Electronic transition in organic compounds

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The discussion focuses on the understanding of different types of pi to pi* transitions in organic compounds, specifically the K-band, B-band, and E-band. K-bands are associated with pi to pi* transitions in non-cyclic conjugated systems like ethylene. B and E bands are linked to the group representation conventions of the D6h point group, with B bands representing transitions from the A1g ground state to singly degenerate excited states (B1u and B2u), and E bands corresponding to transitions into the doubly degenerate E1u state. The conversation also highlights that benzene has two E bands (E1 and E2) due to its symmetrical structure, which can change with the introduction of functional groups that disrupt this symmetry. The discussion clarifies that R-bands refer to n to pi* transitions. Overall, the exchange provides a clearer understanding of these transitions and their significance in aromatic compounds.
Eureka99
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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.
Thanks in advance
 
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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.
 
TeethWhitener said:
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.
Really? I didn' t know that, but I guess I'll have to try to understand it anyway for the exam :sorry:
Thanks for the response :)
 
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.)
 
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TeethWhitener said:
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.)
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!
 
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