Cyclic Hydrocarbons

The secular equations, in the Huckel approximation for a cyclic hydrocarbon are

$$\beta c_{i,r-1}+(\alpha-\epsilon_i)c_{ir}+\beta c_{i,r+1}=0$$ (72)

 where each orbital is expanded in the $\pi$ atomic orbitals $\phi_i=\sum_r^Nc_{ir}\pi_r$, and cyclic symmetry implies c_ir=c_i,r+N. We show by substitution that two solutions are possible

$$c_{ir}=\cos\frac{2\pi ir}{N},\ \ {\rm or}\ c_{ir}=\sin\frac{2\pi ir}{N}$$ (73)

 Substituting the first into the secular equation yields

$$\beta[\cos\frac{2\pi i(r-1)}{N}+\cos\frac{2\pi i(r+1)}{N}]+(\alpha-\epsilon_i)\cos\frac{2\piir}{N}=0$$ (74)

 Using $\cos(a+b)=\cos a\cos b-\sin a\sin b$, and cancelling $\cos\frac{2\pi ir}{N}$ gives

$$\alpha-\epsilon_i=-2\beta\cos\frac{2\pi i}{N}$$ (75)

 The argument similarly follows for $\sin\frac{2\pi ir}{N}$. Note that each solution obeys the periodic boundary condition. Thus each energy level appears to be degenerate. However for i=0. the lowest orbital only has $\cos$ solution, and $i=\frac{N}{2}$ only has the $\cos$ solution if N is even. Thus

for N=4, the energy levels are
 
 

 
 
 
 
 
 

for N=6, the energy levels are

There is another theoretical model for a hydrocarbon, called the mobius hydrocarbon, for which the periodic condition is c_ir=-c_i,r+N. For this problem the solutions are

$$c_{ir}=\cos\frac{2\pi (i+\frac{1}{2})r}{N},\ \ {\rm or}\ c_{ir}=\sin\frac{2\pi (i+\frac{1}{2})r}{N}$$ (76)

 with energy levels given by

$$\alpha-\epsilon_i=-2\beta\cos\frac{2\pi (i+\frac{1}{2})}{N}$$ (77)

 for N=4, the energy levels are for N=6, the energy levels are

The interesting observation is that the regular N=4 hydrocarbon gives the TS orbital energy level diagram for the disrotatory ring opening of cyclbutene, and the mobius N=4 hydrocarbon gives the TS orbital energy level diagram for the conrotatory ring opening.

1998-09-23