Module 2 : Molecular Structure
Lecture 9 : Polyatomics
  9.6

Bonding in butadiene and benzene

It is instructive to study the bonding in unsaturated compounds as it gives good insights into the nature of MOs. It is convenient to study the bonding independent of the sigma bonds which can be treated as constituting the "rigid" framework of the molecule. In ethene, for example, the bond is in the plane perpendicular to the molecular plane. In butadiene, all the carbons and hydrogens lie in a plane and all the bonds in the plane are bonds. The bonds are in a plane perpendicular to this plane and are formed from the four p orbitals (one on each of the four carbon atoms) perpendicular to the plane. The four bonds that can be constructed from the four p orbitals (represented as , , and in the following equation) are shown in the following figure (fig 9.10).

 

Figure 9.10 (a) The molecular orbitals of butadiene
 
 

Figure 9.10 (b) Top view of the MOs and (c) The energy level diagram.

The equations for the 4 orbitals are as follows.

= 0.37 + 0.60 + 0.60 + 0.37

(9.6 a)
= 0.60 + 0.37 - 0.37 - 0.60 (9.6 b)
= 0.60 - 0.37 - 0.37 + 0.60 (9.6 c)
= 0.37 - 0.60 + 0.60 - 0.37 (9.6 d)
 
Observe that has no node perpendicular to the molecular plane. , and have one, two and three nodes respectively. The energies of and are above the energies of the p orbitals and hence they are antibonding. Although butadiene is often represented as CH2=CH--CH=CH2 , wherein the two bonds are shown as equivalent and distinct, the correct bonding picture is that there are two electrons in and two in which are not degenerate (not having the same value of energy).