[oelimtrila7]

So for the title when it comes to predicting the product that forms here why wouldn't the Ethoxide ion attack the double bond carbon. The major product that forms here is determined to be the attack on the allylic carbon but from an orbital interaction perspective isn't that an attack on the sigma-antibonding? I though just from looking at the picture the attack would be on the pi-antibonding since that would be the LUMO preferred by the SP3 oxygen (pi-antibonding being a better LUMO than sigma-antibonding). Or does the attack on the double-bonded carbon not involve an attack on an empty pi-antibonding orbital?

[Babcock_Hall]

I am not going to be of much help from a molecular orbital perspective.  What about analyzing the question using the conceptual framework from introductory organic chemistry?

[Marko]

From an orbital perspective, I think you can go by the same reasoning as why in general S_N2 reactions don't occur on double bonds: there is no constructive overlap possible with the antibonding pi orbital, so there is no gain in energy when the nucleophile approaches the alkene.

So, the energy of the antibonding pi orbital may be lower than of the antibonding sigma orbital between the allylic carbon and Cl, but there is no gain in using it to form a new bond with the nucleophile.

The general rule is that electrons flow from the HOMO of the nucleophile to the LUMO of the electrophile. It doesn't specify that electrons from from the (highest) HOMO to the lowest LUMO.

[rolnor]

On allylic halides nucleophiles can atack both the allylic carbon and the alkene-carbon, then you get an allylic shift. If you have a bromo on the 3-position this might hinder an atack by steric hindrans but I am not sure in this case.

https://en.wikipedia.org/wiki/Allylic_rearrangement

[hollytara]

If we are discussing a concerted allylic shift, then the leaving group has to be oriented so the sigma bond is coplanar with the pi bond, since as the bond to the LG breaks, a new pi bond is made to the central Carbon.  The pi bond also breaks while the nucleophile comes in.  Since this also has to be coplanar, the antibonding pi orbital would be in the correct symmetry while sigma antibonding on the double bonded carbon is perpendicular..

[rolnor]

Would not the eletrons in the ethoxide nucleophile go to/attack a orbital with as low energy as possible, antibonding orbitals are generally high in energy?

[hollytara]

But if it is a concerted SN2' allylic shift, there has to be conservation of orbital symmetry. 

So a direct SN2 on the allyl CH2 will follow a sigma symmetry: nucelophile lone pair to sigma antibonding to sigma bonding

But the allylic SN2' reaction has to follow pi symmetry and use the pi antibonding orbital. 

If it is SN1' and a free allyl or allylic cation forms, there is no antibonding orbital to accept electrons.