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1.5 Aromaticity
Hückel aromaticity: (4n+2) π-electrons ------ special stability
1.5.1 Aromatic Systems
n = 1 n = 0 n = 3 n = 4
Where does this special stability come from?
1.5.2 Antiaromatic System
Hückel antiaromaticity: 4n π-electrons ------ destabilisation
n = 1 n = 2 n = 3 n = 4
More reactiveeg.)
(review)Subustituents change the overall energy, energy and polarization of FMO.
where k is the number of the atom along the sequence of n atoms
Orbital Energy
FrostCircle
HOMO LUMO
Cyclobutadiene (4π)
Less stabilization than a pair of conjugated double bond
2 x (1.618 + 0.618)β = 4.472β4β
No net stabilization
Same as two isolated ethylene(2β)
Cyclobutadiene (4π)
Antiaromatic
more reactive
high HOMOlow LUMOBiradical?
Jahn-Teller Distortion
Square => rectangular (more stable)
Separating the pair (C-1 from C-4 and C-2 from C-3) will reduce the amount ofπ antibonding between them, and hence lower the energy.
Butadiene
1.5.3 Cyclopentadienyl Anion and Cation
Anion: 2 x (2 + 1.236 β = 6.472β
Cation: 2 x (2 + 0.618 β = 5.236β
Pentadienyl: 2 x (1 + 1.732)β = 5.464β>Aromatic (4n + 2)
pKa =16
Pentadienyl: 2 x (1 + 1.732)β = 5.464β<
Anti-aromatic (4n)
Total π bonding energy of benzene
2 x 4β = 8β
hexatriene
2 x (1.802 + 1.247 + 0.445)β = 7βhexatriene
Frost circle
6 π orbitals of benzene
Ψ1~Ψ6
down
down
up
From Hexatriene to Benzene No 2 (2016.5.13)2 Molecular Orbitals and the Structures of Organic Molecules
2.1.1 A Notation for Substituents
2.1 The Effects of π Conjugation
“Thermodynamic Stability”
Houl,JACS1973,95,4092
Benzene ethylene
+
Ψ1
Ψ2∗
Stabilization: 2 x 4β Stabilization: 2 x β
styrene
2.1.2 Alkene-Stabilising Groups2.1.2.1 C-Substituents Stabilization: 2 x 5.21β
HOMO upLUMO down
Styrene
Benzene + Ethylene:2 x 5β
2.1.2.2 Z-Substituents
HOMO sameLUMO down
acrolein
vs. ethylene
Whatishappening?
Allylcationnature=>Lowertheenergyofbutadienenature
The average of two components.
2.1.2.3 X-SubstituentsStabilization by25 kJ/mol
HOMO upLUMO up
NBMO
2 x 0.414β
(116 kJ/mol)
ref) allyl system
The average of two components. 2.1.3 Cation-Stabilising and Destabilising Groups
2.1.3.1 C- and X-Substituents
A molecule having an empty p orbital on carbon will be lowered overall in energy by π conjugation
Weak Lewis Acid Strong Lewis Acid
Electronegativityless π-overlap
(energy, size)
F?
X = F= Cl, Br, I
2.1.3.2 Z-Substituents
Small stabilizationby conjugation
Large destabilization byCoulombic effect
But,
“Conjugation cannot always be relied upon to lower the overall energy.”
2.1.4 Anion-Stabilising and Destabilising Groups
2.1.4.1 C-Substituents
“carbanion”, “enolate”, “C-M”
Similar to allyl anion
π stabilisation
2.1.4.2 Z-Substituents
Stabilizationby conjugation
- -
σ-conjugation (=> 2.2)
Stabilization of carbanion by neighbouring Si, P, S group
Overlap of p- and σ∗YR
anti-periplanar
2.1.4.3 X-Substituents Usually π-destabilising
TMS+BuLi
2.1.5 Radical-Stabilising Groups
2.1.5.1 C-, Z- and X-Substituents All stabilise radicals
2.1.5.2 Captodative Stabilization (Radical)Long-lived radical
Capto: Z (electron capture)Dative: X (electron donate)
Overall energy drop
2.1.6 Energy-Raising Conjugation
E2 > E1
Destabilization
Twisted conformation to minimize the lone pair conjugation
(Coulombic effect)
OO
HH
HO
OH
H O
O H
Both the diketone and the enediolate are destabilised systems
2.2 Hyperconjugation-σ Conjugation
2.2.1 C�H and C�C Hyperconjugation (���)
Anti-periplanar
E2 > E1
2.2.1.1 Hyperconjugation of C�H Bonds with C�H Bonds
“Overlapofσ bondswithσ bondsorporbitals”
Klyne-Prelog System
1JC-H
122Hz
126Hz
Perlin effect
In cyclohexane and in six-membered rings having one or more heteroatomsof the first row attached to the carbon of interest,1JC–H is always larger for
an equatorial hydrogen than for an axial hydrogen.
C-H hyperconjugation is stronger than C-C hyperconjugation
2700-2800cm-1
(ref: normally2800-2900cm-1)
longer
shorter
2.2.1.1 Hyperconjugation of C�H Bonds with Lone Pairs
Bohlmann bandsIR:C-H Weakeningbond
β-hydrideelimination
2.2.1.3 Stabilisation of Alkyl Cations by Hyperconjugation
“Alkyl substituents stabilise carbocations”
2E stabilisation
Because the two p-type orbitals�πz and πY' have the same energy, the interactions in the two conformations are equal (EA = EB). We can expect that the barrier to rotation about the C---C bond of the ethyl cation will be small. (Hybridisation is not necessary!)
Ex) Solvolysis (SN1)
Stabilization of an empty p orbital
Favorable conformation
A
B
C
D
E
F
π-bonding in C-C 2.2.1.4 Bridging in Carbocations
Two electrons, two bonds system
Transitionstructureforthe[1.2]-HshiftWagner-Meerwein typerearrangement
H-C-C (θ)
θ
Rapid interconversions make all the methylene carbons equivalent
not but
2.2.1.5 Stabilisation of a π Bond by Hyperconjugation
Alkenes prefer to be more rather than less substituted by alkyl groups.
10kJ/molstabilizationperalkyl
E1 > E2
2.2.3 Negative Hyperconjugation: Conjugation with a negative charge
With Anion
2.2.3.1 With Cation2.2.3.2 With Anion
With Cation
smallstabilization
largestabilization
2.2.3.3 The Anomeric Effect
Exo Anomeric effect: Gauche effect
RO-C-X
FCH2OH CH3OCH2Cl diazaacetal
2.3 The Configurations and Conformations of Molecules
2.3.1 Restricted Rotation in π-Conjugated System
2.3.1.1 One π Bond 2Eπ =~280kJ/mol
Configuration ���� high energy barrier > rtConformation���� low energy barrier < rt
Stabilization of TS is necessary
“Push-Pull”
Coalescence at-90 �C
Experimental activation barrier: 2.3.1.2 Allyl and Related Systems
(116 kJ/mol)E = 2 x 0.414β
Cation: 140 kJ/molRadical: 63 kJ/molAnion: 85 kJ/mol
Rotation barrier
1,3-disubstituted
t1/2=10min
-10�C
t1/2=10min
35�C74kJ/mol101kJ/mol
R=D:66kJ/mol
R=CH3:60kJ/mol
M=Li:45kJ/mol
M=K:70kJ/molM=Cs:76kJ/mol
Thompson,JACS 1979,101,5459
M+ M+
Lower the rotation barrier++
Allyl radical
η3
η1 :weaken
:strengthen
Azomethin ylides85kJ/mol
84kJ/mol
Draw enolate ions with the charge on oxygen
5-25 kJ/mol more stable, small barrier for rotation
Mono-substituted amide, ester, ether
s-trans s-cis
Anomeric effect
2.3.1.3 Dienes28kJ/mol
16kJ/mol
12kJ/mol
λmax
214nm
λmax
253nm
1. Explain why silylamines are weaker bases than ammonia and why hexaphenyldisiloxane (Ph3Si)2O is linear.
2. Explain why it is easy to remove a proton from the methyl group attached to the boron atom in the trialkylborane 2.105.
Report No2 (5/13)