Weeks #9 and #10 (Due Sunday, March 14 and 21(see below))

 

 

1.     (Due 3/14) Complete the tutorial “3-Cyano-4-methylcyclohexenyl Radical” Chapter 3 of Spartan Tutorial and Users Guide.

a.     Answer any questions in the tutorial.

Methyl group is equatorial (Although picture in book displays as axial)

b.     Record the lowest energy from the MMFF conformer analysis

E = -19.2568 kcal/mol

c.     Record the final energy of the AM1 equilibrium geometry

E=33.885 kcal/mol (This was not in the original output but had to be calculated separately)

d.     Record the single point energy from the Hartree-Fock calculation

E= -361.1706 hartrees

e.     Based on reading the output estimate the time required to perform a conformational analysis using the AM1 method?

f.       

For a 500Mhz PowerBook G4 a single energy calculation using AM1 required* .1s CPU and 1.5s Wall time. So to calculate the 9 conformers as was done with MMFF it would take .9s and 13.5s. Hartree-Fock method?

For single point energy calculation: 54.1s and 1:06.6s. So for conformers would require 8:51.9s and 9:59.4s

 

*The cpu and wall time in the original output are for the AM1 geometry which involves several energy calculations during optimization. Hence to find the time for an AM1 single point calculation I had to do separate calculation

 

2.     (Due 3/21) Use Dr. Gilbert’s HMO program to model the same system (excluding the sp³ hybridized carbons). When describing the nitrogen assume it is of the (C=N-) type, so there will only be 5 pi electrons, not 7. Compare the electron densities from the HMO program with the SOMO and discuss.

Numbering:

 

C1-C2-C3-C4-N

 

HMO electron density is given by the diagonal elements of the following matrix. For example, the electron density on carbon 1 is .7783 (Note the sum of these diagonal elements must give the total # of electrons)

 

 

 

 

The SOMO predicts the electron density for the free electron to be on C1,C3 and N while HMO predicts more overall electron density on C2 versus C1 and comparable to C3. Keep in mind the HMO electron density includes contributions from all occupied orbitals.

The breakdown of each MO in terms of the AO basis is just a single column of the ‘PI MO ENERGIES’ output(See arrow).

 

 

 

 

Looking at the contribution from the third highest energy MO from HMO (where the lone electron will lie), the relative magnitude of the coefficients**2 are consistent with the SOMO except nitrogen has a lower value than the cyano-carbon.

 

Sources of these differences include the simplified basis set used in Huckel, neglect of next nearest neighbor interactions and comparing this sub-structure to the SOMO wrought from the full molecule in Spartan.