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)