In a molecule that has both an alkene (C=C) and carbonyl (C=O) that are NOT conjugated, such as pent-4-en-2-one, how do you determine which π* anti-bonding orbital is the LUMO?

Essentially, I'm trying to reconcile two statements that I thought were true but apparently contradict here:

  1. Increased electronegativity of an atom lowers the energy of its molecular orbitals. This lead me to think that π* C=O is the LUMO because the higher electronegativity of oxygen would lower the energy of the anti-bonding orbital (as well as the bonding orbital).
  2. The lower the energy of the bonding MO, the higher the energy its anti-bonding MO is. This leads me to think that π* C=C would be the LUMO because π C=C is the highest energy bonding orbital, so π* C=C would be the lowest energy anti-bonding orbital.

Any help you could provide would be GREATLY appreciated!!

  • 1
    $\begingroup$ My guess is that the HOMO is the C=C pi bond whereas the LUMO is the C=O pi* bond. This would make sense as the electrophiles attack C=C first, and nucleophiles attack C=O first. Note that one molecule can have only one HOMO and only one LUMO. $\endgroup$
    – S R Maiti
    May 26, 2021 at 20:11

1 Answer 1


I think here you are missing the point on what exactly a HOMO and a LUMO are. They are descriptions of the energy splitting that occurs when quantum wave functions constructively, or destructively, interfere with eachother when in proximity. Because this system is not conjugated, the alkene and the carbonyl with both INDIVIDUALLY have their own HOMO and their own LUMO. In the case of both of these systems, the HOMO will be the pi bonding orbital and the LUMO will be the pi* anti-bonding orbital. Now, when it comes to which of these LUMOs will be lower in energy, it honestly will depend on the compound. Electronegative atoms, electron withdrawing groups, etc. narrow the energy gap between bonding and antibonding orbitals, making the LUMOs more reactive. So in this particular example, the carbonyl will probably have a larger pi* coefficient and thus be lower in energy. For more complex systems, reactivities need computer algorithms, clever thinking, or just experimentation to determine which energies are lower.

  • $\begingroup$ There is a contradiction in terms, here. Namely and literally a molecule can't have two highest (lowest) (un)occupied molecular orbitals. Beside this remark, what you say is pretty much useful to OP and stress that we should not be assertively pedantic when it comes to molecular orbitals.... Often this type of questions receive comments like "the orbital are those of the whole molecule" and stuff like that. But reality counts. $\endgroup$
    – Alchimista
    May 26, 2021 at 10:53

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