651.06 - 1 -

1

Chemical Bonding (A-B) I. Valence Bond Approach

II. Molecular Orbital Approach

I. Valence Bonds In the chemical bond between A & B, the two electrons involved in the bond are

indistinguishable and are permitted to interact equally with both nuclei.

A. Hybridization

The four electron orbitals are replaced by equivalent hybrid orbitals where the

number of equivalent hybrid orbitals is determined by the number of ligands to

carbon.

CH4 4 ligands sp3 hybridization

CH2CH2 3 ligands sp2 hybridization with the

nonhybridized p providing additional

overlap

HCCH 2 ligands sp hybridization with two non-

hybridized p orbitals providing

additional overlap

Hybridization, in turn predicts bond length, bond strength and bond angle. This

approach to viewing chemical bonding is useful because of the relative constancy

of the bond properties from molecule to molecule.

ex

is a regular hexagon with all C-C-C bond angles = 120o

O

HH has an H-C-H angle of 118o

651.06 - 2 -

2

Bond Length

Bond Bond Length (Å)1 Representative Compound

C-C (sp3-sp3) 1.53

C-C (sp3-sp2) 1.51

C-C (sp3-sp) 1.47

C-C (sp2-sp2) 1.48

C-C (sp2-sp) 1.43

C-C (sp-sp) 1.38

C=C (sp2-sp2) 1.32

C=C (sp2-sp) 1.31 C

C=C (sp-sp) 1.28 C C

C C 1.18

C-H (sp3-H) 1.09 H3C H

C-H (sp2-H) 1.08 H

C-H (sp-H) 1.08 H CN

C-O (sp3-O) 1.43 HO CH2CH3

C-O (sp2-O) 1.34

HOCH

O

C=O (sp2-O) 1.21

CH2

O

C=O (sp-O) 1.16

O C O

O-H 0.962 H3CO H

651.06 - 3 -

3

Bond Bond Length (Å) Representative Compound

O-H 1.052 O

CHOH

C-N (sp3-N) 1.47 H2N CH3

C-N (sp2-N) 1.38 H2N CH

O

C=N (sp2-N) 1.28 HN CHCH3

C N 1.14 HC N

N-H 1.032 R3N H

N-H 1.012

H3CN H

H

C-S (sp3-S) 1.82 HS CH3

C-S (sp2-S) 1.75 PhS SPh

C-S (sp-S) 1.68 H3CS CN

C=S (sp-S) 1.67

S-H 1.332 H3CS H

C-F (sp3-F) 1.40

C-F (sp2-F) 1.34

C-F (sp-F) 1.27

C-Cl (sp3-Cl) 1.79

C-Cl (sp2-Cl) 1.73

C-Cl (sp-Cl) 1.63

C-Br (sp3-Br) 1.97

C-Br (sp2-Br) 1.88

651.06 - 4 -

4

Bond Bond Length (Å) Representative

Compound C-Br (sp-Br) 1.79

C-I (sp3-I) 2.16

C-I (sp2-I) 2.10

C-I (sp-I) 1.99

C-Si 1.862 protecting groups for oxygen

O-Si 1.632 protecting groups for oxygen

C-P 1.872 H3C P(CH3)2

in general3

C-C 1.54

C=C 1.36

C C 1.20

C-H 1.08

C-O 1.42

C=O 1.22

1 J. March, Advanced Organic Chemistry, 4th Ed., Wiley & Sons, New York, 1992. 2 Handbook of Chemistry and Physics, 72nd Ed., CRC Press, D. R. Lide (Ed.), Boca Raton, 1991. 3 F. C. Carey and R. J. Sundberg, Advanced Organic Chemistry, Part A, 3rd Ed., Plenum Press, New York, 1990.

651.06 - 5 -

5

Bond Strength

Bond Bond Strength (kcal/mol)

Bond Bond Strength (kcal/mol)

O-O 343 Br-H3 87

I-I 363 C-O1 88

Br-Br 453 C-Si2 90

C-I 521 N-H1 93

Cl-Cl 573 C-H1 98

C-S 611 Cl-H3 102

C-Sn 632 H-H3 103

C-Br 661 C-F3 108

I-H 713 O-H1 110

C-N 721 C=N1 143

C-Cl 791 C=C1 148

S-H 821 C=O1 177

C-C 841 C C 1 200

H-CH2Ph 853 H-CH2CH3 1003

F-CH3 1083 H-CHCH2 1063

H-OCH3 1023 H-CCH 1323

H-NH2 1033 H-C6H5 1113

H-NHCH3 923 CH3-CH3 903

H-CN 1243 H2C=CH2 1723

H-CHCH3OH 933 HC CH 2303 1 J. March, Advanced Organic Chemistry, 4th Ed., Wiley & Sons, New York, 1992. 2 Handbook of Chemistry and Physics, 72nd Ed., CRC Press, D. R. Lide (Ed.), Boca Raton, 1991. 3 F. C. Carey and R. J. Sundberg, Advanced Organic Chemistry, Part A, 3rd Ed., Plenum Press, New York, 1990.

651.06 - 6 -

6

Bond Angle

Atoms Compound Carbon Hybridization Angle (o)

H-C-H ethane2 sp3 109

H-C-H ethylene3 sp2 117

H-C-C acetylene3 sp 180

O-C=O formic acid2 sp2 124

H-C-O formic acid2 sp2 118

C-C=O propynal2 sp2 123

C-O-C dimethyl ether2 - 110

C-O-C diphenyl ether1 - 124 1 J. March, Advanced Organic Chemistry, 4th Ed., Wiley & Sons, New York, 1992. 2 Handbook of Chemistry and Physics, 72nd Ed., CRC Press, D. R. Lide (Ed.), Boca Raton, 1991. 3 G. M. Loudon, Organic Chemistry, 2rd Ed., Benjamin/Cummings, Menlo Park, California, 1988.

651.06 - 7 -

7

B. Resonance

1. When one Lewis structure is inadequate for describing a molecular structure:

The “true” molecular structure is derived from a quantum mechanical

summing of each contributing structure.

2. The amount a particular Lewis acid structure contributes to the ground state of

a hybrid structure is dependent on its stability. The greater contribution is

made by the more stable structure.

3. Enhanced stability is associated with electron delocalization.

Resonance concept indicates that the valence bond approach to chemical bonding

has difficulty in treating certain types of chemical structures.

H

H

H

HH

H

cyclopropane

Each carbon has to bond to two other hydrogens and two other carbons. This

requires an sp3 hybridization.

However, to account for angle strain, each C-C bond is given more p character

while the C-H bond is given more s character

% S character

C-C 17

C-H 33

Reactivity: Br2

Br Br

BrNaCN

no reaction

H

O

H

O

H

O

651.06 - 8 -

8

Other deficiencies of Valence Bond theory:

• Why does the SN2 mechanism proceed with stereochemical inversion?

• Why is benzene so much more stable than other conjugated polyenes?

• The stereoselectivity of the Diels-Alder reaction:

+

CO2Et

CO2Et CO2Et

CO2Et

H

H

However, another model exists that can explain these phenomena…

II. Molecular Orbital Theory MO Theory - Bonding electron pairs are not localized between specific atoms.

Orbitals are extended over entire molecule.

Ψ - is the wave function which describes the interaction of the electron with other

electrons and nuclei of the molecule

Molecular orbitals are the linear combination of atomic orbitals

Ψmc = c1φ1 + c2φ2 + .... cnφn

φ - basis functions (usually atomic orbitals)

s p c1 - n - weighting coefficients which indicate the contribution of each atomic

orbital to the molecular orbital.

this is referred to as the LCAO-MO approximation

651.06 - 9 -

9

A. Hydrogen HA - HB

ΨMO = ϕ1SA - ϕ1SB

ΨMO = ϕ1SA + ϕ1SB

The number of molecular orbitals formed always equals the number of atomic

orbitals used.

!1SA !1SB

"E

"E

H H

H H

HHEnergy

651.06

10

B. Hybrid orbitals - Methane

1) orient methane with the carbon at the origin of a coordinate system

H1 is in the +x, +y, +z quadrant

H2 is in the -x, -y, +z quadrant

H3 is in the +x, -y, -z quadrant

H4 is in the -x, +y, -z quadrant

2) linearly combine the atomic orbitals

χsp31 = ϕ2s + ϕ2px

+ ϕ2py + ϕ2pz

χsp32 = ϕ2s - ϕ2px

- ϕ2py + ϕ2pz

χsp33 = ϕ2s + ϕ2px

- ϕ2py - ϕ2pz

χsp34 = ϕ2s - ϕ2px

+ ϕ2py - ϕ2pz

ϕ2s + ϕ2px + ϕ2py

+ ϕ2pz

+z

-z

-x

+x

-y+y

+z

-z

-x

+x

-y+y

+z

-z

-x

+x

-y+y

+z

-z

-x

+x

-y+y

= χsp3

+z

-z

-x

+x

-y+y

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11

C. Bond Orbitals

ΨMOσ1 = χsp3

1 - ϕ1S1

C

H

C H

C H

ΨMOσ1 = χsp3

1 + ϕ1S1

others

ΨMOσ2 = χsp3

2 + ϕ1S2

ΨMOσ3 = χsp3

3 + ϕ1S3

ΨMOσ4 = χsp3

4 + ϕ1S4

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12

D. MO Treatment of Cyclopropane H

H

H

HH

H

each carbon bonded to 4 other atoms, valence bond indicates sp3 but, H - C - H bond is ~120o therefore sp2!

The Walsh hybridization model for cyclopropane. All carbons are sp2 hybridized, two of the sp2 orbitals form

σ bonds to hydrogen with the third in the center of the ring. The p orbitals lie in the plane of the ring.

Above are the energy level diagram of the hybrid orbitals of cyclopropane and a representation of it's

molecular orbitals. The "σ type" orbitals arise from the overlap of sp2 hybrid orbitals with σ2 and σ3 being

degenerate and antibonding. The "π type" orbitals are made from p orbitals with π1 and π2 degenerate and

π3 being antibonding. The resulting electron density of cyclopropane is shown on the right.

MO treatment of cyclopropane explains chemistry of molecule much better than

valence bond treatment because chemistry of molecule is more typical of alkenes

than that of alkanes.

Br Br

Br+ HBr

+ Br2

!1

!2 !3

"1 "2

"3

Energy

non-bonding level

H

H

H

H

H

H

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13

E. Construction of Molecular Orbitals

1) Symmetry

Only orbitals of the same symmetry can interact

a. constructed hybrid orbitals (oxygen is sp3) OHH

b. constructed bond orbitals

OHH

OHH

ϕ1 ϕ2

Question: How much bonding which holds a water molecule together is due

to the interaction between ϕ1 and ϕ2?

Problem: Question can not be answered until ϕ1 and ϕ2 are symmetry

adapted.

Molecules posses elements of symmetry. All molecular orbitals which

describe a structure must be symmetric or antisymmetric with respect to

each of the molecule’s elements of symmetry.

OHH

O

HH

OHH

axis of symmetry plane of symmetry plane of symmetry

ϕ1 and ϕ2 are neither symmetrical or asymmetric relative to C2 axis

ϕ1 and ϕ2 are neither symmetrical or asymmetrical relative to σ plane

ϕ1 and ϕ2 are symmetrical relative to σ' plane

651.06

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To adapt:

ΨA = ϕ1 - ϕ2

O

HH

symmetric to C2 axis symmetric to σ plane symmetric to σ‘ plane

ΨB = ϕ1 + ϕ2

O

HH

antisymmetric to C2 axis antisymmetric to σ plane symmetric to σ‘ plane

Bonding orbitals are now symmetry adapted

Note: Orbitals do not have to be entirely symmetric or asymmetric relative to the symmetry elements of the molecule. ΨB is twice antisymmetric and once symmetric

ΨB = ϕ1 + ϕ2

ΨA = ϕ1 - ϕ2

O

HH

O

HH

O

HH

O

HH

!1 !2

651.06

15

Because ϕ1 and ϕ2 both possess two electrons, both higher and lower energy

levels will be occupied. No stabilization. Overlap between ϕ1 and ϕ2

contributes little bonding which holds the water molecule together.

III. Perturbation

!E

!E

"#

"+

$2

$1

"# = $2 - %$1

"+ = $1 + %$2 % = (0-1)

Working Example of PMO Theory

A. Hyperconjugation [Why are tertiary carbocations more stable?]

Enthalpy (ΔHo) for Ionization of Alkyl Chlorides (Gas Phase)

R Cl R

+Cl-

+

R ΔHo, kcal/mol CH3 229

CH3CH2 190

CH3CH2CH2 193

CH3CH2CH2CH2 193

(CH3)2CH 168

(CH3)3C 153

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B. PMO Treatment

H C

H-C bond is symmetric relative to symmetry element of the molecule

P orbital is antisymmetric relative to symmetry of the molecule P

C C

Ha

Hb

Hc

H

H

C CHc

H

H

ϕa ϕb

C

Ha

Hb

Hc

H

HC

C

Hb

Hc

H

H

Ha

C

ΨS = ϕa + ϕb ΨA = ϕa - ϕb

C C

Ha

Hb

Hc

H

H

C C

Ha

Hb

Hc

H

H

651.06

17

Qualitative MO Theory

Effect of electronegativity on MO energy levels:

nC

nN

nO

Bond formation:

Look at reaction of Lewis base with Lewis acid: B: A

How do we show this in MO theory? By mixing their MOs!

C C

Ha

Hb

HcH

H

C C

Ha

Hb

HcH

H

C C

Ha

Hb

HcH

H

! " #$A

!

$A

C C

Ha

Hb

HcH

H

$A + #!

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18

B:

A

A B

A B

New sigma bond is formed by mixing the filled sp3 orbital on B with the empty orbital on

A

Oxidation and reduction involve the removel/addition of one or more electrons from/to

the appropriate orbital

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Energy Levels in MO Theory

HΨ = EΨ H = Hamiltonian energy operator Ψ = electron amplitude function (wavefunction) E = the total energy of the system consisting of both kinetic and potential energy HΨ is non-commutable (i.e. ΨHΨ ≠ HΨ2) Ψ2 is the probability of finding an electron

!2

"#

#

$"#

#

$"#

#

$ dxdydz = 1 short form: !2

" d# = 1

HΨ = EΨ ΨHΨ = EΨ2

!"!d# = E !2

$$ d# % E =!"!d#$

! 2d#$

- solve for H2+

LCAO Approximation: Ψ = c1χ1 + c2χ2

E =(c1!1 + c2! 2)H(c1!1 + c2!2 )d"#

(c1!1 + c2!2 )2d"#

E =c1!1Hc

1!1+ c

1!1Hc

2!2+ c

2!2Hc

1!1+ c

2!2Hc

2!2[ ]d"#

c1

2!1

2 + 2c1c2!1!2+ c

2

2!2

2( )d"#

Substitutions:

H11= !

1" H!1d#

H12= H

21= !

1" H!2d# = !

2" H!1d#

H22= !

2" H!2d#

S11= !

1

2d#"

S12= !

1" !2d#

S22= !

2"2d#

E =c1

2H11+ 2c

1c2H12+ c

2

2H22

c1

2S11+ 2c

1c2S12+ c

2

2S22

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20

To find minimum value of the energy (as required by the variation principle), E is differentiated relative to each coefficient to give the secular equations:

!E

!c1

= c1H11" ES

11( ) + c

2H12" ES

12( ) = 0

!E

!c2

= c1H12" ES

12( ) + c

2H22" ES

22( ) = 0

Determining the energy levels then requires solving the secular determinant derived from the secular equations

H11! ES

11H12! ES

12

H12! ES

12H22! ES

22

= 0

Simplifying Assumptions:

Sik is the overlap integral = !i

"#

#

$ !kd%

where i = k there is complete overlap and Sik = 1 where i ≠ k the orbitals are orthogonal and Sik = 0 Hii= !

i" H!id#

Hii = α Hik= !

i" H!kd#

Hik = β

651.06

21

Energy Levels in MO Theory Solutions

1) solve for E: H11! ES

11H12! ES

12

H12! ES

12H22! ES

22

= 0

now becomes

! " E # " 0

# " 0 ! " E= 0

! " E #

# ! " E= 0

! 2 " 2!E + E2 " # 2 = 0

E2" 2!E + !

2" #

2( ) = 0

using quadratic formula

E =2! ± 4! 2 " 4 ! 2 " # 2( )

2

roots: E = α + β E = α − β 2) solve for c1 and c2 plug E back into secular equation to solve for c1 and c2 c1 ( α - E ) + c2 ( β ) = 0 c1 ( β ) + c2 ( α - E ) = 0

c1

c2

=!"

# ! E or

c1

c2

=!" + E

#

when E = α + β c1 = c2 when E = α − β c1 = - c2

651.06

22

Solutions (cont’d) 3) normalization remember !

2

" d# = 1 Ψ = χ1 + χ2

!2

" d# = $1+ $

2( )"2

d#

!2

" d# = $1"2d# + $

2"2d# + 2 $

1" $2d#

!

2

" d# = 1+ 1+ 0 = 2 to normalize divide through by 2:

!2

" d# = 12" $

1

2d# + 1

2" $2

2d# + 2 1

2" $1$2d#

!2

" d# = 12+ 12+ 0 = 1

for E = α + β, Ψ1 = c1χ1 + c1χ2

normalized Ψ1 = 1

2c1

c1χ1 + 1

2c1

c1χ2

for E = α − β, Ψ2 = c1χ1 - c1χ2 normalized Ψ2 =

1

2c1

c1χ1 - 1

2c1

c1χ2

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23

IN GENERAL Ψj = cj1χ1 + cj2χ2 + ... + cjiχi The secular equations will be:

cji(H11 - S11Ej) + cj2(H12 - S12Ej) + ... + cji(H1i - S1iEj) = 0

cji(H21 - S21Ej) + cj2(H22 - S22Ej) + ... + cji(H2i - S2iEj) = 0

.

.

.

cji(Hi1 - Si1Ej) + cj2(Hi2 - Si2Ej) + ... + cji(Hii - SiiEj) = 0

651.06

24

Huckel π Electron Method

1) Hik, Sik conditions are same as before

2) Only π electrons are considered

3) Basis functions consist of n p orbitals, one on each of the carbon atoms of the π system.

Ψ = c1χ1 + c2χ2 χ1 χ2 where χ are p orbitals and not atoms

step 1

secular equation: c1(H11 - S11E) + c2(H12 - S12E) = 0

c1(H21 - S21E) + c2(H22 - S22E) = 0

secular determinant:

H11! S

11E H

12! S

12E

H21! S

21E H

22! S

22E= 0

" ! E #

# " ! E= 0

step 2

c1(α - E) + c2(β) = 0 c1 = c2 for E = α + β

c1(β) + c2(α - E) = 0 c1 = -c2 for E = α − β

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25

step 3

Ψ1 = 1

2c1

c1χ1 + 1

2c1

c1χ2 Ψ2 = 1

2c1

c1χ1 - 1

2c1

c1χ2

step 4

draw energy diagram

E2 = α − β antibonding

!2

= 1

2

" #

$ % &1 '

1

2

" #

$ % &2 1

2 12

E1 = α + β bonding

!1

= 1

2

" #

$ % &1 + 1

2

" #

$ % &2 1

2 12

E2

E1

!2

!1

-"

+"

#

651.06

26

Butadiene

!1 !4!2 !3

Ψ = c1χ1 + c2χ2 + c3χ3 + c4χ4

1) set up secular equations and solve the secular determinant

c1(H11 - S11E) + c2(H12 - S12E) + c3(H13 - S13E) + c4(H14 - S14E) =0

c1(H21 - S21E) + c2(H22 - S22E) + c3(H23 - S23E) + c4(H24 - S24E) =0

c1(H31 - S31E) + c2(H32 - S32E) + c3(H33 - S33E) + c4(H34 - S34E) =0

c1(H41 - S41E) + c2(H42 - S42E) + c3(H43 - S43E) + c4(H44 - S44E) =0

H11! S

11E H

12! S

12E H

13! S

13E H

14! S

14E

H21! S

21E H

22! S

22E H

23! S

23E H

24! S

24E

H31! S

31E H

32! S

32E H

33! S

33E H

34! S

34E

H41! S

41E H

42! S

42E H

43! S

43E H

44! S

44E

= 0

! " E # 0 0

# ! " E # 0

0 # ! " E #

0 0 # ! " E

= 0

x 1 0 0

1 x 1 0

0 1 x 1

0 0 1 x

= 0 = x4 - 3x2 + 1

solve for x: x2=3 ± 9 ! 4

2=3 ± 5

2" x = ±

3 ± 5

2

x = ± 0.618, ± 1.618

651.06

27

substitute

E4 = α -1.618β

Ε3 = α - 0.618β

E2 = α + 0.618β

E1 = α + 1.618β

2) Plug E’s back into secular equations and solve for coefficients

Ψ4 = 0.372χ1 - 0.602χ2 + 0.602χ3 - 0.372χ4

Ψ3 = 0.602χ1 - 0.372χ2 - 0.372χ3 +0.602χ4

Ψ2 = 0.602χ1 + 0.372χ2 - 0.372χ3 - 0.602χ4

Ψ1 = 0.372χ1 + 0.602χ2 + 0.602χ3 + 0.372χ4

E4 = ! - 1.618"

#4

E3 = ! - 0.618"

#3

#2

#1

E2 = ! + 0.618"

E1 = ! +1.618"

non-bonding level

651.06

28

Mobile Bond Order (π bond order)

for butadiene: a qualitative assesment of bonding

Ψ1 c1 c2 c3 c4 c2 - c3 strong bonding

c1 - c2 , c3 - c4 fair bonding

Ψ2 c1 - c2, c3 - c4 fair bonding

c2 - c3 antibonding

Pij = Σncicj Pij = mobile bond order

n = # of electrons in an occuppied orbital

ci , cj = normalized coefficients for atoms i & j

P12 = (n1c1c2)Ψ1 + (n2c1c2)Ψ2

= 2(0.372)(0.602) + 2(0.602)(0.372)

= 0.896

P23 = (n1c2c3)Ψ1 + (n2c2c3)Ψ2

= 2(0.602)(0.602) + 2(0.372)(-0.372)

= 0.448

P34 = (n1c3c4)Ψ1 + (n2c3c4)Ψ2

= 2(0.602)(0.372) + 2(0.372)(0.602)

= 0.896

for the σ bond, add 1.0 to each calculated Pij:

H2C CH

CH

CH2

1.896 1.8961.448

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29

π Bond Energy:

E! = ni

" Ei Eπ = π bond energy

n = # of electrons in each orbital Ei = energy of each orbital

for butadiene

Eπ = 2(α + 1.62β) + 2(α + 0.62β)

= 4α + 4.48β

Resonance Energy:

EHDE = Eπ - Eloc EHDE = Huckel delocalization energy (or resonance energy) Eπ = π bond energy Eloc = π bond energy of the component, isolated bonds

for butadiene:

EHDE = 4α + 4.48β - 4(α + β)

= 0.48β

for benzene:

Eπ = 2(α + 2β) + 4(α + β) = 6α + 8β

EHDE = 6α + 8β - 6α − 6β = 2β

Experimental value for resonance energy of benzene = 36 kcal/mol

β = -18 kcal/mol for carbon systems

651.06

30

Simplify

For linear polyenes of formula CnHn+2:

E = α + mjiβ where mj = 2cos (jπ / n+1)

j = mo #

n = # of carbons

# of molecular orbitals = # of basis orbitals ≈ in Huckel treatment to # of p

orbitals

coefficient crj of atom r in the jth molecular orbital:

crj =2

n +1sin

rj!

n +1

" #

$ %

For cyclic conjugated systems (Frost’s Circle)

Ej = ! + 2" cos2#J

N

$ %

& ' J = 0, 1, 2, ..., n-1 the orbital number

counting counter-clockwise from lowest vertex

N = ring size

J = 0

J = 1

J = 2J = 3

J = 4

! + 2"! + 0.618"

! # 1.618"

651.06

31

Aromaticity

cyclobutadiene benzene

Huckel Molecular Orbital Treatment

cyclobutadiene DE = 2(α + 2β) + 2α − 4(α + β) = 0

benzene DE = 2(α + 2β) + 4(α + β) − 6(α + β) = 2β

Experimental Data:

cyclobutadiene is extremely reactive

35

K

O

medium pressuremercury lamp

7 K, argon matrixguest - host ratio1 : 100 - 500

O

+

651.06

32

4n + 2 rule (Huckel’s Rule)

Species is aromatic if:

1) cyclic array of p orbitals

2) fully conjugated

3) planar

4) contains 4n + 2 π electrons

Physical measurements relevant to aromaticity:

1) bond lengths

aromatic molecules show bond lenths of 1.38 - 1.40Ao and are uniform

around the ring

2) diamagnetic anisotropy

in a magnetic field, π electron circulation leads to a magnetic field which

opposes the externally applied magnetic field

H0 nuclei within cone are shielded, nuclei on periphery of cone are

deshielded

3) thermodynamic characteristics

(a) heats of hydrogenation

49.8 kcal/mol

28.6 kcal/mol

stabilization associated with benzene = 3(28.6 kcal/mol) - 49.8 kcal/mol = 36 kcal/mol

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(b) hydride & proton dissociation constants

Apply criterion for aromatic systems:

2e-

6e-

6e-

6e-

N

HN O S

O

6e-

6e-

6e-

6e-

6e-

Terms

aromatic - as discussed

antiaromatic - 4n system which is strongly destabilized relative to conjugated but

noncyclic equivalent

vs.

nonaromatic - 4n system which is isolable but not possessed of any stabilization relative

to the noncyclic equivalent

vs.

homoaromatic - 4n + 2, stabilized, cyclic conjugated system which is formed by

bypassing a saturated carbon atom

Ha

Hb

HaHb

Ha is 5.8 ppm upfield of Hb ⇒ ring current

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[10] annulene

cyclodecapentaenes

10e-

4n + 2 system, cyclic but does not possess aromatic stabilization

all cis bonds are strongly destabilized due to angle strain (ref. 1a-e)

H

H

trans, cis, trans, cis, cis - minimun angle strain but severe

nonbonded repulsion between the two internal hydrogens (ref. 1a-d, f)

h! +

NMR does not indicate ring current, nonaromaticity is due to

nonplanarity (ref 1g, h)

Belted Aromatics (ref 1i-l)

HbHa

placing unsaturated carbon in system relieves the irresolvable

conflict between angle strain and nonbonded steric interactions (ref 1m-

o)

Ha and Hb shifts ⇒ ring current

x-ray indicates no significant alternation in c-c bond lengths around the ring

now: 4n + 2 cyclic fully conjugated planar (ref 1p) ∴ aromatic further examples of belted aromatics

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HN

(ref 1q-t)

O

(ref 1q, u)

[14] annulene

HHHH

(ref 1v-y)

[18] annulene

H

H

H

H

H

H

(ref 1z, aa-ac)

C18

(ref 1ad, ae)

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Dissociation Constants

Cl Cl

ClCl

+ SbCl5-

ClCl

Cl

+ SbCl6-

H

H

H

H

H

H

H

H3C H

CH3H3C CH3H3C

CH3

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References: 1 a) Top. Nonbenzenoid Aromat. Chem 1973, 1, 121 (review) b) Acc. Chem. Res. 1972, 5, 272 (review) c) Snyder, Nonbenzenoid Aromatics, vol. 1; Academic Press: NY, 1969, pp. 63-116 (rev) d) Garratt, Aromaticity; Wiley: NY, 1986, pp.113-147 (rev) e) JACS 1971, 4966 f) J. Chem. Phys. 1952, 1489 g) Croat. Chem. Acta. 1977, 49, 441 h) JACS 1971, 4966 i) Pure Appl. Chem. 1982, 54, 1015 (rev) j) Isr. J. Chem. 1980, 20, 215 k) Chimia 1968, 22, 21 l) Angew. Chem. Int. Ed. Engl. 1967, 6, 385 m)Angew. Chem. Int. Ed. Engl. 1964, 3, 228 n) Angew. Chem. Int. Ed. Engl. 1964, 3, 642 o) Tetrahedron Lett. 1966, 1569 p) Acc. Chem. Res. 1988, 21, 243 q) Angew. Chem. Int. Ed. Engl. 1964, 3, 642 r) JACS 1964, 3168 s) JACS 1967, 6310 t) Chem. Commun. 1967, 1039 u) Tetrahedron Lett. 1965, 3613 v) JACS 1964, 521 w) JACS 1972, 471 x) JACS 1973, 3893 y) Helv. Chim. Acta. 1995, 78, 679 z) JACS 1962, 274 aa) JACS 1972, 8644 ab) Helv. Chim. Acta. 1974, 57, 2276 ac) Helv. Chim. Acta. 1982, 65, 1885 ad) JACS 1990, 4966 ae) Nature 1994, 369, 199