Principles of Organic Chemistry
lecture 19, page 1
LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS
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BENZENE MOs
o Collecting all the molecular orbitals for benzene and arranging them
according to energy and putting in the electrons gives the following
energy diagram.
6
1
5
2
4
3
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What is the resonance energy of the benzene ring?
o Imagine putting three walls between the double bonds to isolate them.
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What would be the pi-energy of this version of benzene?
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E = −6|β|.
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Where does this result come from?
The pi-energy of benzene is
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E = 2(−2|β|) + 4(−|β|) = −8|β|
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The difference, and therefore, the resonance energy of
benzene is −2β.
Aromaticity
o 4N+2
o filled atomic orbitals –paired
Principles of Organic Chemistry
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lecture 19, page 2
lead to stability
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Note the electronic configuration of C6H6.
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It is like electronic configuration of the mono-atomic noble
gasses.
o filled molecular orbitals
o aromatic species
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need to be cyclic and fully conjugated
H
H
H
H
H
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H
H
H H
H
H
H H
H
H
1,3,5-hexatriene above: is fully conjugated but acyclic; and 1,3,5cycloheptatriene is cyclic but not totally conjugated.
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There is a methylene unit insulating the p orbitals at 1 and 6
from parallel interaction.
Frontier Molecular Orbital Theory
o reactivity and coefficients
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these are important for anions and cations
o kinetic results versus thermodynamic results
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How does this relate to MO calculations?
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Electronic changes are instantaneous compared to changes in
atomic displacement.
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Let’s talk about the protonation of 1,3-butadiene
1.618
0.
37
2
0.
60
2
0.618
60
2
0.
37
2
-0.618
0.
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H
o
-1.618
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electron density functions as c2 so .602/.372 = 2.6.
Principles of Organic Chemistry
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lecture 19, page 3
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The electron density is heavily weighted toward the
termini.
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Where is the molecule going to like protonate in the first
elemental event?
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What should we consider if we are going to talk about the
stability of species on the reaction pathway in the
protonation of l,3-butadiere?
Frost-Hückel diagrams
o Only valid for fully conjugated systems.
o −2βcos(m2π/n) = Em, n = # atoms, m (integer) = 0, 1, . . . ≤ n/2
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for cyclic conjugated molecules
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for benzene n=6
2π/3
π/3
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for cyclobutadiene
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for linear conjugated molecules
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for allyl
θ
•
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for butadiene
Principles of Organic Chemistry
lecture 19, page 4
θ
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o The frost method gives the energies straight away!
o −2βcos(mπ/[(n+1)) = Em, n = # atoms, m (integer) = 1, . . . n
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IN BOTH fully conjugated cyclic systems and linear systems we
are looking form the projection of the radius of the circle onto the
energy axis.
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Don’t commit these equations to memory. Establish an operational
definition and work them out every time.
o Can you envision how to use these with the secular determinants to
solve more complex systems?
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You can get the energies with these Frost diagrams and then use
the method of expansion by cofactors to get the MO coefficients!
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The cofactor process is simplified by the fact that you are
not looking for roots of polynomials!
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You are looking for ratios of expressions as a function of
energies.
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The energies are calculated directly above for many
interesting systems.
Odd/Even Alternate and non-alternate hydrocarbons.
o For the operational definition, in conjugated molecules, mark the
conjugated atoms so that the number of marked atoms is maximized.
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An alternate fully conjugated molecule is one in which alternate
atoms can be marked with no marks adjacent to each other.
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H
C
Examples.
H
H
H
C
*
*
*
*
*
*
*
*
*
*
*
*
*
*
Principles of Organic Chemistry
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lecture 19, page 5
The fully conjugated molecule is non-alternate if two un-marked
atoms or two marked atoms are adjacent after the making
operation.
H
C
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so azulene at right is non-alternate.
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some more examples of non-alternate conjugated
hydrocarbons:
H
C
*
H
C
*
*
*
*
*
*
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H
H
*
*
*
H
H
Odd alternate conjugated molecules have an unequal number of
marked versus unmarked atoms.
example
*
H
H
H
*
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H
*C
* C
H
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H
*
H
*
Even alternate conjugated molecules have the same number of
marked versus unmarked atoms.
*
*
H
*
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*
All linear and branched structures are by definition
alternate.
*
*
*
*
*
•
*
and see the benzyl cation above.
Now here is the utility in all this: Large structures lead to difficult secular
determinants. However the following rules allow some prediction regarding the
characteristics of the HOMOs of these orbitals.
o Alternate, conjugated molecules dispose orbital energies symmetrically
about zero.
Principles of Organic Chemistry
lecture 19, page 6
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The non-branched acyclic even alternate conjugated molecules
dispose orbital energies above and below zero, but not at zero.
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The odd alternate, conjugated molecules dispose orbital energies
such that at least one orbital is non-bonding and has value 0 β.
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We would have had a good head start on allyl with this
knowledge.
o The non-alternate simple cyclic systems like cyclopropenyl,
cyclopentenyl, cycloheptenyl dispose orbital energy dissymmetrically
about zero
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Non-alternate bicyclic systems and non-alternate cyclic branched
systems loose this property. Symmetrical systems of this variety
can dispose orbital energies symmetrically about the zero line.
What can we say about trimethylenemethane?
o Is it alternate? If it is alternate, is it odd or even?
o We want to maximize the number of marks on the structure.
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When we maximize the number of marks, the nodes in the
NBOs (non-bonding orbitals) occur at the unmarked atoms.
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The sum of the coefficients of the atomic orbitals on the starred
atoms directly linked to ONE unmarked atom is zero.
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Example
H
4*
H
H
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H
1
*
H
2
C2 +C3 + C4 = 0
H
3
*
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How should the orbitals of trimethylenemethane disposed in
energy?
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How many molecular orbitals are there? (four)
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It is an odd alternate conjugated system.
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After you put one molecular orbital at the zero line you
have three left.
o There is no way to symmetrically dispose these
remaining three orbitals about zero without putting
another at 0β.
o If you carefully workout the secular determinant
you will find roots 0, 0, SQRT(3), −SQRT(3)
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Can you do this?
Principles of Organic Chemistry
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lecture 19, page 7
There are two unique ways to symmetrically satisfy C2
+C3 + C4 = 0.
o C2 = −C3 and C2 = C3.
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Think about how this relates to our solution of the benzene
molecule when we used mirror planes to simplify the
secular determinant.
These two solutions lead to C4 = 0 and C4 = −2C2. After you
normalize you get:
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Ψ2 = (1/SQRT(6))χ2 + (1/SQRT(6))χ3 − (2/SQRT(6))χ4 .
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Ψ3 = (1/SQRT(2))χ2 − (1/SQRT(2))χ3 .
o This structure may look so esoteric that you think it is not important to
consider it.
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If we add to electrons to trimethylenemethane and change the C
atoms to N atoms we get guanidinium cation.
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If we change the C in the middle to a boron atom and the
atoms connected to oxygen atoms we can think about an
inorganic structure.
H
•
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H
N
H
H
N
N
H
H
H
H
O
H
B
O
O
H
H
C
H
H
C
H
H
the pi systems in these structure are isoelectronic.
o Their MO’s will be similar with perturbations by
the electronegativity of the N, B and O atoms.
o Your author does the benzyl example for you.
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You should read it, page 200-204.
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I don’t agree with the analysis that resonance structures predict
+2/5 on C1.
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You need atomic coefficients to predict charge.
Principles of Organic Chemistry
lecture 19, page 8
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I don’t agree that the benzyl system was the only system
for which we get different results when we do Hückel MO
theory versus resonance theory.
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Remember we found bond order differences in simpler allyl
system.