The SN2 reaction is a bimolecular nucleophilic substitution reaction that occurs in one step. The nucleophile performs a backside attack on the carbon to which the leaving group is attached. If the carbon is asymmetric, inversion of stereochemistry is observed.
The SN2 mechanism proceeds through a concerted mechanism. All that means is that it proceeds in one step and there’s no intermediate. As the nucleophile (Nu–) performs a “backside attack,” the leaving group (LG) dissociates. In other words, the nucleophile makes a bond and breaks the bond of the leaving group to the carbon holding it.
In this generic reaction, the nucleophile is attacking the carbon holding the leaving group, which causes the nucleophile to dissociate. Notice that the net charge of the reaction stays the same. There’s no intermediate in this reaction; it goes straight from reagents to products.
As the nucleophile forms a bond to the carbon, the leaving group’s bond is broken. This is called the transition state, and it’s indicated from the double dagger (≠) that’s generally placed at the top right of the box it’s included in. Notice that both the nucleophile and leaving group have partial negative charges. That makes sense because the nucleophile is donating an electron while the leaving group is accepting an electron.
Now that we know what’s happening in the reaction, let’s look at the reaction-coordinate diagram:
In this diagram, there are really only three parts: the reagents, the transition state, and the products. The transition state is the point in the reaction with the highest energy level, and the difference in energy between the reagents and transition state is called the activation energy (often abbreviated as Ea). Remember that the rate-determining step is the step that has the highest activation energy in the reaction.
The nucleophile in SN2 reactions is generally anionic. Sometimes the negative charges aren’t shown explicitly because a cationic metal is used to stabilize the negative. A great example of this is NaCN. CN is negatively charged, and Na is positively charged. It’s an ionic bond that readily dissociates in the presence of a good electrophile or leaving group. Group 1 atoms like Na, Li, and K are a dead giveaway!
Atoms or molecules that can easily hold a negative charge are generally good leaving groups. What’s a good way to know if a molecule can be a good leaving group? Check its conjugate acid’s pKa!
When a leaving group dissociates from the substrate, it gains an electron. If it can hold the resulting negative charge “comfortably,” it’s a good leaving group. The bromine in the reaction above is a good leaving group because the pKa of its conjugate acid HBr is -9, which means it can hold a negative charge well.
Not all leaving groups are created equal! Even if you’ve got a bromine like in the example above, its degree greatly influences reactivity toward SN2 reactions.
Methyl (0º) leaving groups are very reactive toward SN2 reactions, and tertiary (3º) are not reactive toward them at all. Secondary leaving groups risk competing with E2 reactions. Try to imagine a nucleophile trying to overcome the sterics of even three methyl groups, let alone three phenyl groups. It’s almost impossible for it to squeeze through; SN1 reactions are much more likely to happen in that case.
See how hard it is for the nucleophile to bypass all the R-groups to get to the carbon holding the tertiary halogen (X)? A tertiary halide with just methyl groups is the best-case scenario, and that’s already enough to stop an SN2 from occurring.
Let’s look at this oversimplified coordinate diagram of a theoretical tertiary SN2 reaction relative to secondary, primary, and methyl reactions.
The activation energy is so ridiculously high that it’s just not going to happen. Other reactions that have lower activation energies will happen instead.
SN2 reactions function best in polar aprotic reactions. Basically that means polar solvents that don’t have any acidic protons. Acetone, DMSO, and dimethyl chloride are commonly used polar aprotic solvents.
The reaction rate of SN2 reactions depends on the concentrations of both the nucleophile and the leaving group. That’s why it’s called a bimolecular nucleophilic substitution reaction. Here’s the rate law:
So, what happens to the rate if we double the concentration of the nucleophile? The leaving group? Both?
Let’s go ahead and react an achiral, secondary alkyl halide with NaCN.
All that happens is that the –CN attacks the secondary carbon bonded to the Br, and the formation of that bond causes the C-Br bond to break. The NaBr salt that forms is an inorganic product, and it can generally be ignored.
Now, wait a minute. See how I specified that the alkyl halide above was achiral? What happens when the leaving group is at a chiral center like in the following example?
In this case, the mechanism looks the same, but there’s one key difference. The iodine was on a wedged bond, but the resulting alcohol is on a dashed bond. That’s inversion of stereochemistry! In the very first paragraph, I said that the nucleophile performs a “backside attack,” and this is the result. Remember that the iodine is facing toward us, so that means the hydroxide attacked from the back and forms a bond on dash.
P.S. In case you’re wondering how we know if an anion will work as a nucleophile vs base, check out the Big Daddy Flowchart! It’ll also help you determine if a mechanism will be SN1, SN2, E1, or E2.