k )
k
 p 1  0.3721  0.6022  0.6023  0.3724
 p 2  0.6021  0.3722  0.3723  0.6024
1  2  0.3722  2  0.6022  1.000
 2  2  0.6022  2  0.3722  1.000
 3   4  1.000
5.4.3 Population analysis and molecular diagrams of conjugated system
ii. Bond order --- the strength of the p-bond between atoms i and j
pij   nk ci (k )c j (k )
k
nk: number of electrons in the kth occupied MO.
For 1,3-butadiene, the two occupied MOs are
 p 1  0.3721  0.6022  0.6023  0.3724
 p 2  0.6021  0.3722  0.3723  0.6024
p12  2  0.372  0.602  2  0.602  0.372  0.896  p34
p23  2  0.6022  2  0.3722  0.448
p12  p34  2 p23
• The p-bond order of C1=C2 or C3=C4 is twice of the C2-C3 pbond order.
After-class Problem
As shown in the last few pages, the two occupied HMOs (1 & 2)
of 1,3-butadiene are delocalized and expressed as
 1  0.3721  0.6022  0.6023  0.3724 E1    1.618
 2  0.6021  0.3722  0.3723  0.6024 E2    0.618
Let us consider the linear combinations of these two delocalized
HMOs, which can be expressed as
 1  ( 1   2 ) / 2  [0.974(1  2 )  0.23(3  4 )] / 2
 2  ( 1   2 ) / 2  [0.23(2  1 )  0.974(3  4 )] / 2
Please prove the two wavefunctions, 1 and 2, have equal energy.
Please figure out what they represent.
iii. Free valence index --- the relative magnitude of the
residue bonding ability of the ith atom
In butadiene, the 1,4-sites
Fi  Pmax   Pij ;
Pmax  3
are more reactive than the
j  i 1
2,3-sites!
F1  3  P12  3  0.896  0.836  F4
F2  3  P12  P23  3  0.448  0.896  0.388  F3
Example: 1,4 addition reaction of butadiene
H2C=CH-CH=CH2 + Br2  BrH2C-CH=CH-CH2Br
iv. Molecular diagram.
0.836
0.388
0.388
0.836
Pij
0.896
i
0.448
0.896
H 2C
CH
CH
1.000
1.000
1.000
CH2
1.000
Fi
Another example:
Please derive the secular determinant of
trimethylenemethane.


  c1 
  E

 
E
0
0  c2 
 
p-MO
 0 Variation


 

c3
0
E
0
theorem

 
  cii
c

0
0


E

 4 
i

Then the secular determinant is
E




E
0
0
0

0
E
0

0
0
E
x
( - E )

x 1 1 1
1 x 0 0
0
1 0 x 0
1 0 0 x
5.4.4 HMO treatments of cyclic conjugated polyenes (CnHn)
e.g., p-conjugation in Benzene
6
   cii
i 1
Variation
Theorem

0
0
0
  c1 
  E

 
 E

0
0
0  c2 
 
 0

 E

0
0  c3 

   0
0

 E

0  c4 
 0
 0
 c 
0
0



E


 5 
 
0
0
0

  E  c6 

1
(1  2  3  4  5  6 )
6
 6 (b2 g ) 
1
 5 (e2u )  (2  3  5  6 )
2
1
 4 (e2u ) 
(21  2  3  24  5  6 )
12
1
 3 (e1g )  (2  3  5  6 )
2
1
 2 (e1g ) 
(21  2  3  24  5  6 )
12
1
 1 (a2u ) 
(1  2  3  4  5  6 )
E6 = -2
E4=E5= -
E2=E3= +
E1 = +2
General HMO solutions for cyclic conjugated polyenes (CnHn)
E    2
E 
E    2
• When n=4m+1 or 4m+3, the system has a singly occupied
HOMO, being radicaloid.
• When n=4m, the system has two degenerate singly-occupied
HOMOs (non-bonding), thus being diradicaloid.
• Only when n= 4m+2 can a cyclically conjugated system has fully
occupied HOMOs and be chemically stable. (fulfilling the Hückel
rule of aromaticity!)
§5.5 Graphical method to predefine the
coefficients of HMOs for conjugated systems
Background
•
The AO coefficients of HMOs for p-conjugated organic
systems correlate with their geometries, displaying “quasi-
 1  0.3721  0.6022  0.6023  0.3724
e.g., 1,3-butadiene  2  0.6021  0.3722  0.3723  0.6024
 3  0.6021  0.3722  0.3723  0.6024
 4  0.3721  0.6022  0.6023  0.3724
periodicity”.
•
Accordingly, a graphical method was developed by Q.E.
Zhang et al. to predefine the AO coefficients of HMOs for
such type of systems.
n
5.5.1 Principle for linear conjugated [n]polyenes.
   Cii
i 1
According to the Variation theorem, we have secular equations:
C1(-E) + C2=0;
C1 + C2(-E) + C3 = 0
C2 + C3(-E) + C4 = 0;
……
Ck-1 + Ck(-E) + Ck+1 = 0 ; ……;
whose matrix form is

0
  E

E 
 
 





 
 0
0
0

 0
0
0

0
0



0
0


E
0

Cn-1 + Cn(-E) = 0
 C1 


  C2 
  
0

  
  Cn 1 




  E  Cn 
0
0


let
x  (  E ) /    2cos  E    x    2 cos
-2cos C1+C2=0
C1(-E)+C2=0
C1+C2(-E)+C3 =0
C2+C3(-E)+C4=0
……
Ck-1+Ck(-E)+Ck+1=0
 -E

 -2cos
C1 - 2cos C2+C3 =0
C2 - 2cos C3+C4=0
……
Cn-1 - 2cos Cn=0
……
Cn-1+Cn(-E)=0
C2 = 2cos C1,
Ck+1  Ck-1 = 2cos Ck
Cn-1 = 2cos Cn
(cyclic formula)
C3 = 2cos C2  C1
… , Ck+1 = 2cos Ck  Ck-1 , …
For a linear [n]polyene, Cn+1 = 0, C0 = 0)
Ck 1  Ck -1  2cosCk  sin1  sin 2  2 sin
C 2  2cosC1 and
let
1   2
2
cos
1   2
2
C k 1  2cosC k - C k-1
C1  sin
C 2  2cossin  sin2
C 3  2 cos C2  C1  2 cos  sin 2  sin   sin3
C 4  2 cos  sin 3  sin 2  sin4 ,
.......
thus :
C1  sin , C 2  sin2 , C 3  sin3 ,..., Ck  sin k ,..., C n  sinn
Ci  sin  sin 2 sin 3
sin( n  1) sin n
Thus, get   get E = +2cos and {Ck} simultaneously!
So we don’t have to solve algebraically the n-order secular
determinant!
Ci 
sin  sin 2 sin 3
sin( n  1)  sin n
 sin n  0 & sin( n  1 )  0 (boundary condition)
mp
( n  1 )  mp   
 p ( m  1,2,3,..., n)
n 1
mp
Higher m,
 Em    2 cos
(m  1,2,3,..., n)
higher E.
n 1
kmp
2 n
kmp
 m  A k sin(
)
k sin(
) (m  1,2,3,..., n)

n 1
n  1 k 1
n 1
k 1
n
Normalization coefficient
• For linear polyenes having even orbitals，there are n/2 bonding
orbitals and n/2 antibonding orbitals.
• For linear polyenes having odd orbitals，there are (n-1)/2 bonding
orbitals, (n-1)/2 antibonding orbitals and a non-bonding orbital.
Frontier MOs of [n]polyene: CnHn+2
mp
2 n
kmp
Em    2 cos
, m 

sin
(
) (m  1,2,3,..., n)

k
n 1
n  1 k 1
n 1
a) When n=odd, SOMO, m = (n+1)/2,  SOMO  p / 2, ESOMO  
 SOMO  A(1  3  5  ...) Non-bonding!
b) When n=even, HOMO, m= n/2; LUMO, m=(n+2)/2.
 HOMO
np
p
p
( n  2 )p p
p

 
 LUMO 
 
2(n  1) 2 2( n  1 )
2( n  1 ) 2 2( n  1 )
Coefficients of AO’s:
HOMO
LUMO
n = 4l+2:
Cn  C1 , Cn 1  C2 , ...
Cn  C1 , Cn 1  C2 ,...
n = 4l:
Cn  C1 , Cn 1  C2 ,... Cn  C1 , Cn 1  C2 , ...
which reflect the symmetry of HOMO and LUMO!
Example: butadiene
p  MOs :
(n=4)
4
  A cii
sin  sin 2 sin 3 sin 4 
i 1
 sin 5  0 & sin   0,
m
1

p/5
2 cos  1.618
E
  1.618
  mπ / 5 (m  1,2,3,4)
2
2p/5
0.618
  0.618
3
3p/5
- 0.618
 - 0.618
4
4p/5
- 1.618
 - 1.618
2
mp
2mp
3mp
4mp
 m
(sin
1  sin
2  sin
3  sin
4 )
5
5
5
5
5
 1  0.3721  0.6022  0.6023  0.3724
 2  0.6021  0.3722  0.3723  0.6024 HOMO
 3  0.6021  0.3722  0.3723  0.6024 LUMO
 4  0.3721  0.6022  0.6023  0.3724
E1    1.618 ,  1  0.3721  0.6022  0.6023  0.3724 symmetric
E2    0.618 , 2  0.6021  0.3722  0.3723  0.6024 asymmetric
E3    0.618 , 3  0.6021  0.3722  0.3723  0.6024 symmetric
E4    1.618 , 4  0.3721  0.6022  0.6023  0.3724 asymmetric
Major sym. element :
trans-butadiene: C2
cis-butadiene: CS
Note: These p–MOs are either
symmetric or asymmetric with
respect to the C2 (or CS)
operation!
Can we obtain the p–MOs
based on their symmetries?
5.5.2 Symmetry classification
a. [n]polyenes with even-number carbon atoms: CnHn+2
(n/2)
1
(n/2)
2
Symmetric MOs:
1
2
 C1  C1'
C2  C2' ...,
Cn / 2  C( n / 2 )'
3
3
(n  1)
1
1
(n  1)

cos
 cos  cos  , cos  cos  cos
2
2
2
2
2
2
& Ck 1  Ck 1  2Ck cos 
（Cyclic formula）
Let coefficients of central atoms (1 & 1) be cos(  / 2 )
3


3


 C2  C2'  2 cos cos   cos  (cos  cos )  cos  cos
2
2
2
2
2
2
3

5


5
C3  C3'  2 cos cos   cos  (cos  cos )  cos  cos
2
2
2
2
2
2
..., C( n / 2 )  C( n / 2 )'
( n  1)
 cos
2
Symmetric MOs:
1
1
n 1
n3
n3
n 1

 cos  , cos  cos
 cos
cos
 cos
2
2
2
2
2
2
(n/2)
1
1
(n/2)
Boundary condition: cos n  1   0
2
n 1
2m  1
2m  1
n


p ( m  0 ,1,2 ,...,  )   
p
2
2
n 1
2
2m  1
n
sym
 E m    2 cos
p ( m  0 ,1,2 ,...,  )
n 1
2
A total of n/2 values
Asymmetric MOs:
& Ck 1  Ck 1  2Ck cos 
 C1  -C1 , C2  C2 ,...,
Let coefficients for central atoms be  sin(  / 2 ), sin(  / 2 )
n3
n 1
Ci  sin n  1  ,  sin n  3 
sin
 , sin

2
2
2
2
Then coefficients for terminal atoms are  sin n  1  and sin n  1 
2
2
The boundary condition:
n 1
n 1
sin
 0 
  mp
2
2
2mp
n
A total of n/2 values.
 
(m  1,2,..., )
n 1
2
2mp
n
 E    2  cos
(m  1,2,..., )
n 1
2
2m  1
n
p (m  0,1,2,...,  )
Symmetric MOs Esym    2 cos
n 1
2
2m
n
E



2

cos
p
(
m

1
,
2
,...,
)
Asymmetric MOs asym
n 1
2
Thus, the lowest n/2 MOs (<p/2) are bonding and doubly occupied!
n= 4k
n/2= 2k
(n+2)/2= 2k+1
n= 4k+2
n/2= 2k+1
(n+2)/2= 2k+2
LUMO
Symm.
Asymm.
HOMO
Asymm.
Symm.
…
E
(n+2)/2
E(n+2)/2
E4
E3
E2
E1
…
…
En/2
n/2
4
3
2
1
Asymm.
Symm.
Asymm. • Even-numbered MO’s are asymmetric.
Symm. • Odd-numbered MO’s are symmetric.
2m  1
n
p (m  0,1,2,...,  )
Symmetric MOs Esym    2 cos
n 1
2
2m
n
E



2

cos
p
(
m

1
,
2
,...,
)
Asymmetric MOs asym
n 1
2
Thus, the lowest n/2 MOs (<p/2) are bonding and doubly occupied!

n = 4k FMO symmetry
Asymm.
n
n
p
m
4 2( n  1)
Symm.
n n2
p
m
4 2(n  1)
HOMO
Symm.
n2
m
4
LUMO
Asymm.
m
HOMO
LUMO
E
E asym
n
   2 cos
p
2( n  1)
Esym
n2
   2 cos
p
2(n  1)
n = 4k+2
n2
4
n
n
p Esym    2 cos
p
2( n  1)
2(n  1)
n2
p E asym    2 cos n  2 p
2(n  1)
2( n  1)
C2 or v
Example: hexatriene n=6
Symmetric MO
Asymmetric
cos
5
( )
2
3
( )
2
1
( )
2
1
( )
2
3
( )
2
5
( )
2
5
[ ]
2
3
[ ]
2
1
[ ]
2
1
[ ]
2
3
[ ]
2
5
[ ]
2
sin
7
2m  1
cos


0



p (m  0,1,2)
Symmetric MO:
2
7
m  0,1 ( E  α ,  bonding)
2m  1
E    2 cos
p ( m  0,1,2)
m2
( E  α, anti - bonding)
7
Asymmetric MO: sin 7   0    2m p (m  1,2,3)
2
7
2m
E    2 cos
p
7
( m  1,2,3)
m 1
m  2,3
bonding
anti - bonding
Three doubly occupied MOs, two sym. + one asym.
b. [n]polyenes with odd-number carbon atoms
Symmetric MOs:
1
cos
n 1
 cos n  3 
2
2
0
1
[(n-1)/2]
cos  , 11, cos 
n 1
n3

cos
 cos
2
2
[(n-1)/2]
 Ck 1  Ck 1  2Ck cos  & C0  1, C1  C1' ,..., C [( n 1 ) / 2 ]  C [( n 1 ) / 2 ]'
 C1  C1'  cos   C2  2C1 cos   1  cos 2
 ..., C( n 3 ) / 2
n 1
n3
 cos

  C( n 1 ) / 2  cos
2
2
Boundary conditions:
2m  1
n 1
n 1
n 1
2m  1
p ( m  0 ,1,2 ,...,
)
 0 

p  
n 1
2
2
2
2
2m  1
n 1
 E sym    2  cos     2 cos
p ( m  0 ,1,2 ,...,
)
n 1
2
cos
Asymmetric MOs:
sin
[(n-1)/2]
1
n 1
n3
 sin

2
2
sin , 0,
0 - sin  sin
0
[(n-1)/2]
1
n3
n 1
  sin

2
2
 Ck 1  Ck 1  2Ck cos  & C0  0 , C1  -C1' ,..., C [( n 1 ) / 2 ]  -C [( n 1 ) / 2 ]'
Let C1  -C1  sin  C2  2C1 cos   sin2
...  C [( n 3 ) / 2 ]  sin[(n - 3) / 2 ]  -C [( n 3 ) / 2 ] 
 C [( n 1 ) / 2 ]  sin[(n - 1) / 2 ]  -C [( n 1 ) / 2 ] 
Boundary condition: sin n  1  0    2m p ( m  1,2 ,..., n  1 )
n 1
2
2
E asym
2m
   2  cos     2 cos
p
n 1
n 1
( m  1,2 ,...,
)
2
MOs of [n]polyenes with odd-number carbon atoms
…
2m
n 1
p (m  1,2,...,
)
Asymmetric MO’s: E asym    2 cos
n 1
2
Symmetric MO’s: Esym    2 cos 2m  1 p ( m  0,1,2,..., n  1)
n 1
2
• Even-numbered MO’s are asymmetric.
• Odd-numbered MO’s are symmetric.
E
ne ,  (n-1)/2 MOs doubly occupied, and one
E(n+3)/2
(n+3)/2
(n+1)/2
(n-1)/2
E(n+1)/2
non-bonding SOMO!
…
…
E(n-1)/2
E2
E1
p
n = 4k+1
SOMO
2 Asym.
1 Sym.
(n+1)/2= 2k+1
Sym.
 = p/2
ESOMO = 
n = 4k+3
(n+1)/2= 2k+2
Asym.
 = p/2
ESOMO = 
FMOs of [n]polyenes with odd-number carbon atoms
n 1
n3
cos
 , cos
 cos , 1, cos 
2
2
Symmetric:
n 1
n3
θ , - sin
θ  sin θ , 0, sinθ
Asymmetric:  sin
2
2
n3
n 1
cos
 , cos

2
2
- sin
n3
n 1
θ , - sin
θ
2
2
nep,  (n-1)/2 MOs doubly occupied, and one non-bonding SOMO!
SOMO
n = 4k+1
symmetric
n = 4k+3
asymmetric
E  α  2 β cos θ   p / 2, E  
  p / 2, E  
  p / 2, E  
n
ψ SOMO  A ( φ1  φ3  φ5  ...)
k 1
Simplified diagram of SOMO:
n=4k+1
n=4k+3
Example: ( 2)
n=5
[2]
(1)
( 0)
(1)
( 2)
[1]
[ 0]
[1]
[2] asymmetric
symmetric
2m  1
cos3  0   
p (m  0,1,2)
6
m  0，  p/6， Es1    1.732 
2m  1
E  α  2 β cos
π
m  1，  3p/6，Es 2  
6
(m  0,1,2 )
m  2，  5p/6，Es 3    1.732 
Symmetric MOs:
1 1
3
3
1
Ψs1 
( φ1 
φ2  φ3 
φ4  φ5 ) Bonding
3 2
2
2
2
1
Ψs 2 
( φ1  φ3  φ5 )
3
Non-bonding
1 1
3
3
1
Ψs 3 
( φ1 
φ2  φ3 
φ4  φ5 ) anti-bonding
3 2
2
2
2
Example: ( 2)
n=5
[2]
Asymmetric MOs:
(1)
( 0)
(1)
( 2)
[1]
[ 0]
[1]
[2] asymmetric
sin 3  0 & sin   0   
m
E    2 cos p
3
symmetric
mp
( m  1,2)
3
( m  1,2)
m  1,  p / 3, Eas1    
bonding
m  2 ,  2p / 3, Eas 2     anti - bonding
1
Ψas1  ( φ1  φ2  φ4  φ5 ) Bonding
2
1
Ψas 2  ( φ1  φ2  φ4  φ5 ) Anti-bonding
2
Example: ( 2)
n=5
[2]
(1)
( 0)
(1)
( 2)
[1]
[ 0]
[1]
[2] asymmetric
symmetric
1 1
3
3
1
s 3 
( 1 
2  3  4  5 )
3 2
2
2
2
  3
1
as 2  ( 1  2  4  5 )
2
-
1
s 2 
( 1  3  5 )
3
1
as1  ( 1  2  4  5 )
2
1 1
3
3
1
s1 
( 1 
2  3  4  5 )
3 2
2
2
2


  3
c. Cyclic conjugated molecules
Example 1： benzene ---solution I (n=6 !)
Boundary condition:
2
3
( )
2
i) Symmetric MOs
2k  1
Ck  Ck '  cos

2
1
( )
2
n
(k  1,2,.. )
2
3
4
1
5
( )
2
2 cos Ck C k 1Ck 1
1
( )
2
6
5
v
5
( )
2
2 cos C6 C 5C1
5
5
3
2 cos  cos   cos   cos  
2
2
2
7
5

cos   cos   0   2 sin 3 sin  0
2
2
2
 sin 3  0    mp / 3 (m  0,1,2)
 2 cos   2, 1,  1
3
( )
2
θ
cos θ
2 cos θ
E
θ
0
0 ,
1
2
α  2β
60 ,
1/2
1
α β
120
- 1/2
-1
α- β
E

α  2β
ψ1  N ( φ1  φ2  φ3  φ4  φ5  φ6 )

60
α β
1
( )
2
2
3
( )
2
4
1
5
( )
2
6
1
( φ1  φ2  φ3  φ4  φ5  φ6 )
6
ψ2  N ( 0 
3
1
( )
2
5
3
( )
2
5
( )
2
Bonding
3
3
3
3
φ2 
φ3  0 
φ5 
φ6 )
2
2
2
2
1
 ( φ2  φ3  φ5  φ6 )
2
1
1
1
1
ψ  N ( φ1  φ2  φ3  φ4  φ5  φ6 )
120 α  β 3
2
2
2
2
1

( 2φ1  φ2  φ3  2φ4  φ5  φ6 )
6
Bonding
Antibonding
ii) Asymmetric MOs
Boundary condition： 2 cos Ck C k 1Ck 1
5
3
5
2 cos  sin   sin   sin 
2
2
2
7
3
3
5
sin   sin   sin   sin 
2
2
2
2
7
5

sin   sin   0  sin 3 cos  0
2
2
2
m
 sin 3  0    p (m  1,2,3)
3
1
[ ]
2
3
[ ]
2
3
2
4
1
5
[ ]
2
1
[ ]
2
5
6
3
[ ]
2
5
[ ]
2
v

60 ,
120 ,
cos  1/2
- 1/2
2 cos  1
-1
E
   
4
5
180
-1
-2
 - 2
6
θ
60
E
α β
120 α  β
180 α  2 β
1
(21  2  3  24  5  6 ) Bonding
6
1
 5  (2  3  5  6 )
Anti-bonding
2
1
6 
(1  2  3  4  5  6 )
Anti-bonding
6
E
6
 -2
4 
1
[ ]
2
5
3
2
4
1
 -
3
[ ]
2



 +
 +2
2
3
4
1
5
[ ]
2
1
[ ]
2
6
5
5
[ ]
2
3
[ ]
2
Example 2： benzene ---solution II (n= 7! )
i) Symmetric MOs
Boundary condition:
2 cos Ck C k 1Ck 1
(0)
(1) 6
v
1
2 (1)
 2 cos  cos 3  cos 2  cos 2
3
cos 4  cos 2  2 cos 2  cos 4 - cos 2  0 (2) 5
( 2)
4
  sin 3 sin  0  sin 3  0
(3) (3)
m
7
   p ( m  0 ,1,2 ,3  )
Similar to the linear
3
2
polyene (n=7) case
by supposing the

0 ,
60
120
180
two terminal atoms
cos  1
1/2
- 1/2
-1
overlapping!
2 cos  2
1
E
  2   
1
2
bonding
-1
 -
3
-2
  2
4
anti - bonding
ii) Asymmetric MOs
 sin 3   sin 3 and sin   0
Question: Why do
m
we exclude the
 sin 3  0    p (m  1,2  7/2)
case of m = 3?
3

60 ,
cos 1/2
2 cos 1
E

1
bonding
120
- 1/2
-1
 -
2
anti - bonding
Please derive the MOs after class!
v
[ 0]
1
[1] 6
2 [1]
5
[2]
3 [ 2]
4
[3] [3]
Complicated p-conjugation: e.g., benzyl radical
How to get the coefficient of atom 7?
1) Symmetric MOs:
The secular equation centred at atom 6 :
C6 (   E )  C4   C5   C7   0
let - 2cos  (   E ) / 
C7  2 cos C6  C4  C5
 2 cos  cos 3  2 cos 2
 cos 4  cos 2
Boundary condition: 2 cos C  C 
7
6
2 cos  (cos 4  cos 2 )  cos 3  2 cos   0,  (3  2 )1 / 2
2) Asymmetric MOs:
The coefficients of atoms 1,6 and 7 that are located in the reflection
plane should be 0. Then sin 3  0    mp / 3 (m  1,2)
 2 cos  1
Remarks on HMO treatment of system with complicated pconjugation,
1) The use of symmetry can simplify the process. e.g., naphthalene
2) Secular equations pertaining to the multiply connected atoms
should be carefully considered!
Example: Please derive the HMOs of trimethylenemethane (44)
without directly solving its secular determinant.
§5.6 Symmetry of molecular orbital and
symmetry rules for molecular reactions
Brief introduction:
• Frontier Molecular orbital (FMO) Theory proposed by K. Fukui in
early 1950s.
• The Principle of Orbital Symmetry Conservation proposed by R.B.
Woodward (1965 Nobel Prize) and R. Hoffmann in 1965.
• Owing to their aforementioned contributions, Fukui and Hoffmann
were awarded Nobel Prize in Chemistry in 1981.
5.6.1
Frontier molecular orbitals Theory (1951)
• A deeper understanding of chemical reactivity can be gained by
focusing on the frontier orbitals of the reactants.
• We need to consider only two frontier molecular orbitals (FMOs),
the HOMO and LUMO, to predict the structure of the product.
Highest occupied molecular orbital (HOMO)
Lowest unoccupied molecular orbital (LUMO)
•The FMO theory can be regarded as a natural extension of
MO theory that facilitates us to obtain qualitative
understanding of chemical reactivity.
Reaction condition
1. During the course of a chemical reaction, the orbitals that are most
readily accessible for interaction are the frontier orbitals.
2. When two molecules approach each other, the symmetry of the
HOMO of one molecule must be compatible with that of the
LUMO of the other molecule, i.e. orbitals with the same sign
will overlap. This forms a transition state which is relatively stable
and is a symmetry-allowed state.
3. The energy levels of the interacting HOMO and LUMO must be
comparable ( < 6 eV).
4. When the HOMO and LUMO of two molecules overlap, electrons
are transferred from the HOMO of one molecule to the LUMO of
the other molecule. The direction of the transfer should be in line
with the electronegativities and be consistent with the weakening
of the original bond.
H 2  I 2  2HI
example A :
H H
H
e
H
2HI
I
I
I
I
H 2 : ( 1s ) 2 ( 1s *)0
Symmetryincompatible
overlap=0
I 2 : ( 5 p ) 2 (p 5 p ) 4 (p 5 p *)4 ( 5 p *)0
LUMO(H2)
e
HI
e
 2I
I  H 2  HI  H
 HI
H 2  I 2  2 HI
（b）
HOMO(I2)
Electronegativityincompatible
Real mechanism
I2
HOMO(H2)
（a）
LUMO(I2)
I
e
H2
I
H 2 : ( σ1s )2 ( σ1s*)0
I 2 : ( σ5 p )2 ( π5 p )4 ( π5 p*)4 ( σ5 p*)0
Such a three-step
mechanism meets
all requirements!
Example B：C2H4 + H2  C2H6
C2H4
+
+
+
-
-
-
-
+
p
HOMO
H 2 : ( 1g ) 2 ( 1u ) 0
-
p*
LUMO
HOMO
Without catalyst
LUMO
HOMO
+
-
-
+
+
LUMO
With TM catalyst
+
+
C2H4
-
-
HOMO
H2
-
+
LUMO
Overlap=0
incompatible in symmetry
+
-
-
+
-
+
-
+
C2H4
H2
e
Example C,
N 2  O2  2 NO
×
*
N 2 : KK ( 1σ g )2 ( 1σ u )2 ( 1πu )4 ( 2σ g )2 ( 1π g*)0
*
O2 : KK ( σ 2 s )2 ( σ 2 s )2 ( σ 2 pz )2 ( π 2 px )2 ( π 2 py )2 ( π 2 px*)1( π 2 py*)1
HOMO (N2)
SOMO (O2)
overlap=0
Incompatible
symmetry!
Though both N2 and O2 are abundant in
the atmosphere of the Earth, no such
reaction can occur without catalyst!
LUMO (N2)
e
SOMO (O2)
Unreasonable
electron transfer!
Example D: C4 H 6  C2 H 4
Diels-Alder Reaction
A Nobel-prize-wining Rxn!
[4+2]-p-electrons

Transition state
• We can illustrate the HOMO-LUMO interactions by way of the
Diels-Alder reaction between ethylene and 1,3-butadiene.
• We need consider only the p electrons of ethylene and 1,3butadiene, and ignore the framework of  bonds in each
reactants.
p-MOs of 1,3-Butadiene
p-MOs of ethylene
LUMO
LUMO
HOMO
HOMO
v
v
Correlation of FMOs

v
v
LUMO
e
HOMO
HOMO
e
LUMO
FMOs of two reactants can overlap effectively!
• Symmetry-allowed and thermally activated.
Example E:
A thermally "forbidden" reaction
H2C
H2C
+
CH2
CH2
• The dimerization of ethylene to give cyclobutane does
not occur under conditions of typical Diels-Alder
reactions(heating). i.e., it’s thermally forbidden.
• Why?
A thermally "forbidden" reaction
2H2C
CH2
• HOMO-LUMO mismatch of
two ethylene molecules in
their ground state precludes
single-step formation of two
new  bonds!
The addition reaction is thermally forbidden, but allowed
between two photo-excited ethenes.
Example F: 2,4,6-Octatriene Thermal Electrocyclic Reactions
p-MOs
E
v
6
5
4
3
LUMO
+
+
-
v
HOMO
disrotary
v
2
1
This reaction is thermally allowed with the
stereochemistry of its product governed by
the symmetry-allowed disrotary.
-
+
-
Photochemical Electrocyclic Reactions
Example G: 2,4,6-octatriene
6
p-MOs
C2
5
4
LUMO
3
HOMO
hv
HOMO
C2
conrotary
LUMO
2
1
This photochemical cyclic reaction occurs with the stereochemistry
of its product being governed by the symmetry-allowed conrotary.
Stereochemical Rules for the Electrocyclic Reactions
of [n]polyenes (n =even)
Number of
Electron pairs
(double bonds)
Thermal Reaction
Photochemical
Reaction
Even (i.e., n= 4k)
conrotatory
disrotatory
Odd (i.e., n= 4k+2 )
disrotatory
conrotatory
Why ?
[n]Polyenes with even-number pp-orbitals.
n 1
n3
cos
θ , cos
θ
2
2
Symm.
θ
θ
cos , cos
2
2
Upon v
Asym.
v
n 1
n3


 sin
 ,  sin
  sin , sin
2
2
2
2
HOMO
v / C2
n=4k (k=1,2,,..): Asym / Sym
Nature of FMOs
n=4k+2 (k=1,..): Sym / Asym
cos
n3
n 1
θ , cos
θ
2
2
n3
n 1
sin
 , sin

2
2
LUMO
v / C2
Sym / Asym
Asym / Sym
Thermal reaction
(HOMO-based)
Photo-reaction
(LUMO-based)
n=4k
Conrotary (C2)
Disrotary (v)
n=4k+2
Disrotary (v)
Conrotary (C2)
Stereochemical rules for electrocyclic reactions
[1,n] sigmatropic shift : stereochemical rules
[1,5]--shift
[1,7]--shift
Suppose the transition state of such reaction is a combination of H
atom and a [n]polyene with odd-number pp orbitals.
[n]Polyenes: n=odd
ci
   cii
sin( n  1) , sin n
sin  , sin 2 ....
  mp/(n  1); E    2 cos (m  1,2,....n)
i
If n =odd, SOMO, m = (n+1)/2,  somo  p / 2, Esomo  
c1  sin( p / 2)  1; c2  0; c3  -1; ..., cn  sin( np / 2)
n
  SOMO  A (1  3  5  ...) 
k 1
Sigmatropic shift
Symmetry of SOMO
v
C2
TS symmetry
Mode
n= 4k+1
Sym. Asym.
v
Suprafacial
n= 4k+3
Asym. Sym.
C2
Antarafacial
Note that the H-transfer demands both ends of SOMO of
[n]polyene (n=odd) to overlap effectively with the H 1s orbital.
• The sigmatropic H-transfer demands both ends of SOMO of such
[n]polyene (n=odd) can overlap with the H 1s orbital. i.e.,
n=4k+1
Suprafacial mode
n=4k+3
antarafacial mode
TS requires CS-symmetry!
TS requires C2-symmetry!
QM-predicted transition states of [1,n]-sigmatropic shift
[1,5]-sigmatropic shift
[1,7]-sigmatropic shift
2. Woodward-Hoffmann rules --- principle of orbital
symmetry conservation ( 分子轨道对称性守恒原理)
Proposed originally to predict the stereochemistry of
pericyclic reactions based on orbital symmetry.
i. There is a one to one correspondence between the MO’s of
the reactant and the product.
ii. The symmetry of the correlated orbitals is the same.
iii. The correlated orbitals should have comparable energies.
iv. The correlation lines for orbitals with the same symmetry do
not intersect.
Orbital Symmetry Conservation
Example A:
Electrocyclic Reactions of butadiene
p MOs of butadiene
Orbital Symmetry Conservation
v
C2
A
S
S
A
LUMO
A
S
HOMO
S
A
Orbital Symmetry Conservation
v
v
v
conserved
B
A
A
B
A
A
Disrotary, h
C2
C2
C2
conserved
B
A
A
B
B
B Conrotary, 
B
A
B
A
1. During the disrotatory cyclic
reaction, the systems have a
common v symmetry.
2. Thus the symmetries of all
involved molecular orbitals
subject to this operation
should be maintained
before/after the reaction.
3. In this regard, the HOMO
and LUMO of the reactant
are correlated with the
LUMO and HOMO of the
product by conservation of
orbital symmetry.
Energy/orbital correlation diagram 4. Thus the reaction should be
photo-promoted.
1. During the conrotatory cyclic
reaction, the C2 symmetry is
conserved.
2. The symmetries of all
involved molecular orbitals
subject to this operator
should be maintained
before/after the reaction.
3. The HOMO and HOMO-1 of
the reactant are correlated
with the HOMO-1 and
HOMO of the product. So
do the unoccupied MO’s.
Energy correlation diagram
4. Thus the reaction occurs for
the reactant at its ground
state.
v
C2
 4S
A * 4 A
A*
 3A
S p * 3 S
Ap
 2S
A
p
2 A
S
1A
S

1 S
S
Conrotary mode
Thermal reaction
Disrotary mode
Photochemical reaction
The MO correlation diagrams of (a) conrotary and (b) disrotary
cyclic reactions of butadiene to cyclobutene
p
*
Example B:
Cycloaddition of ethylene:
a photochemical reaction
2H2C
CH2
y
z
reactants
xz
x
product
yz
x
In photochemical
reaction mode
• Conserved symm.
Operators: xz and yz.
• The HOMO and
LUMO of reactants are
correlated with the
xz
LUMO and HOMO of
yz
the product, respectively.
• Thus excited states of
the reactants should be
involved for the reaction
to occur.
FMO theory
Symmetry of MOs
vs.
Orbital symmetry
conservation principle
• Viability of reactions
• Stereochemical rules of reactions.
• FMOs of reactants
• All correlated valence MOs
of reactants and products
• Symmetry of the FMOs
• Need to figure out the
symmetry operators conserved
in the reactions.
determines everything.
• Simple, and widely applicable
• Useful, yet a bit complicated
The reaction of [Et4N]I and tetracyanoethylene (TCNE) forms
[Et4N]2[TCNE]2, which possesses [TCNE]22 with an intradimer
C-C bond distance of 2.827(3) Å (CrystEngComm, 2001, 47,1).
Please analyze the unusual intradimer bonding.
 The LUMO of neutral TCNE is the p*
MO of the C=C moiety.
 The SOMO of [TCNE]:
 So the SOMOs of two [TCNE] are
symmetry-compatible and can
effectively interact with each other!
Spin conservation
• A chemical reaction occurs with conservation
of spin.

g

g
2 H 2 ( Σ )  O2 ( Σ ) 

2 H 2O( A1 )
Δ
×
1
3
1
C2 H 4 ( s  0 )  C4 H 6 ( s  0 ) 

C6 H10 ( s  0 )
Δ
• Please make use of the FMO theory to assess
whether the ene reaction can occur
thermally?
Hint: In the transition state of this reaction, the propene can be
regarded as an allylic anion and a proton.
Symmetry allowed!
Summary of Chapter 5
Key points/concepts
1. Hybrid orbital theory and VSEPR
(Mostly Qualitative Theory/model!)
2. HMO, HMO treatment of p-conjugated systems
3. Graphical method to predefine coefficients of HMO of pconjugated systems.
(Semi-quantitative Theory)
4. Symmetry rules for molecular reactions.
(Qualitative Theory)
Spatial Arrangements of Electron Pairs in
Terminal Atoms in ABn
..
..
cos(   )  cos  cos   sin  sin 
sin(   )  sin  cos   cos  sin 


cos   cos   2 cos
cos
2
2


cos   cos   2 sin
sin
2
2


sin   sin   2 sin
cos
2
2


sin   sin   2 cos
sin
2
2
[n]Polyenes: Nature of their FMOs
Ci
sin( n  1) , sin n
sin  , sin 2 ....
  mp/(n  1); E    2 cos (m  1,2,....n)
nep, bonding MOs, p/2 ; If n being an even number,
HOMO: m = n/2  np / 2( n  1)
LUMO: m = (n/2)+1   ( n  2)p / 2(n  1)
Nature of HOMO c1  sin   sin[ np / 2( n  1)]  cos[p / 2( n  1)]
n

n2
p
p
cn  sin n  sin
p  sin  p  [ 
]
2( n  1)
2 2( n  1) 
2
n

p
p
p
c

sin
p

[

]


cos
 c1 Asymmetric
If n= 4k n


2 2(n  1) 
2(n  1)
2
If n= 4k+2 c  sin  n p  [ p 

n
2
symmetric
2
p

p
]  cos
 c1 symmetric
2(n  1) 
2(n  1)
asymmetric
[n]Polyenes: Nature of their FMOs
sin( n  1) , sin n
sin  , sin 2 ....
Ci
  mp/(n  1); E    2 cos (m  1,2,....n)
LUMO
m = (n/2)+1
Nature of LUMO
  (n  2)p / 2(n  1)
c1  sin   sin[( n  2)p / 2( n  1)]  cos[p / 2( n  1)]
n

n(n  2)
p
p
cn  sin n  sin
p  sin  p  [ 
]
2( n  1)
2 2( n  1) 
2
If n= 4k
 np p

p
p
cn  sin 
[ 
]  cos
 c1
symmetric
2 2(n  1) 
2(n  1)
 2
If n= 4k+2 cn  sin  n p  [ p 

2
symmetric
2
p

p
]   cos
 c1 asymmetric
2(n  1) 
2(n  1)
asymmetric
[n]Polyenes: Nature of their FMOs
sin  , sin 2 ....
Ci
sin( n  1) , sin n
  mp/(n  1); E    2 cos (m  1,2,....n)
nep, bonding MOs, p/2 ; If n being an odd number,
SOMO m = (n+1)/2   p / 2
Nature of SOMO c1  sin   sin( p / 2)  1; c 2  0; c3  -1;
If n= 4k+1
If n= 4k+3
n
..., cn  sin n  sin p
2
np
cn  sin
 sin( 2kp  p / 2)  1  c1 symmetric
2
np
cn  sin
 sin( 2kp  3p / 2)  1  c1 asymmetric
2
symmetric
asymmetric
HMO Treatment: Linear [n]polyenes (CnHn+2)
n
   cii
i 1
Trial p-MO

n
n

...
0  c1 
  E
 
Variation  
  E ...
0  c2 

0




Theorem
...
...
...
...
...

 
...
   E  cn 
 0
• The n MOs differ from each other
in energy (non-degenerate MOs!).
• The lowest n/2 (n = even) or
(n+1)/2 (n= odd) MOs are
occupied.
{Ej}, {ci(j)}
(j =1, 2, …, n)
n
 ( j )   ci ( j )i
i 1
• For linear polyenes having even orbitals，there are n/2 bonding
orbitals and n/2 antibonding orbitals.
• For linear polyenes having odd orbitals，there are (n-1)/2 bonding
orbitals, (n-1)/2 antibonding orbitals and a non-bonding orbital.
e.g., Hexatriene - 66
4
  A cii
i 1
sin  sin 2 sin 3 sin 4 
 sin 7  0 & sin   0;
sin 5 sin 6
mp
 
(m  1,2,...,6)
7
mp
Em    2  cos
( m  1,2,...,6) m= 3, HOMO, bonding!
7
m=4, LUMO, antibonding!
2 6
Kmp
 m＝  (sin
)K
7 K 1
7