# HOMO Odd, LUMO Even: Why does the affected orbital change depending on the carbon? How are carbons numbered?

While looking at how substituents affect the absorption spectra of molecules, I came across this:

Substituents alter the emitted wavelengths of these molecules by affecting the electron densities. Electron Withdrawing Groups (or EWGs) withdraw electron density, thereby making the arrangement more stable. The energy level of the orbital affected subsequently lowers, causing a typically-noticeable red shift in the maximum wavelength. Alternately, Electron Donating Groups (EDGs, for short), add to the electron densities of the molecule. This causes the affected orbital to be less stable, and requires more energy to obtain. The orbital affected depends on the placement of the substituent. A useful mnemonic device in this matter is HOLE: HOMO Odd, LUMO Even. This explains that the affected orbital alternates between odd and even placements.

Why does the orbital affected depend on the placement of the substituent? I thought substituents that caused bathochromic shifts just extended conjugation and hence reduced the HOMO-LUMO gap? Is this odd-even rule only applicable to azulene, or can it be applied to other chromophores?

Also, why are the carbons numbered the way they are? Here's a picture with the numbering. Shouldn't the carbons with the question mark also be counted?

• This is more general than extending conjugation. It is a kind of rule that applies always. Then is matter of the extent of the bathochromic ecc effect. Eg compare benzene and chlorobenzene – Alchimista Oct 26 '17 at 14:56
• The numbering question is unrelated. It could make for a question of its own. Therefore, I am voting to close as too broad (asking too many distinct questions at once) – Jan Oct 27 '17 at 4:17
• I intended for the numbering question to add on to the even-odd question, since if it's numbered differently, then an odd carbon might become an even carbon etc. From Tyberius' answer, I now understand that the even-odd behaviour is due to the shape of the HOMO and LUMO rather than an intrinsic property of odd/even carbons. I'm sorry that the question was phrased too broadly. – oswinso Oct 27 '17 at 7:02
• It's not too broad in my opinion (and others, too), don't worry about this. If the answer by Tyberius helped you, please consider accepting it. – Martin - マーチン Oct 27 '17 at 12:31

Notice the LUMO has no density on carbons $1$ or $3$ and (while it is a little difficult to see in this image) little density on carbons $5$ and $7$. Similarly the HOMO has little to no density on any of the even carbons. So, if we want to affect the energy of these orbitals, we have to add a substituent to a carbon that has electron density from that orbital. The odd-trend is likely not a general rule, but if you know the HOMO and LUMO reside mainly on different carbons, you can assume that selective functionalization of those carbons will affect the orbital of interest. I say "assume" because you should always keep in mind that the orbital picture is an approximation and that once you make a substitution you have to check to make sure you haven't perturbed the system enough where your new compound doesn't have frontier orbitals that look like this anymore.