physical chemistry

[farmacista]

i hope someone can help me...how can i determine the symmetry (irreducible representation) of benzene orbitals? for the doubly degenerated pi orbitals which have two nodal plane...after C6 rotation orbitals change sign on two carbon and remain the same on four carbon...i know they have E1g symmetry so the character for the irreducible representation is 1 but why? i can't find any explanation on the book or on internet http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/orbitals/molecular/benzene.png a picture of the orbitals

 

[Pharmokinetictal]

Man it's been a while since I took physical Chem. I only do neurochem now and drug development, but I'm thinking...maybe this is a dumb answer...isn't resonance stabilization and uncertainty principle part of it? Any structure that can form a stable resonance, including enols and enolates, are in constant equilibrium with their resonance states. The carbons change because it's an aromatic ring and therefore the pi bonds go in circles indefinitely. So if say...resonance via aromaticity?

 

[farmacista]

well in conjugated dienes or alpha beta unsaturated ketones i can easily determine orbital symmetry, my problem is when i have to evaluate the orbital symmetry of a molecule of the D6h point group (benzene)...the point group is what makes the difference between molecules i'm able to determine orbital symmetry or not (in order to then evaluate the allowed and forbidden electronic transition) but thanks anyway

 

[Pharmokinetictal]

Benzene doesn't have orbital symmetry. I think that's the answer. It has its identity but no symmetry. In other words, the point group D6h has no symmetry. Nor does it have chirality.
http://symmetry.otterbein.edu/tutorial/benzene.html

 

[Pharmokinetictal]

This is far more advanced p.chem than I've taken. But all I knowww from some reading is that the p electrons are obviously delocalized above and below the planar ring and no particular electron in these orbitals is more excitable than another right? And based on aromatic stacking, only two of the carbons would be involved with either T stacking or staggered stacking. And regarding substitution, any carbon is as nucleophilic as another. Yeah? Sorry I'm probly wasting your time but it interests me now too. Good luck!

 

[farmacista]

well if you're interested the 6 pi electron are delocalized on the ring but are in 3 different orbitals, one (p1) has one nodal plane (the ring plane) the other two (p2 and p3) are degenerated and have two nodal plane (the ring plus another that either pass between 2 opposite carbon or 2 opposite C-C bond) and then we have three anti-bonding pi orbitals of which p4 and p5 are degenerated. they have 4 different types of symmetry (indicated with a label, a for an orbital symmetric respect to the principal axis of rotation, b for an antisymmetric one etc.) on which depends if a particular electronic transition between a bonding and an anti-bonding orbital is allowed or forbidden. here i was wondering why the p2 and p3 orbital has this particular label (E1g) i understand that is E because the orbital are degenerated (two dimension irreducible representation)

 

[Jabberwocky]

wow, I used to know this stuff 10 years ago, but haven't used it since. Might find the following link on symmetry operations (see notation) helpful:

https://en.wikipedia.org/wiki/List_of_character_tables_for_chemically_important_3D_point_groups
"g and u subscripts denote symmetry and antisymmetry, respectively, with respect to a center of inversion. Subscripts "1" and "2" denote symmetry and antisymmetry, respectively, with respect to a nonprincipal rotation axis. Higher numbers denote additional representations with such asymmetry.
Single prime ( ' ) and double prime ( '' ) superscripts denote symmetry and antisymmetry, respectively, with respect to a horizontal mirror plane σh, one perpendicular to the principal rotation axis."

 

[farmacista]

we give a value of 1 if a orbital doesn't change sign after a certain symmetry operation and -1 if it changes sign, and in benzene the p2/p3 orbitals which divides the ring in 2 part changes sign for one third after rotation around the C6 axis (C6 means that if you rotate the molecule by 60 degrees you have a benzene ring that is not distinguishable from the original) so i was thinking 2/3 - 1/3= 1/3 but apparently it's 1 for C6 on the p2/p3 orbitals...

 

[sekio]

I think you can model molecular orbitals in Chem3d, it may take some work though.

 

[clubcard]

Anyone else remember having to balance salts and the teacher always ending the module with the most complex salt with a single element anion and single element cation. I seem to remember it was 19:23 i.e. 2 prime numbers.... Boring, I know, but always a fun one to throw at students... if that's you're humour (sad, I know).