# Electronic transition in organic compounds

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1. May 31, 2017

### Eureka99

Hello everybody :)
I have a problem understanding the different types of pi -> pi * transition of organic compounds. I can't understand what's the difference between K-band, B- band, and E-bands, and I cannot find any explanation on the internet. Also I don't understand why benzene has two E bands ( E1 and E2), and Toluene only one; I also tried to see the pi molecular orbitals of different aromatic compounds, to see what pi -> pi* transitions are possible, but I found the MO's only of benzene. If anyone can explain to me the difference between the three types of band, maybe with a couple of examples, I would be very grateful.

2. May 31, 2017

### TeethWhitener

As far as I can tell, these are just really antiquated ways of talking about $\pi \rightarrow \pi^*$ and $n \rightarrow \pi^*$ transitions of various conjugated and aromatic systems.

3. May 31, 2017

### Eureka99

Really? I didn' t know that, but I guess I'll have to try to understand it anyway for the exam
Thanks for the response :)

4. May 31, 2017

### TeethWhitener

It looks like R-bands are "radical" bands, or $n \rightarrow \pi^*$. K bands are $\pi \rightarrow \pi^*$ bands in conjugated, non-cyclic systems like ethylene or 1,3-butadiene. B and E bands refer to the group representation conventions from the character table for the $D_{6h}$ point group that benzene is a member of. B bands are transitions from the $A_{1g}$ ground state into singly degenerate $B_{1u}$ and $B_{2u}$ excited states, and E bands are transitions into the doubly degenerate $E_{1u}$ state.

(Note: the B and E bands are the only ones I'm sure of. These are still standard. The R and K bands were just what I could glean from various old papers.)

5. Jun 1, 2017

### Eureka99

Oh well, now it makes sense! It also makes sense why when the benzene has functional groups that make him lost its symmetry, it loses as well the E- bands. Thank you very much, that was very helpful!

6. Jun 2, 2017

7. Jun 2, 2017