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An Introductory to the Physics of Organic Chemistry

Building Bridges to Knowledge

Photo of Camp Fire Michaelle Cadet

The information in this paper is generally reserved for a more advanced chemistry student. It is included as a stand-alone paper, because it introduces the beginning organic chemistry student to the physics and mathematics inextricably associated with organic chemistry. This paper presents three concepts of physical organic chemistry that may help students understand the connection between physics and chemistry. Organic chemistry is not magical, but follows a logical sequence of well thought out scientific principles including mathematical formulations that give reasons and

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validity to chemical reactions and predictions of physical parameters.

Students taking a beginning course in organic chemistry are not burden with mathematics and physics; however, mathematics is the language of chemistry, and, with physic, can explain some important aspects of the behavior of organic molecules. Generally, these concepts and principles are reserved for more advanced students. Nevertheless, some exposure to the physics of organic chemistry may help students develop an appreciation for organic chemistry as a logical discipline. In addition, this paper serves to enhance the underlying principles of the discipline.

The paper will discuss three important concepts that give insight into the physics of organic chemistry. These concepts are the Hückel Molecular Orbital Theory (MOT); pericyclic reactions, and an alternant method for ascertaining electron densities for delocalized charges in hydrocarbons systems.

Hückel Molecular Orbital Theory

The concepts and principle of Molecular Orbital Theory (MOT) have approximately eighty years of history. The Schrödinger equation discussed in the paper titled “Basic Principles in Organic Chemistry” is difficult to solve for the “p” atomic orbitals and the π molecular orbital frame. The Hückel MOT along with Valence Bond Theory are relatively simple mathematical models for obtaining eigenvalues and eigenfunctions for π molecular orbitals of conjugated organic molecules. Even with these simple mathematical approaches, the

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operations are generally beyond the mathematical abilities of many students taking an introductory organic chemistry course.

Using Hückel’s approach for treating sections of quantum mechanical equations that are difficult to solve as parameters, and expressing the energies of molecular orbitals with those parameters will avoid solving complicated mathematical expressions.

The Hamiltonian operator for the Schrödinger equation is the sum of the kinetic and potential energies. The Hamiltonian operates on the eigenfunction (wavelength) to give an eigenvalue (energy) of the electron in an atomic or molecular orbital and the eigenfunction back again. Using the Hückel approximations, the eigenfunction, Φ, can be obtained through the linear combination of atomic orbitals to form molecular orbitals. The equations for these approximations can be represented by equation 21.1.

Equation 21.2

where j = 1,2,3,…,n; i = 1,2,3,…,n; Φi equals the atomic orbitals; and cji equals the coefficients for each atomic orbital.

For example, a molecule consisting of two atomic orbitals can be described by the following two molecular orbitals.

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The above diagram illustrates the linear combination of atomic orbitals Φ1 and Φ2, that form molecular orbitals Ψ1 and Ψ2.

The two equations that lead to mathematical expressions for the two molecular orbitals would be represented by equations 21.2 and 21.3.

Equation 21.2

Equation 21.3

Once the equations have been established, the best setof values for cji are obtained by the variation principle. The variation principle states that any wave function other than the correct wave function will yield a value for the energy of the orbital that is numerically greater than the true value. This can be mathematically accomplished in the following manner.

1 11 1 12 2ψ = C Cφ φ+

ψ 2 = c11φ1 + c12φ2

1 1Hψ = Eψ

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To make the eigenfunction as good an approximation as possible relevant to the true eigenfunction, the parameter ci should be selected such that E is made as small as possible. Minimizing E with respect to each possible coefficient does this. This can be done by equations analogous to equation 21.4.

Equation 21.4

Equation 21.4 can be rearranged to equation 21.5

Equation 21.5

1 1 1 1ψ Hψ = ψ Eψ

1 1 1 1ψ Hψ dτ = ψ Eψ dτ∫ ∫2

1 1 1ψ Hψ = Eψ∫ ∫

1 1

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ψ Hψ dτ = E

ψ dτ∫∫

1 1true value2

1

ψ Hψ dτ = E and E E

ψ dτ≥∫

∫

i i j j

i i j j

(C ) H ( C ) dτE =

(C )( C ) dτi j

i

φ φ

φ φ

∑ ∑∫∑∫

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Substitution of

Would give equation 21.6

Equation 21.6

The expression for the derivative of equation 21.6 with respect to ci would give rise to equation 21.7

Equation 21.7

i j i ji j

i j i ji j

C C H dτE =

C C dτ

φ φ

φφ

∑∑ ∫∑∑ ∫

ij i j

ij i j

H for H dτ

and

S for dτ

φ φ

φφ

∫

∫

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where i = 1,2,3,…,n

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Equation 21.7 is the resulting variation equation

Now, let’s apply the variation equation to an organic compound, the allyl carbocation, +CH2CH=CH2. Since there are three carbon atoms involved in this “pi” system, there are three equations leading to the solution of the three molecular orbitals in the allyl system Where j = 1,2, and 3.

These systems can be diagramed in the following manner with their associated eigenfunctions or wavelengths.

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The equations are solved by separating each element ”Hij + ESij” and placing them in a matrix. The matrix is converted into a determinant that is set equal to zero.

Once the values for E, Hij ; and Sij are obtained, they can be substituted back into the equations to solve for cij, i.e., c11; c12; c13; c21; c22; c23; c31; c32; and c33.

The Solution Using the Hückel MOT Method

The Solution Using the Hückel MOT Method doesn’t use numerical method values for Hij and Sij. The integrals are treated as the

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parameters α and β with the following assumptions:

Hii = α , the coulomb integral, i.e., energy of an electron in a carbon 2p atomic orbital

Hij = β , if “i” is adjacent to “j,” then Hij = β; if “i” is not adjacent to “j,” then Hij = 0

Sii = 1; Sij = 0 for i ≠ j

Sigma bonds are localized; therefore, they can be considered as a rigid framework for the π system.

All Hii values are equal; however, if a heteroatom atom is in the system, than a different value must be used instead of α.

The β resonance integral, the energy of interaction between two atomic orbitals on adjacent atoms, i.e., “i” and “j” are equal. This is a major assumption.

If i=j then Sij = 1. This is justifiable, because

,

the probability of finding an electron within a volume element.

Returning to the allyl system,

2dτ 1iφ =∫

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and using the assumptions above, the matrix would be

Each element in the matrix can be divided by β

Substituting x where

would give a matrix for the energy of the three energy levels of the allyl framework

The solution to this matrix is

α - Ex =

β

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X3 – 2x = 0

x(x2 -2) = 0

x = 0 and x2 -2 = 0

Therefore,

x = 0; x = 1.414; and x = -1.414

β is inherently negative; the three energy levels for the three molecular orbitals would be:

E= α -1.414β

E = α

E= α + 1.414β

A similar approach can be used for 1,3-butadiene where the matrix would be a 4 by 4 matrix.

1,3-butadiene

The determinant for the butadiene system would be:

α - Ex =

β

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The solution to this matrix would be x4 -3x +1=0

The four energy levels for butadiene are

E= α - 1.62 β

E= α -0.62 β

E= α + 0.62 β

E= α + 1.62 β

Butadiene has four electrons in the bonding molecular orbitals; therefore, the total π energy would be 2(α + 0.62 β) + 2 (α + 1.62 β) = 4α + 4.48 β

The delocalization energy resulting from electron interacting in the 1,3-butadiene system is equal to the localization energy minus the π energy which is equal to the number of electrons times (α + β). The π energy is 4(α + β) = 4α + 4 β , and the localization energy is 4α + 4.48 β; therefore, the delocalization energy (4α + 4.48 β)- (4α + 4) = 0.48 β. The energy is given in terms of Hij where “i” is adjacent to “j.”

In the case of the allylic system

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Allylic system

The localization energy is the energy of the electrons when the electrons are localized, i.e., they are not involved in resonance. This energy is given by equation 21.8

Equation 21.8

Where nπ equals the number of electrons in the double bonds, and ni equals the number of unpaired electrons.

The energy of the resonance stabilized π system is Eπ and is given by equation 21.9.

Equation 21.9

The stabilization energy resulting from electron delocalization is Eπ - Elocalization; therefore, for

The allyl carbocation

localization π iE = n (α+β) + n α

π j jj

E = n E∑

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The stabilization energy is 2(α + 1.414 β) - 2(α + β) = 2 α + 2.828 β - 2 α – 2 β= 0.828 β

The allyl radical

The stabilization energy is 2(α + 1.414 β) + α - 2(α + β) – α = 2 α + 2.828 β - 2 α – 2 β = 0.828 β

The allyl carbanion

The stabilization energy is 2(α + 1.414 β) + 2α - 2(α + β) – 2α = 2 α + 2.828 β - 2 α – 2 β = 0.828 β

A method is needed to determine cj in the equation

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If the molecular orbitals are normalized, i.e.,

then

Since

Then

for the allyl systems, i.e., the carbocation, the radical, or the carbanion, there are three molecular orbitals in these systems and the energies of the these molecular orbitals were previously determine in reference to Hii and Hij. Recall that the eigenfunctions (the wave functions) for these systems are

j jiψ c ii

φ=∑

2ψ dτ=1∫

2

) dτ = 1i iicφ⎛ ⎞

⎜ ⎟⎝ ⎠∑∫

2 2 dτ = 1i iic φ∑ ∫

2dτ=1iφ∫

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The matrix

Implies that the coefficients would have the following three equations

ci1x + ci2 = 0

ci1 + ci2x + ci3 = 0

ci2 + ci3x = 0

since

from

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These coefficients can be calculated for the allylic system.

X = ±1.414, and x= 0 for the allylic systems. X equals -1.414 for the bonding molecular orbital, because as previously mentioned β is inherently negative.

Then

(1) -1.414 c1 + c2 = 0

(2) c1 -1.414 c2 + c3 = 0

(3) c2 - 1.414 c3 = 0

From (1) 1.414 c1 = c2

Substituting this value into (3) would give 1.414 c1 - 1.414 c3 = 0, then c1 = c3

Since

c12 + c2

2 + c32 = 1 , then ,

c12 + (1.414 c1)2 + c1

2 = 1

therefore, c1 = 0.5

and c3 = 0.5

and c2 = 0.71

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Consequently,

For x = 0, using an analogous approach as above, c21 = 0.7 and c23 = -0.7; therefore,

The same approach can be used for x = +1.414

With the following results

The energies and the associated wave functions for the three molecular orbitals in the allylic system can be illustrated in the following manner:

1 11 1 12 2 13 3ψ c c cφ φ φ= + +

1 1 2 3ψ 0.5 0.7 0.5 φ φ φ= + +

2 1 3ψ 0.7 - 0.7 φ φ=

3 1 2 3ψ 0.5 - 0.7 0.5 φ φ φ= +

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This is a tedious process; therefore, an alternative method is used. This method is called the cofactor method. The cofactor method doesn’t work for degenerative (equal energy) orbitals. In considering degenerate molecular orbitals, a longer process is required. For example, the cyclopropyl system has two degenerate orbitals.

A pictorial representation of the bonding molecular orbital is

The two pictorial representations of the antibonding orbitals are

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Note that a node (a zero probability of finding the electron) divides c1 and c2 from c3 in the first of the two antibonding molecular orbitals. Also, in the second illustration, there is a node that goes through c2; therefore, in this structure, c2 is zero. One way to determine the molecular orbitals of cyclic molecules is to inscribe a flat geometric figure of the molecule in a circle, and where it touches the circle, a molecular orbital is formed.

The matrix for this system would be:

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The solution to this matrix is x3 - 3 x +2 = 0

The solution to this polynomial is (x-1)( x2 + x – 2) = 0

and

(x-1)(x +2)(x-1) = 0

Therefore x = 1 ; x = -2; and x = 1

This means that the three energy levels (two are degenerate) are

The coefficients for the wave functions can be calculated from

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c1x + c2 + c3 = 0

c1 + c2x + c3 = 0

c2 + c2 + c3x = 0

For x = 1 and knowing that c2 = 0

Then c3 = - c1

for normalization

c12 + 0 + c1

2 = 1

c1 = 0.7

and c3 = -0.7

Therefore, the wave function for the antibonding molecular orbital with a node through c2 would be

c1 + c2 + c3 = 0

c1 + c2 + c3 = 0

c2 + c2 + c3 = 0

c1 = c3 symmetry relationships

antiboding 1 3ψ 0.7 - 0.7 φ φ=

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c1 + c2 + c1 = 0

- 2c1 = c2

Using the normalization equation,

c1 = 0.4

c3 = 0.4

c2 = 0.8

Finally, the values for x = -2 can be calculated.

The three equations would be

-2c1 + c2 + c3 = 0

c1 -2c2 + c3 = 0

c1 + c2 -2 c3 = 0

c1 = c3

c1 = c3 = 0.6

c2 = 0.6

antiboding 1 2 3ψ 0.4 - 0.8 + 0.4 φ φ φ=

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As indicated earlier, the Hückel MOT method is time consuming; therefore a short cut needs to be used as long as degenerate orbitals are not involved. Such a method is called the cofactor method. The cofactor method is a simple ratio. If “i” is an odd number, then the cofactor would be:

If ”i” is an even number, then the cofactor would be:

The top row of the determinant is generally used and identified as “1,” but this is not essential.

Let’s use 1,3-butadiene as an illustration in using the cofactor method.

1,3-butadiene

bonding 1 2 3ψ 0.6 + 0.6 + 0.6 φ φ φ=

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Four 2p atomic orbitals in the π frame of 1,3-butadiene linearly combine with form four molecular orbitals. Two of the molecular orbitals are bonding molecular orbitals and two are antibonding molecular orbitals. The energies of the four molecular orbitals were previously determined.

The determinant for butadiene is

The cofactor method is

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x has four values +1.62; -1.62; +0.62; and -0.62

For x = -1.62

Using

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The eigenfunction for the lowest energy bonding molecular orbital in 1,3-butadiene is

Similar calculations can be done for x = -0.62

To give a wave function with the following coefficients

The solution for the coefficients of wave function for x = +0.62 , an antibonding molecular orbital is

1 1 2 3 4ψ 0.37 0.592 0.60 + 0.37 φ φ φ φ= + +

2 1 2 3 4ψ 0.60 0.37 - 0.37 -60 φ φ φ φ= +

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The solutions for the coefficients of the wave function for x = -1.62, an antibonding molecular orbital is

Following are the summaries of these results:

Including the nodal plane of the carbon skeleton, this bonding molecular orbital has one node. A node is the zero probability of finding the electron within a region of space.

3 1 2 3 4ψ 0.60 - 0.37 - 0.37 +60 φ φ φ φ=

4 1 2 3 4ψ 0.37 - 0.60 + 0.60 -37 φ φ φ φ=

1 1 2 3 4ψ 0.37 0.592 0.60 + 0.37 φ φ φ φ= + +

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Including the nodal plane of the carbon skeleton, this bonding molecular orbital has two nodes.

Including the nodal plane of the carbon skeleton, this antibonding molecular orbital has three nodes.

2 1 2 3 4ψ 0.60 0.37 - 0.37 -60 φ φ φ φ= +

3 1 2 3 4ψ 0.60 - 0.37 - 0.37 +60 φ φ φ φ=

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Including the nodal plane of the carbon skeleton, this antibonding molecular orbital has four nodes.

The magnitude of c is indicative of the contribution of each atomic orbital to the eigenfunction of the molecular orbital and is instrumental in sketching the molecular orbital.

Pericyclic Reactions

Pericyclic reactions involve the making and breaking of bonds in a single concerted transition. This may lead to a cyclic or noncyclic structure. These reactions are not affected by changes in solvents unless the reaction involves charges. Pericyclic reactions can also be affected by the presence of a catalyst.

Following are some examples of pericyclic reactions:

(1) An electrocyclic reaction is a reaction where one π bond is converted into a σ bond or vice versa.

4 1 2 3 4ψ 0.37 - 0.60 + 0.60 -37 φ φ φ φ=

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(2) A cycloaddition reaction is a cyclization reaction where two or more unsaturated molecules (or parts of the same molecule) form a cyclic product. The products of cycloaddition reactions exhibit a decrease in the number of π bonds and an increase in the number of σ bonds.

(3) A sigmatropic reaction is a reaction where a sigma bond is changed to another sigma bond in an uncatalyzed intramolecular process.

(4) A cheletropic reaction is a reaction where one atom on one of the reagents acquires two new bonds.

An example of a pericyclic reaction is the ring closure of (2E,4E)-hexadiene. (2E,4E)-hexadiene undergoes ring closure photochemically (light, hѵ) to form cis-3,4-dimethylcyclobutene. (2E,4E)-hexadiene undergoes ring closure to form trans-3,4-

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dimethylcyclobutene with heat (∆).

(2E,4E)-hexadiene cis-3,4-dimethylcyclobutene

(2E,4E)-hexadiene trans-3,4-dimethylcyclobutene

How can these observations be explained? The best way to explain these transitions is to consider how the reactions proceed. The ring opening and closing reactions occur through a process referred to as either conrotatory closure or disrotatory closure. Conrotatory means that the “p” orbitals resulting in the formation of the sigma bond rotate in a clockwise direction. Disrotatory means that the “p” orbitals, resulting in the formation of the sigma bond, rotate in opposite directions, i.e., one rotates clockwise and the other rotates counterclockwise. However, before a conclusion is made whether

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the closure is conrotatory or disrotatory, the appropriate molecular orbital must be identified that leads to a conrotatory or disrotatory closure. The conrotatory or disrotatory closure must be symmetry allowed, i.e., the atomic orbitals forming the sigma bond must have the same phase.

The π molecular orbital backbone for (2E,4E)-hexadiene can be illustrated in figure 21.1.

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Figure 21.1 π Molecular orbitals for (2E,4E)-hexadiene

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The highest occupied molecular orbital (HOMO) is the molecular orbital that needs to be considered in determining if the ring closure to form the cyclobutene structure is thermally allowed. Note that when the ψ2 molecular orbital is used to form the four member-ring, cyclobutene, the phases of the ψ2 “p” atomic orbitals on carbon atoms 2 and 5 must rotate clockwise (figure 21.2) to have the appropriate symmetry for the formation of the sigma bond. This is illustrated in figure 21.2.

Figure 21.2 Conrotatory rotation of the ”p” orbitals to form the sigma bond

The “p” orbitals rotate in the same direction, conrotatory, to form the desired product. The conrotatory rotation of the molecular orbitals results in the product of the thermally allowed ring closure of (2E4,E)-hexadiene forming trans-3,4-dimethylcyclobutene

Photochemical reactions occur by way of free radical processes. If an electron in the bonding molecular orbital of (2E,4E)-hexadiene is excited to the lowest unoccupied molecular orbital (LUMO). A diagram with the electron promoted to LUMO is illustrated in figure 21.3 .

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Figure 21.3 An electron is excited to the lowest unoccupied molecular orbital (LUMO), ψ3.

The lowest unoccupied molecular orbital (LUMO) is the molecular orbital that is considered in determining if the ring closure to form the cyclobutene structure is photochemically allowed. Note that when the ψ3 molecular orbital is used to form the four member-ring, cyclobutene, the phases of the ψ3 “p” orbitals on carbon atoms 2 and 5 are in phase to form a sigma bond, but must rotate

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counterclockwise (figure 21.2) to have the appropriate symmetry for bond formation. This is illustrated in figure 21.4.

Figure 21.4 Disrotatory(one clockwise and the other counterclockwise) rotation of the “p” orbitals to form the sigma bond

These observations lead to some interesting generalizations. Systems that have 4q electrons where q equals I,2,3,…undergo pericyclic reactions that follow a conrotatory pathway. If the number of “p” orbitals forming the π frame molecular orbitals is equal to 4q where q is equal to 1,2,3…., then the pericyclic reaction is conrotatory. (2E,4E)-hexadiene has four electrons in the four “p” orbitals that form the four molecular orbitals of the π frame for (2E,4E)-hexadiene; therefore, 4q = 4 and q = 1. A 4q system will close in a conrotatory manner with heat. This process has been traditional referred to as 4q∆con. The “4q∆con” phraseology means that the 4q systems will proceed with heat in a conrotatory manner.

For a photochemical ( hѵ) reaction, the closure reaction proceeds in a disrotatory manner in order to be symmetry allowed.

The formation of the sigma bonds in Figures 21.2 and 21.3 indicate that the bonds are formed through conrotatory or disrotatory

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processes, and these rotations determine the stereochemistry of the attached substituents. If the rotation is conroratory, the substituents, A, B, C, and D would have the following designated orientations.

Conrotatory rotation must occur through a symmetry allowed orientation

Disrotatory rotation must occur through a symmetry allowed orientation

Note the stereochemistry in each process. 4q+2 Systems undergo ring closure in an opposite manner, i.e., ring closure takes place photochemically (in light) in a conrotatory process, and in the presence of heat by a disrotatory process.

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As indicated earlier, both 4q and 4q +2 reactions occur with conservation of orbital symmetry. So orbital formation must be symmetry allowed. If they are symmetry forbidden, then the reaction is very difficult to occur or will not occur. Substituents will not affect ring closure; however, if steric crowding prevents conrotatory closure, then the reaction will take place by a nonconcerted radical cleavage or by a symmetry forbidden process.

The molecular orbitals for the π frame of (2E,4Z,6E)-octatriene are indicated in figure 21.5.

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Figure 21.5 The π molecular orbitals of the (2E,4Z,6E)--octatriene system

Careful observations indicate that ψ1 has zero nodes, excluding the nodal plane of the carbon atoms; ψ2 has one node, excluding the

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nodal plane of the carbon atoms; ψ3 has two nodes, excluding the nodal plane of the carbon atoms; ψ4 has three nodes, excluding the nodal plane of the carbon atoms; ψ5 has four nodes, excluding the nodal plane of the carbon atoms; and ψ6 has five nodes, excluding the nodal plane of the carbon atoms.

The highest occupied molecular orbital (HOMO) of this system is the ψ3 molecular orbital. In order for the ψ3 “p” orbitals to have the correct phase when forming the sigma bond of the substituted cyclohexane ring system, one orbital must rotate clockwise and the other orbital must rotate counterclockwise. This is a disrotatory rotation as illustrated in figure 21.6.

Figure 21.6 Disrotatory (one clockwise and the other counterclockwise) rotation of the “p” orbitals to form the sigma bond

The lowest unoccupied molecular orbital (LUMO) of this system is the ψ4 molecular orbital. In order for the ψ4 “p” orbitals to have the correct phase when forming the sigma bond of the substituted cyclohexane ring system, the “p” orbitals must rotate clockwise. This is a conrotatory rotation as illustrated in figure 21.7.

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Figure 21.7 Conrotatory rotation of the “p” orbitals to form the sigma bond

The following matrix is a review of the ring formations of pericyclic reactions.

System Reaction Condition Rotation

4q heat (∆) conrotatory

4q+2 heat (∆) disrotatory

4q photochemical (hѵ) disrotatory

4q+2 photochemical (hѵ) conrotatory

As indicated previously, the 4q + 2 systems undergoing pericyclic reactions proceed in the opposite manner as the 4q systems. Systems with 4q + 2 electrons where q equals I,2,3,…undergo pericyclic reactions that follow a disrotatory pathway for thermally allowed reactions. If the number of “p” orbitals forming the π frame molecular orbitals is equal to 4q + 2 where q is equal to 1,2,3…., then the pericyclic reaction is disrotatory under thermal conditions. (2E,4Z,6E)--Octatriene has six electrons in the “p” orbitals that form the six molecular orbitals of that system; therefore, 4q + 2= 6 and q = 1. A “4q + 2” system will close disrotatorially with heat. This

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process has been traditionally referred to as (4q+ 2)∆dis. The “(4q+2)∆dis” phraseology means that 4q +2 systems will proceed with heat in a disrotatory fashion.

Ring opening occurs in an analogous manner as ring closure. If ring closure is conrotatory, then ring opening will be conrotatory. If the ring closure is disrotatory, then ring opening will be disrotatory.

4+2 and 2+2 Cycloaddition Reactions

Theoretically, butadiene reacts with ethylene (ethene), a 4 + 2 cycloaddition reaction, by a thermal process to form cyclohexene.

This reaction is referred to as the Diels-Alder reaction (paper on Dienes, Building Bridges to Knowledge). As mentioned in the paper on Dienes, Building Bridges to Knowledge, the reaction is much more effective when the dienophile contains electron releasing groups and the diene contains electron withdrawing groups.

The Diels Alder reaction is a thermally allowed reaction, because the symmetry of the HOMO of butadiene is in symmetry with the LUMO of ethylene. This phase symmetry is referred to as being suprafacial. If the symmetry of the reactants exhibited antarafacial characteristics, then the reaction is thermally forbidden.

This phenomenon is represented by figure 21.8

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HOMO of butadiene is suprafacial with the LUMO of ethene; therefore, the reaction is thermally allowed (figure 21.9).

Figure 21.9 Interaction of HOMO of butadiene with LUMO of ethylene, a suprafacial arrangement, resulting in the thermal formation of cyclohexene.

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Reactants that are not suprafacial, but antarafacial will not undergo pericyclic reactions thermally, but will undergo pericyclic reactions photochemically.

For example, ethene will react with itself, a 2+2 cycloaddition reaction, to form cyclobutane photochemically, not thermally. Figure 21.10 shows that the HOMO molecular orbital has an antarafacial arrangement with the LUMO of another ethene molecule; therefore, a free radical reaction where an electron is promoted to the LUMO will provide the correct symmetry for reactivity through a photochemical process for the formation of cyclobutane.

Figure 21.10 the molecular orbitals of the two ethylene molecules

The HOMO of one ethylene is antarafacial to the LUMO of the other ethylene molecule as indicated in Figure 29.11.

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Figure 21.11 The antarafacial arrangement of the HOMO of one ethylene molecule with the LUMO of another ethylene molecule; therefore, the reaction between these two molecules is thermally not allowed, but photochemically allowed.

If an electron is excited, a photochemical process, to the LUMO molecular orbital, than the appropriate phases for bond formation would exist for the formation of cyclobutane. This is illustrated in figure 21.12.

Figure 21.12 The photochemical allowed formation of cyclobutane, and where the phases of the orbital will have the correct orientation for formation of the sigma bonds in cyclobutane.

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The formation of cyclobutane from ethene occurs photochemically. This photochemical process involves the excitation of an electron from LUMO to HOMO. The following illustration demonstrates the correct symmetry for forming sigma bonds in cyclobutane.

Correlation diagrams

Correlation diagrams are another way of determining if a reaction is thermally allowed. The Diels-Alder reaction of 1,3-butadiene with ethene is thermally allowed. This can be determined using a correlation diagram. The diagram involves examining the molecular orbitals of the reactants with the molecular orbitals of the products. The molecular orbitals of the reactants are labeled as symmetrical or anti-symmetrical. The same is done for the products. If there are no cross overs between the bonding molecular orbitals and the anti-bonding molecular orbitals, then the reaction is thermally allowed. If there is a crossover between the bonding and anti-bonding molecular orbitals, then the reaction is thermally forbidden. The thermally forbidden reaction may be photochemically allowed. Figure 21.13 is representative of this methodology.

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Figure 21.13 Correlation diagram of reaction of 1,3-butadiene with ethene

None of the symmetry allowed transitions cross the dotted line representing the separation from bonding and antibonding molecular orbitals; therefore, the reaction is symmetry allowed and thermally executed.

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A similar diagram can be drawn for a reaction between two ethylene molecules. In order for symmetries to match, a line has to be drawn that crosses the boundary that separates the antibonding molecular orbitals from the bonding molecular orbitals. This means that the reaction is asymmetrical (anti-symmetrical); therefore, it is thermally forbidden. The reaction can occur photometrically (light, hѵ). The symmetry relationship (correlation diagram) between two ethene molecules the reaction of two ethene molecules is represented in figure 21.14.

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Figure 21.14 Correlation diagram of reaction between ethene and ethene

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Alternant Method for Ascertaining Electron Densities for Delocalized Charges in Hydrocarbons

An alternant hydrocarbon is a planar conjugated hydrocarbon where the carbons can be divided into starred sets such that no starred carbon is adjacent. The unstarred carbon atoms are also not adjacent. The Alternant Hydrocarbon Method (AHM) can be used to simplify calculations required to determine the coefficients of the wavelengths (eigenfunctions) that describe the molecular orbitals of cyclic and alycyclic hydrocarbons.

This section will use the Alternant Hydrocarbon Method to determine the electron density on hydrocarbon with a positive charge or negative charge that is delocalized throughout the molecule. This is done by alternately starring the carbon atoms. The numerical values on the starred carbon atoms are indicative of the extent to which that carbon atoms sustain a positive charge or negative charge.

For a carbocations, the lower the numerical value, the more positive the starred site. For carbanions, the higher the numerical value, the more negative the starred site. The unstarred carbon atoms would have an electron density of 1.0.

A value equal to 1.0 indicates that the positive charge or negative charge does not reside on that carbon atom. Once the carbon atoms are starred, then the values of the starred carbons must be zero around the unstarred atoms.

For example, the benzyl carbocation has the following resonance

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structures:

benzyl carbocation

The positive charge is delocalized throughout the molecule, but is more prevalent on one of the carbon atoms. AHM can be used to ascertain this information by way of the following process.

The structure is drawn, and the carbon atoms are starred in an alternate manner to generate the maximum number of stars.

An arbitrary lettering system is established on the starred carbon atoms in a manner that gives a sum of zero on the unstarred carbon atoms.

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-c + (-c) +2c = 0

-c +c = 0

-c +c = 0

The sum of square of the values is equal to one, i.e., the system is normalized.

Therefore, (2c)2 + (-c)2 + (-c)2 + c2 = 1

c = 0.38 and 2c = 0.76

The electron density equals 1-xi2 for carbocations.

The carbon with x = 2c would have an electron density value of 0.42 (1-.58); the carbon atoms with c would have an electron density of 0.86

The electron density at each non-stared position is 1.0

The result would be:

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The positive charge would be more prevalent on the carbon atom with the lowest electron density value.

If the intermediate is a carbanion, the electron density at each carbon atom would be 1+xi

2 . For benzyl carbanion,

benzyl carbanion

The resonance structures for benzyl carbanion are:

The values would be identical for c, but the electron densities would change, because the electron density for carbanions is 1+xi

2 ; therefore, the carbon with a x = 2c would have an electron density value of 1.58(1+.58); the carbon atoms with c would have an electron

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density value of 1.14.

The result would be:

The negative charge would be more prevalent on the carbon atom with the highest value for the electron density.

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Problems

An Introduction to the Physics of Organic Chemistry

1. Using Hückel Molecular Orbital Theory, calculate the eigenvalues and eigenfunctions corresponding to the six energy levels of benzene.

2. Calculate the charge densities on each carbon atom of the following carbocation.

Draw the resonance structures for this carbocation

3. Complete the following reactions, and give rationales for your

ψ1

ψ2 ψ3

ψ4 ψ5

ψ6

+

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answers.

(a)

(b)

(c)

(d)

4. Construct a correlation diagram for the following reaction.

hν

H H

hν

2 Δ

H

H

CH3

CH3

Δ

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From your results, determined if this reaction is thermally allowed or photochemically allowed.

+ + +

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Solutions to Problems An Introductory to Physics of Organic Chemistry

1. Using Hückel Molecular Orbital Theory, calculate the eigenvalues and eigenfunctions corresponding to the six energy levels of benzene.

ψ1

ψ2 ψ3

ψ4 ψ5

ψ6

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Divide each element in the matrix by

Let

Then the matrix becomes

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The six by six matrix can be solved in the following manner:

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These matrices can be converted into the following smaller matrices.

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Since the equation is a six degree equation, there are six values for x,

+2, -2, +1, +1, -1, and -1

and since

The eigenvalues, E, from the lowest energy to the highest energy would be

The wave functions for benzene cannot be determine using the cofactor method, because the “p” system of benzene has degenerate orbitals. The eigenfunctions for the six “p” molecular orbitals would be:

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(1)

Excluding the node at the separation of the “p” orbitals at the plane that includes the carbon atoms and hydrogen atoms, the lowest bonding molecular orbital for benzene have zero nodes.

The wavefunction for the bonding molecular orbital with energy equal to α + 2β (the molecular orbital with the lowest energy) is

(2) and (3)

The next two wavefunctions are degenerate, i.e., their eigenvalues (energy values) are equal.

Excluding the node at the separation of the “p” orbitals at the plane that includes the carbon atoms and hydrogen atoms, these two degenerate bonding molecular orbitals have two nodes each. They have a higher energy than the molecular orbital with an energy value symbolized by α + 2 β.

ϕ1 = c11φ1 + c12φ2 + c13φ3 + c14φ4 + c15φ5 + c16φ6

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The wavefunction for the bonding molecular orbital with energy equal to α + β (where c2 and c3 are 0) is

The wavefunction for the bonding molecular orbital with energy equal to α + β (where c2 and c3 are not 0) is

(4) and (5)

The next two wavefunctions are also degenerate, i.e., their energy values are equal. Excluding the node at the separation of the “p” orbitals at the plane that includes the carbon atoms and hydrogen atoms, theses two antibonding molecular orbitals have four nodes

ϕ2 = c21φ2 - c23φ3 - c24φ4 + c26φ6

ϕ3 = c31φ1 + c32φ2 + c33φ3- c34φ4 - c35φ5 - c36φ6

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each, and can be represented in the following manner:

The wavefunction for the antibonding molecular orbital with energy equal to α - β (where c2 and c3 are 0) is

The wavefunction for the antibonding molecular orbital with energy equal to α - β (where c2 and c3 are not 0) is

(6)

ϕ4 = c41φ1 - c43φ3 + c44φ4 - c46φ6

ϕ5 = c51φ1 - c52φ2 + c53φ3 + c54φ4 - c55φ5 + c56φ6

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The highest molecular orbital is the antibonding molecular orbital that has six nodes (excluding the node at the separation of the “p” orbitals at the plane that includes the carbon atoms and hydrogen atoms). This molecular orbital formed from the “p” orbitals can be represented in the following manner:

The wavefunction for the antibonding molecular orbital with the highest energy equal, α - 2β, is

The next step is to calculate the values for the coefficients cij. The eigenvalue coefficients can be calculated using the matrix

ϕ6 = c61φ1 - c62φ2 + c63φ3 − c64φ4 + c65φ5 - c66φ6

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Consequently, this matrix can be used to calculate the coefficients for the eigenfunction with energy α + 2 β where x = -2.

-2c11 + c12 + c16 = 0

c11 -2c12 + c13 = 0

c12 - 2c13 + c14 = 0

c13 - 2c14 + c15 = 0

c14 - 2c15 + c16 = 0

c11 + c15 - 2c16 = 0

Normalization of the coefficients is represented by

c112 + c12

2 + c132 + c14

2 + c152 + c16

2 = 1

These data can be used to obtain the values of the coefficients in the eigenfunction with energy α + 2 β where x = -2.

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c11 = c12 = c13 = c14 = c15 = c16

c112 + c11

2 + c112 + c11

2 + c112 + c11

2 = 1

c11 = (1/6)0.5 = 0.41

c11 = c12 = c13 = c14 = c15 = c16 = 0.41

Coefficients for the eigenfunction with energy α + β where x = -1 and c22 and c25 equal 0 can be calculated from

-c21 + c26 = 0

c21 + c23 = 0

- c23 + c24 = 0

c23 - c24 = 0

c24 + c26 = 0

c21 - c26 = 0

Therefore, c21 = c26 ; c21 = -c23 ; c23 = c24 ; and -c24 = c26

c212 + c23

2 + c242 + c26

2 = 1

c212 + c21

2 + c212 + c21

2 = 1

c21 = (1/4)0.5 = 0.50

c23 = -0.50; c24 = 0.50; c26 = -0.50

ϕ1 = 0.41φ1 + 0.41φ2 + 0.41φ3 + 0.41φ4 + 0.41φ5 + 0.41 φ6

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Coefficients for the eigenfunction with energy α + β where x = -1 and c22 and c25 do not equal 0 can be calculated from

-c31 + c32 + c36 = 0

c31 -c32 + c33 = 0

c32 - c33 + c34 = 0

c33 - c34 + c35 = 0

c34 - c35 + c36 = 0

c31 + c35 - c36 = 0

c31 = -c34; 2c31 = c35; c31 = -c36; c31 = c33; and c32 = 2c31

normalization

c312 + c32

2 + c332 + c34

2 + c352 + c36

2 = 1

c312 + 4c31

2 + c312 + c31

2 + 4c312 + c31

2 = 1

c31 = (1/12)0.5 = 0.3

c32 = 0.6; c33 = 0.3; c34 = -0.3; c35 = 0.6; and c36 = -0.3

Coefficients for the eigenfunction with energy α - β where x = 1 and c42 and c45 equal 0 can be calculated from

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c41 + c46 = 0

c41 + c43 = 0

c43 + c44 = 0

c43 + c44 = 0

c44 + c46 = 0

c41 + c46 = 0

c41 = -c46; c41 = -c43; c43 = -c44; c44 = -c46

normalization

c412 + c43

2 + c442 + c46

2 = 1

c412 + c41

2 + c412 + c41

2 = 1

c41 = (1/4)0.5 = 0.50

The coefficients for the eigenfunction with energy α- β where x = 1 and and c52 and c55 do not equal 0 can be calculated from

c51 + c52 + c56 = 0

c51 + c52 + c53 = 0

c52 + c53 + c54 = 0

c53 + c54 + c55 = 0

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c54 + c55 + c56 = 0

c51 + c55 + c56 = 0

Normalization of the coefficients is represented by

c512 + c52

2 + c532 + c54

2 + c552 + c56

2 = 1

These data can be used to obtain the values of the coefficients for the eigenfunction with energy α - β where x = 1.

c51 = c54; 2c51 = -c55; c51 = c56; c51 = c53; and c52 = -2c51

c512 + 4c51

2 + c512 + c51

2 + 4c512 + c51

2 = 1

c11 = (1/12)0.5 = 0.30

The coefficients for the eigenfunction at the highest energy level, the antibonding molecular orbital with energy α - 2β, can be obtained from the following sequence of equations.

2c61 + c62 + c66 = 0

c61 + 2c62 + c63 = 0

c62 + 2c63 + c64 = 0

c63 + 2c64 + c65 = 0

c64 + 2c65 + c66 = 0

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c61 + c65 + 2c66 = 0

Normalization of the coefficients is represented by

c112 + c12

2 + c132 + c14

2 + c152 + c16

2 = 1

These data can be used to obtain the values of the coefficients in the eigenfunction with energy α - 2 β where x = 2.

c61 = c63 = c65

c62 = c64 = c66

c61 = - c62

c61 2 + c61

2 + c612 + c61

2 + c612 + c61

2 = 1

c61 = (1/)0.5 = 0.41

c61 = c63 = c65 = 0.41

c62 = c64 = c66 = - 0.41

2. Calculate the charge densities on each carbon atom of the following carbocation.

+

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The electron density equals 1-xi2 for carbocations; where xi equals

the value of nai. For example, if ai is 0, then Xi is 0, and the electron density would be 1-(0)2 or 1. If nai is 3a, then Xi is 0.66, the electron density would be 1-(66)2 or 0.56. Therefore, the electron density at each carbon atom for

(3a)2 + (-2a)2 + (2a)2 + (-a)2 + a2 + (-a)2 = 1

9a2 + 4a2 + 4a2 + a2 + a2 + a2 = 1

a2 = 120

a = 0.22

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would be:

The lower the numerical value for the electron density, the greater the opportunity for the positive charge to be at that position. A value of 1.0 means that the positive charge doesn’t delocalize to that position.

Think about this:

Why do you think the carbon atoms with electron densities of 0.95 on the ring attached to the ring containing the carbon atom with an electron density equal to 0.56 have such high electron density values?

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Draw the resonance structures for this carbocation

Notice that the positive charge doesn’t appear on any carbon atom with an electron density equal to 1.0.

Think about this:

Why would the first resonance structure where the carbocation resides on the carbon attached to the aromatic ring system have the lowest electron density value (0.56)?

3. Complete the following reactions, and give rationales for your answers.

(a)

hν

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(b)

(c)

This 2+2 reaction is not thermally allowed, but photochemically allowed. Since there is a crossover between the bonding molecular orbitals and the antibonding molecular orbitals, then the correlation diagram suggests that the reaction must occur photochemically not thermally.

80

(d)

Δ

81

This is a 4q system; therefore, the ring system closes in a conrotatory manner.

4. Construct a correlation diagram for the following reaction.

From your results, determined if this reaction is thermally allowed or photochemically allowed.

Antibonding Molecular Orbitals

+ + +

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Bonding Molecular Orbital

No crossover between the bonding molecular orbitals to the antibonding molecular orbitals; therefore, the reaction is thermally allowed.