Download Hydroperoxide ion P.9 is much less basic than hydroxide ion P.10

No 1 (2016.5.06)
有機反応化学 （ Advanced Organic Reaction)（H28年度前期�古田）
Text: Molecular Orbitals and Organic Chemical Reactions
Ian Fleming (Wiley, 2009)
1. Chap.1: Molecular Orbital Theory (5/6)
2. Chap.2: Molecular Orbitals and the Structures of Organic Molecules (5/13)
[Question]
Hydroperoxide ion P.9 is much less basic than
hydroxide ion P.10. Why, then, is it so much more
nucleophilic?
3. Chap.3: Chemical Reactions (5/20)
4. Chap.4: Ionic Reaction- Reactivity (5/27)
5. Chap.5: Ionic Reaction- Stereochemistry (6/03, 10)
6. Chap.6: Thermal Pericyclic Reactions (6/24)
7. Chap.7: Radical Reactions (7/15)
8. Chap.8: Photochemical Reactions (7/22)
6/17（集中講義）、7/01（休講）、08（休講）、補講日程未定
参考書：
大学院講義�有機化学I, II （東京化学同人）
�マーチ有機化学
Modern Physical Organic Chemistry (E.V.Anslyn & D. A. Dougherty)
有機反応機構の書き方（奥山�格：丸善）
�������演習で学ぶ有機反応機構（化学同人）
[Question]
Why does diazomethane P.15 add to methyl acrylate
P.16 to give the isomer P.17 in which the nitrogen end of
the dipole is bonded to the carbon atom bearing the
methoxycarbonyl group, and not the other way round
P.14?
1. Molecular Orbital Theory
1.1 The Atomic Orbitals of a Hydrogen Atom
φ: wave function
φ2dτ: the probability of finding the electron in the volume dτ
∫φ2dτ = 1
E1s = -13.60 eV
1.4 Å ~ 90 % , 2 Å ~ 99%.
Pictureofelectrondistribution
The van der Waals radius at 1.2 A has no theoretical
significance-it is an empirical measurement from solid
state structures, being one-half of the distance apart of
the hydrogen atom in a C-H bond and the hydrogen atom
in the C-H bond of an adjacent molecule.
Bohr radius
0.529 Å
H
N
P
O
S
F
Cl
Br
I
(Å)
1.2
1.5
1.9
1.40
1.85
1.35
1.80
1.95
2.15
1.2 Molecules Made from Hydrogen Atoms
1.2.1 The H2 Molecule
σ = c1φ1 + c2φ2
1.1
σ: the molecular orbital
φ1: the atomic 1s wave function on atom 1
φ2: the atomic 1s wave function on atom 2
c1, c2 : a measure of the contribution which the atomic
orbital is making to the molecular orbital
σ2 = (c1φ1 + c2φ2)2 = (c1φ1)2 + (c2φ2)2 + 2c1φ1c2φ2
1.2
c1c2 > 0: the electron population between the two atoms is increased
c1c2 < 0: a low electron population in the space between the nuclei,
By making a bond between two hydrogen atoms, we create two new orbitals,
σ and σ*, which we call the molecular orbitals; the former is bonding and
the latter antibonding. A lowering of energy is equal to twice the value of Eσ.
antibonding
bonding
S12 = ∫φ1φ2dτ
S12: overlap integral
1.3
Energy of an electron in a bonding molecular orbital
Bonding (σ) and Antibonding (σ*) Orbitals
Energy of an electron in an antibonding molecular orbital
α: Coulomb integral
β: resonance integral
H: Hamiltonian
Coeficient (C1, C2)
In each orbital the sum of the squares of all the c–values must equal one,
since only one electron in each spin state can be in the orbital. Since |c1 |
must equal |c2 | in a homonuclear diatomic like H2, the values of c1 and c2
in the bonding orbital must be, namely 1/√2 = 0.707:
If all molecular orbitals were filled, then there would have to be one electron
in each spin state on each atom, and the sum of the squares of all the c values
on any one atom in all the molecular orbitals must also equal one. Thus the
σ*-antibonding orbital of hydrogen will have c-values of 0.707 and –0.707,
because these values make the whole set fit both criteria.
1.2.2 The H3 Molecule
Symmetry rule: combine the orbitals that have the same symmetry
Triangle H3
.
H3+,H3 stable
H3 unstable
Straight line H3
1.2.3 The H4 'Molecule'
(tetrahedral)
Symmetry-(xz, yz)
less bonding
the same energy as 1s
less bonding
Two H2 molecules do not combine to form an H4 molecule.
1.3 C-H and C-C Bonds
C: (1s)2(2s)2(2p)2
E2p = -10.7 eV
1.3. 1 The Atomic Orbitals of a Carbon Atom
C: (1s)2(2s)2(2p)2
E2s = -19.50 eV (vs. E1s = -13.60 eV for H)
node
0.9 Å
1.5 Å
Effective overlap
Atomic Orbital Energies
1.3.2 Methane (CH4)
Atomic Orbitals
H4 + C
Symmetry-(xz, yz)
Molecular Orbitals of CH4
1.3.3 Methylene (:CH2)
3 orbital interaction
H2 + C
Symmetry
xz
yz
two-fold z (H-C-H)
Linear combination of 3 atomic orbitals
Ic1I=Ic3I
bent form (more stable)
SSS
SSS
SSS
1.3.4 Hybridisation
Because of the presence of the inner sphere in the 2s orbital, the
nucleus is actually inside the back lobe, and a small proportion
of the front lobe reaches behind the nucleus
(Review)
CH4
1.3.5 C-Cσ Bonds and π Bonds: Ethane (C2H6)
Energy level and electron distribution
sp3 hybridization
C:8e
H:6e
Since C-C single bonds are typically about 1.54 Å long, the overlap
integral at this distance for π bonding is a little less than half that for
bonding. π bonds are therefore much weaker.
The C-C bond would have been seen as the bonding overlap of sp3 hybridised
orbitals on carbon with each other which would have used different
proportions of s and p orbitals, and would have been labelled sp3.
1.3.6 Ethylene: C=C π-bond
β = 140 kJ/mol (= 1.45 eV = 33 kcal/mol) => 2β = 280 kJ/mol
(π bond energy)
β
Electron in the Box
1.4 Conjugation-Hückel Theory
Allyl System
resonance
σ-framework
Orbital Energy
where k is the number of the atom along the sequence of n atoms
Place dummy atoms at both ends
From 3 orbitals to 4 orbitals?
NBMO
2 x 0.414β
(116 kJ/mol)
We cannot create four orbitals from three, because we cannot use the
p orbital separately twice.
1.4.2 Butadiene
Frontier Orbitals
HOMO: the highest occupied molecular orbital
LUMO: the lowest unoccupied molecular orbital
SOMO: the singly occupied molecular orbitals
allyl cation
LUMO:ψ2
HOMO:ψ1
LUMO
allyl anion
allyl radical
LUMO:ψ3*
HOMO:ψ2
SOMO:ψ2
HOMO
SOMO
giving the coefficient Cjr
for atom j in molecular
orbital r of a conjugated
system of n atoms (so that
j and r = 1, 2, 3, • • ., n)
Conjugated systems are often, but not always, lower in energy than
unconjugated systems.
1.4.3 Longer Conjugated System
HOMO
LUMO
1.5 Aromaticity
6 π orbitals of benzene
1.5.1 Aromatic Systems
Ψ1~Ψ6
Hückel aromaticity: (4n+2) π-electrons ------ special stability
n=1
n=0
n=3
n=4
Where does this special stability come from?
C6H6
hexatriene
Total π bonding energy of benzene
2 x 4β = 8β
Frost circle
hexatriene
2 x (1.802 + 1.247 + 0.445)β = 7β
From Hexatriene to Benzene
1.7 Heteronuclear Bonds, C-M, C-X and C=O
Electronegativity and Atomic Orbital Energies
Pauling s χ
χA − χB =
Δ
23.06
Δ = D(A − B) − D(A − A)D(B − B)
D:bonddissociationenergy
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Ex.NMRmethod
1.7.1 Atomic Orbital Energies and Electronegativity
Inamotoetal,Chem.Lett.1982,1003.
3D Electronegativity Scale
C-H is not strongly polarised.
2.54
2.30
L. C. Allen JACS 1989, 111, 9003
Valence Atomic Orbital Energies (eV)
Atomic Orbital Energies for Hybrid Orbitals (eV)
1.7.2 C-X σ Bonds
H3C-Cl
C-Cl σ bond
Weakerbond?
C
(-10.7eV)
(-13.7eV)
D(C-C):347kJ/mol
D(C-Cl):352kJ/mol
Na-F
Na+ F-
When two orbitals of unequal energy interact, the lowering in
energy is less than when two orbitals of very similar energy
interact.
Ion pair
Orbital overlap is negligible
When orbitals of identical energy interact, the energy lowering
is roughly proportional to S. When they are significantly
different in energy, it is roughly proportional to S2. They are
inversely proportional to the energy difference Ei.
(-5.2eV)
(-18.6eV)
1.7.3 C-M σ Bonds
C-Cl
C-F
C-Li
C-Li
(-5.4eV)
(-10.7eV)
C-Li
CH3Li
HOMO
1.7.4 C=O π Bonds
HCHO
(-10.7eV)
(-15.9eV)
LUMO
Report No1 (5/06)
Back-side attack of nucleophile
1. In the discussion about the H3 molecule, we could have combined the three
atoms in a straight line. Show that there would be less bonding both in σ1 and
σ2* and less antibonding in the σ* orbital.
2. Calculate the coefficients for the frontier orbitals of the heptatrienyl cation.
hint: use equation
or its geometrical equivalent