Principles of Organic Chemistry
lecture 15, page 1
LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS
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The allyl system (cation, anion or radical).
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1. Draw molecule and set up determinant.
1
o
o
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1
H
H
3
1
2
C
C
H
C2 H
3
H
=
X
X
1
1
X
2
3
X 1
0
1
X
1
0
1
X
- 1
0
=
1
1
0
X
+
0
1
X
0
1
=
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X(X2-1) –X = 0 = X3-2X = X(X2-2); X = 0, X = −SQRT(2), X = SQRT(2)
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These are the energies of the orbitals in terms of beta: bonding, −SQRT(2)β; nonbonding, 0; antibonding, SQRT(2)β
discuss the cation anion and radical with regard to the solutions
C C C
C C C
pi MOs of Allyl anion
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C C C
pi MOs of Allyl radical
pi MOs of Allyl cation
The bonding orbital contributes −2·SQRT(2)β to the energy of the molecule.
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The non-bonding orbital contributes nothing to the energy of the molecule.
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It should be relatively easy to put electrons in and take electrons out of the allyl
system.
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In general we observe anionic, radical and cation chemistries for allyl.
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This is similar to the simplest non-bonding orbital of atomic hydrogen.
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This why the chemistry of H is (+), neutral and (−).
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Proton, hydrogen atom (radical) H dot, and hydride.
Does this mean that the allyl anion and cation are really isoenergetic?
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No, there are e/e repulsion terms that are not considered in the Hückel
formulation.
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There are other effects that Hückel does not include.
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Hückel is approximating what the orbital spacing is.
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When we put electrons into the system the Hückel levels don’t
change, but the overall energy of the molecule certainly does.
Principles of Organic Chemistry
lecture 15, page 2
Hückel LCAO is only a first approximation.
o
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You will notice that all the atomic orbitals that make up the MOs are not the same size.
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This is the stuff of Cnm, the weighting constants in the MOs wave function, or the AO
coefficients.
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These AO coefficients can be obtained by expansion of minors of the secular determinant
into cofactors.
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For example:
1
2
3
1
X 1
0
2
1
X
1
3
0
1
X
A11 =
X
1
1
X
•
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o
o
o
o
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A11 = X2-1
likewise
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A12 = −X
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A13 = 1
We can determine the relative ratios of the coefficients.
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c1/c3 = A11/ A13 = X2-1
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c2/c3 = A12/ A13 = −X
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as before, Lets arbitrary set c3 = 1
FROM A11
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For ψ1, X = −SQRT(2), so c11 = 1
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For ψ2, X = 0, so c12 = -1
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For ψ3, X = SQRT(2), so c13 = 1
FROM A12
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For ψ1, X = −SQRT(2), so c21 = SQRT(2)
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For ψ2, X = 0, so c22 = 0
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For ψ3, X = SQRT(2), so c23 = -SQRT(2)
FROM A13
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For ψ1, X = −SQRT(2), so c13 = 1
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For ψ2, X = 0, so c23 = 1
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For ψ3, X = SQRT(2), so c33 = 1
X
A11 (X2-1)
A12 (-X)
A13 (1)
SQRT(2)
1
−SQRT(2)
1
0
-1
0
1
Principles of Organic Chemistry
−SQRT(2)
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1
SQRT(2)
1
Now we can write out the MOs
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ψ1 = 1χ11 + SQRT(2)χ12 + 1χ13
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ψ2 = -1χ21 + 1χ23
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ψ3 = 1χ31 −SQRT(2)χ32 + 1χ33
These are the coefficients; Normalize and you are done. To do this we need to find the
normalization factor to apply to the MO above.
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for ψ1 and ψ3
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SQRT(12+ SQRT(2)2+12) = 2
for ψ2
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lecture 15, page 3
SQRT(12+12) = SQRT(2)
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Thus ½ and 1/ SQRT(2) are the normalization factors for ψ1, ψ3; and ψ2
respectively.
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ψ1 = ½χ11 + 1/(SQRT(2)χ12 + ½χ13
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ψ2 = -1/(SQRT(2)χ21 + 1/(SQRT(2)χ21
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ψ3 = ½χ31 −1/(SQRT(2)χ32 + ½χ33
Bond order and Atomistic Electron Density can be approximated by the
LCAO method.
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The concept of bond order.
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R3C—CR3 has C/C bond order of one.
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R3C=CR3 has C/C bond order of two.
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R3C=O has C/C bond order of two.
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R3C=O—H(+) has C/C bond order < two.
H
O
O
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O
o
O
How might we try to gather experimental evidence for the above shift in bond
order?
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H
H
O
The resonance (second from the right) is more important when the O
atom is protonated.
Likewise one might wonder about the CC bond order in allyl cation.
Bond order: PNij = Sum(Ncicj)
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N is the number of electrons in the MO
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cicj is the product of the coefficients for the bound atoms.
Principles of Organic Chemistry
o
lecture 15, page 4
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We can get these parameters from calculations.
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The better the calculations, the more agreement there is with
experimental results.
Why should this calculation give the bond order?
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Think about interaction between AOs in the molecule.
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It is a zero sum game due to normalization.
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If an orbital grows another has to shrink elsewhere.
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The extent of interaction (bonding) between the p orbital above is a
function of how much volume is gained for the electron occupying both
versus only one orbital.
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The interaction (bonding) scales linearly with the coefficient of the AO
in the MO.
Pi Bond Order in ALLYL
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Pcc(allyl cation) = 2•1/2•1/(SQRT(2) = 1/(SQRT(2) = 0.71
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There are 2 electrons.
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Coefficient 1= 1/2.
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Coefficient 2= 1/(SQRT(2).
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This is the pi bond order. There is also a sigma bond so the actual bond order is
1.71.
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The number is the same for the radical and anion by this method
because these added electrons are in the NBO.
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This is not always the case. When the next electron up is but in a
bonding orbital we get more bond order from more occupation.
Molecular Orbitals of pi systems and their application to organic chemistry
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Instructor compares the C12 bond order in allyl cation with two hypothetical
electronically excited states of allyl cation.
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Instructor talks about the differences in dynamics that one would
expect between ground and excited states.
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[If asked could you predict which electronic state of allyl cation would
be more conformationally stable]
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[Could you do the same thing for ally radical?]
WHAT IF THE COEFFICIENTS JUST DISTRIBUTED e(-) density uniformly?
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ψ1 = 1/(SQRT(3)χ11 + 1/(SQRT(3)χ12 + 1/(SQRT(3)χ13
Principles of Organic Chemistry
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Pcc(allyl cation) = 2•1/(SQRT(3) •1/(SQRT(3) = 2/3 = 0.66
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The real distribution leads to more bonding in the lowest energy orbital than
0.66.
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Instructor draws resonance structures and talks about weighting.
H
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H
C
H
H
H
1 H 3
2
H
1 H 3
2
Instructor explains . . . this effect is not obvious from resonance structures
without a lot of hand waving.
The double bond in ethene has a 66 kcal/mol barrier to rotation.
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H
H
C
Thus, the Hückel LCAO solutions give us a clear picture why conjugation stabilizes
molecular systems!
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Notice that the low-energy orbital is also more spherical.
What is the pi bond order from resonance?
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lecture 15, page 5
The bond is order is 2 in ethene.
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Therefore the rotation about allyl cation according to Hückel should be
0.71• 66 kcal/mol = 47 kcal/mol of the naked gas phase species.
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This is really slow on the molecular time scale!
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Remember that you need a barrier of about 24 kcal/mol between
molecules 1 and 2 to isolate molecule 1 or molecule 2 at room
temperature.
Checking the number against literature values give some discrepancy; the literature
values are interesting.
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Experimental value for the allyl radical is ~15 kcal/mol
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High level calculation for the allyl radical gives ~13 kcal/mol
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Gobbi, A.; Frenking, G., "Resonance Stabilization in Allyl Cation,
Radical, and Anion" J. Am. Chem. Soc 1994, 116, 9275-9286.
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note: the year, this stuff is just being figured out currently.
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It makes you wonder how ‘mature’ Organic Chemistry is if the
fundamental behavior of such simple molecules is unknown.
High level calculation for the allyl cation gives ~39 kcal/mol
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Korth, G.-G.; Trill, H.; Sustmann, R., "[ l-2H]Allyl Radical: Barrier to
Rotation and Allyl Delocalization Energy" J. Am. Chem. Soc 1981,
103, 4483.
High level calculation for the allyl anion gives ~23 kcal/mol
One might hypothesize that the massive difference in electronic distribution
between ground and transition states in the anion and the cation versus the
radical could account for the difference in barrier to rotation.
Principles of Organic Chemistry
•
Demonstrate this on the board.
lecture 15, page 6