Question

(a) Draw all of the pi molecular orbitals for (3E)-1, 3, 5-hexatriene ordering them from lowest to highest in energy. (b) Indicate which are bonding and which are antibonding. (c) Indicate the number of electrons that would be found in each in the ground state for the molecule. (d) Label the HOMO and LUMO.

Answer 1

(a) The molecular orbitals of 1,3,5-hexatriene are represented as 'V function and are ordered from lowest energy to highest energy-ie. from Ψι-Ψ6 where Ψι is of lowest energy and Ψ6 is of highest energy. LUMO Ps LUMO HOMO 4 HOMO Y3 GROUND STATE EXCITED STATE (b) The bonding orbitals are below the dashed line and antibonding orbitals are above the dashed line. This means Vi to V; are bonding orbitals and V4 to Vf are antibonding orbitals. (c) There are 6 pi electrons which occupy the three bonding orbitals in the ground state and 6 pi electrons that occupy the 4 orbitals in the excited state. The arrows indicates the number of electrons (d) The HOMO and LUMO orbitals in ground and excited state are given above in the picture.