Principles of Organic Chemistry
lecture 14, page 1
LCAO APPROXIMATIONS OF ORGANIC Pi MO SYSTEMS
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LCAO linear combination of atomic orbitals
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to construct molecular orbitals.
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H. E. Zimmerman describes the rules for quantum mechanical mixing. Zimmerman, H.
E., Quantum Mechanics for Organic Chemists. 1 ed.; Academic Press: New York, NY,
1975, p 11-12.
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We will dig deeper into the origins of these rules, but for now let’s take them as truths.
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This method is especially applicable to the organic chemistry of pi bonds.
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To a first approximation they can be seen as orthogonal to the sigma frame
work.
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For many reactions the pi-orbitals determine the reactivity of the molecule.
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The electrons in these orbital are often the HOMO because of poor less
efficient parallel overlap in comparison to the head on overlap of the
sigma framework.
We can think of these pi-molecular orbitals as made up of atomic p orbitals.
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The molecular wave function for the MO1 might look like the following,
Ψ1 = c11χ11 + c12χ12 + c13χ13 + c13χ13+ c14χ14+ c15χ15.
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Steps to step up the solution for LCAO -> MO
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1. Draw a Lewis structure of the molecule to be analyzed. Number the atoms to provide
you with a reference point. Write an n x n determinant for n AOs in the molecule.
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2. Label the columns with the atomic orbitals. Label column 1, χ1 and column 2, χ2 etc.
Label the rows in the same manner: row 1, χ1 and row 2, χ2 etc.
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3. For all elements of the determinant, arc, corresponding to a row and a column, when r =
c fill in the square with an X.
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4. The off-diagonal elements are a measure of the interaction between these two atomic
orbitals. The elements are to be filled in with zeros if the two atomic orbitals are not
adjacent. Elements in the determinant corresponding to adjacent orbitals should be filled
in with ones.
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In reality there are varying extents of orbital interaction, for now we consider
only the approximation where all non-vicinal orbital interaction is zero.
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5. Set the determinant equal to zero.
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6. Solve the determinant for X. Each value of X obtained is a molecular orbital energy.
The determinant described above is called the secular determinant.
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7. Determine the coefficients by the method of cofactors.
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8. Normalize the coefficients.
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EXAMPLES
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Single p orbital on Carbon CH3(+)
1
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H
1
H
H
1
X
= 0
Principles of Organic Chemistry
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X=0
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ψ1 = 1χ11
lecture 14, page 2
The energy saved when an electron explores the bigger box (pi-MO) versus the p AO is −β.
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Ethylene H2C=CH2
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The result when two atomic carbon p orbitals combine to make a pi bond (such as
ethylene) is trivial.
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Ψ1 = 1/SQRT(2)·χ1 + 1/SQRT(2)·χ2
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and
Ψ2 = 1/SQRT(2)·χ1 − 1/SQRT(2)·χ2
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The above equation χ1 is a p atomic orbital
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Integrating χ12 over all space x, y, z gives one.
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However if there is one electron in the molecular orbital, and we
integrate over all space we need the total integral to be unity.
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Hence we need coefficients 1/SQRT(2).
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1/SQRT(2)2 + 1/SQRT(2)2 = ½ + ½ = 1
The energy of Ψ1 the bonding orbital is –β.
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For now let’s interpret it to mean the energetic advantage of putting one electron
in two interacting orbitals versus having the electron isolated in one atomic p
orbital
1
2
1
X
1
2
1
X
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H
1 C2
H C
H
H
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X2−1 = 0, X = 1 and X = −1
= 0
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Theses are the energies of the orbitals in terms of the β factor.
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Since X = (α − E)/|β|
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α in this equation is interpreted as the energy of the component atomic
AOs that make up the MOs.
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For now let’s set α to zero.
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If we do this, the solutions of the secular give the energies
of the MOs in terms of β.
The coefficients are expressed in terms of the energy X as a function of the
cofactors of the secular determinant (explained below).
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A11 = X and A12 = −1
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A12 is negative because the cofactors are multiplied by the
term (−1)(i+j) where i and j are rows and columns of the secular
determinants.
We can not say anything about the absolute value of the coefficients.
These can only be expressed in terms of ratios.
Principles of Organic Chemistry
•
lecture 14, page 3
C1/C2 = A11/A12 = −X
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C1 is proportional to A11 and C2 is related to A12.
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Since C2 does not function with X (the solutions of the secular
determinant) let C2 equal 1.
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At this point it would not matter which real number
you chose for C2.
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Below I construct a table of the coefficients:
X
A11 (-X) || cn1
A12 (1) || cn2
-1
1
1
1
-1
1
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ψ1 = 1χ11 + 1χ12
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ψ1 = −1χ11+ 1χ12
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o
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After normalization you would get the same
answer.
So the orbital associated with the X= −1 has C1 and C2 of same sign—
there is no node between the atomic orbitals that constitute the
molecular orbital in this linear combination
Normalize: (sumi Ci2 )0.5 = 1
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ψ1 = 1/SQRT(2)χ11 + 1/SQRT(2)χ12
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ψ2 = 1/SQRT(2)χ11 − 1/SQRT(2)χ12
Normalize by summing the squares of the non normalized orbital coefficients and then
dividing each non-normalized coefficient by the square root of the total.
I check the conditions for normalization below.
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Remember that we have previously discussed that integration over all space of
the atomic distribution function, χ equals zero.
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The cross integral χ1χ2 below equals zero because in the LCAO method we are
completely neglecting the space shared by the p-orbitals.
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This is an approximation that predicates on this cross integral being
small compared to the integral of χ12.
Principles of Organic Chemistry
lecture 14, page 4