Why is this diagram depicting the molecular orbital (MO) basis for a back-side attack the way it is?

Consider:


[![Enter image description here](https://i.stack.imgur.com/Fa4hU.png)](https://i.stack.imgur.com/Fa4hU.png)


The description of this image in my textbook is as follows:



> 
> In order to form a bond, the HOMO (the highest occupied molecular orbital) of one species must interact with the LUMO (the lowest unoccupied molecular orbital) of the other. Therefore, when the nucleophile (Nuc) approaches the alkyl halide to form a new bond, the nonbonding molecular orbital of the nucleophile (its HOMO) must interact with the empty s\* antibonding molecular orbital associated with the C¬Br bond (its LUMO). Figure 9.1a shows that in a back-side attack, a bonding interaction (the interacting lobes are both green) occurs between the nucleophile and the larger lobe of the s\* antibonding MO. But when the nucleophile approaches the front side of the carbon (Figure 9.1b), both a bonding and an antibonding interaction occur, so the two cancel each other and no bond forms. Therefore, an SN2 reaction is successful only if the nucleophile approaches the sp3 carbon from its back side.
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For reference, the nucleophile in the example before this is a hydroxide ion, and the electrophile-leaving group compound
is $\ce{CH3-Br}$.


I suppose where I begin to be confused is why C and Br have an antibonding interaction in the first place. At least, that is what I get from the diagram. Am I interpreting it incorrectly? If my interpretation is correct, why do they have an antibonding interaction?


I was under the impression that C and Br had a normal sigma bond, but maybe it changes in the reaction mechanism? My textbook also says that Hughes and Ingold labeled this a concerted reaction from their experimental evidence, but the only way it makes sense to me at the moment is if a carbocation intermediate is formed.


I think the non-bonding and antibonding interaction is throwing me off. How does the interaction work in more detail? I feel like there is something obvious that I'm missing and it could be fairly simple, or that I am just overcomplicating things.





> 
> I suppose where I begin to be confused is why C and Br have an antibonding interaction in the first place. At least, that is what I get from the diagram. Am I interpreting it incorrectly? If my interpretation is correct, why do they have an antibonding interaction? I was under the impression that C and Br had a normal sigma bond
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(Very loosely speaking) every bonding orbital is accompanied by an antibonding one. In this case, only the bonding orbital is *filled*; the antibonding orbital is initially empty, and that's why there is a covalent bond which can be talked about. (The C–Br bond order is given by (number of bonding electrons − number of antibonding electrons)/2 = (2 − 0)/2 = 1, right? — if you're not familiar with this, I suggest learning a bit about MO theory before coming back to this.)


The diagrams in the book are not intended to state anything about whether the orbitals are filled or not.



> 
> but maybe it changes in the reaction mechanism?
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Well, yes, it does. If electrons from elsewhere (in this case, the nucleophile, or hydroxide ion) flow into the antibonding orbital, then the maths above changes: essentially, you start to *break* the C–Br bond because (again, loosely speaking) the number of antibonding electrons increases. And that's the whole point of the reaction, after all — the C–Br bond is being replaced with a C–OH bond.


(It's a bit of a simplification to talk about integer numbers of electrons, but it'll do for now.)



> 
> My textbook also says that Hughes and Ingold labeled this a concerted reaction from their experimental evidence, but the only way it makes sense to me at the moment is if a carbocation intermediate is formed.
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The MO description doesn't tell you whether it's concerted (i.e. SN2) or stepwise (i.e. SN1). You need physical / experimental evidence to prove whether it's concerted or stepwise, and then the MO analysis follows on from that; not the other way around. The experimental evidence isn't provided here, so there's no way to evaluate this claim.


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