The question is about the specifics of the $\mathrm{S_N2}$ mechanism (nucleophilic bi-molecular substitution).
(Source: Wikipedia)
As we know that in this reaction the nucelophile ($\ce{Nu-}$) approaches the molecule. The main step (According to the molecular orbital theory) of the reaction is that the the nucleophile bonds with its lone pair in the anti-bonding molecular orbital (ABMO) of the central carbon atom. As this happens the leaving group ($\ce{Lg}$) is kicked off by gradual weakening of the $\ce{C-Lg}$ bond. Now the thing that bothers me here is:
Why does carbon prefer to add electrons in its ABMO? As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order whereas, while electrons are pushed into the ABMO it clearly decreases the bond order, which directly indicates the instability of bond formation.
The main question being, what makes the reaction favourable? What is the 'driving force'? Why does it take place when there is a decrease in bond order?
Answer
Why does carbon prefer to add electrons in its ABMO?
Well, preferring is a bit of a strong word. Orbital interactions that are productive towards forming and breaking bonds can only happen between a filled and an empty orbital. Therefore, you should always identify a HOMO (highest occupied molecular orbital) and a LUMO (lowest unoccupied molecular orbital), compare their energies and thereby decide whether they are able to react.
Here, the lowest unoccupied orbital of the carbon compound is typically the $\ce{C-Lg}$ antibonding $\unicode{x3c3}^*$ orbital — because that is the antibonding orbital belonging to the leaving group (I am aware that this is circular reasoning). Its energy is also somewhat favourable with respect to the energy of the filled lone pair orbital of the nucleophile meaning the two can react well.
As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order …
No. Why?
electrons are pushed into the ABMO it clearly decreases the bond order …
To specify: decreases the bond order *of the $\ce{C-Lg}$ bond.
which directly indicates the instability of bond formation.
No, it doesn’t. It only indicates that the $\ce{C-Lg}$ bond is being broken and the leaving group is getting ready to leave.
Why does it take place when there is a decrease in bond order?
You fail to see that while the bond order of the $\ce{C-Lg}$ bond is steadily decreasing the bond order of the $\ce{C-Nu}$ bond is steadily increasing. At any point in time on the reaction coordinate, the sum of both bond orders is $1$.
The main question being, what makes the reaction favourable? What is the 'driving force'?
Two things: the formation of a more stable $\ce{C-X}$ bond (meaning that $\ce{C-Nu}$ is a stronger bond than $\ce{C-Lg}$) and the liberation of a better stabilised anionic species $\ce{Lg-}$ as opposed to $\ce{Nu-}$.
Actually, in many $\mathrm{S_N2}$ reactions only one of those two factors is the driving force. And while we tend to write $\ce{Nu-}$ and $\ce{Lg-}$, we should always remember that these compounds need not be anions. For example, $\ce{H2O}$ is a very good leaving group (after alcohols have been protonated) but not an anion. The same is true for $\ce{N2}$ (from a diazonium salt). And equally, water or ammonia can be nucleophiles although they are neutral compounds.
So it can well be that the reaction is only driven by the liberation of a neutral compound and thereby the removal of charges (as would be the case in the $\mathrm{S_N2}$ reaction of iodide and a protonated alcohol when $\ce{HI}$ is used to generate iodoalkanes). It might also be that the reaction is driven only by the stronger $\ce{C-Nu}$ bond — or by the breaking of the weaker $\ce{C-Lg}$ bond as would be the case in nucleophilic substitutions with tosylate ($\ce{Me-C6H4-SO3-}$) or mesylate ($\ce{MeSO3-}$) as leaving groups.
There is not one same truth for all $\mathrm{S_N2}$ reactions.
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