Reviewing material for the MCAT? Need help studying for orgo? In this post, we will be discussing the mechanism of a particular class of chemical reactions known as bimolecular nucleophilic substitution, or Sn2 reactions. Sn2 reactions are quite common in many organic transformations, and in this blog post, we will not only discuss what they are, but how to depict them using arrow-pushing diagrams.
What are Sn2 reactions?
Breaking down this classification name, we can identify three characteristic features of an Sn2 reaction:
- Bimolecular – two chemical species (nucleophile + electrophile) interact to form a new covalent bond
- Nucleophilic – one of the two chemical species involved, usually electron-rich either through a lone-pair, sigma bond, or pi bond
- Substitution – this suggests that there is one chemical group that will be swapped with another within a particular molecule
In broad strokes, here’s how an Sn2 reaction works: first, a nucleophile denoted Nu- will “attack” an electrophilic (electron-poor) center C opposite of a good leaving group which we will call X. The nucleophile will attack in line, but opposite to the C-X bond, in a process termed “backside attack.” To form a new bond, the nucleophile Nu- will use its lone pair to form a new Nu-C bond, simultaneously breaking the opposite C-X bond. As shown in the diagram below, the result is what we call an “inversion of stereochemistry.”
How do I communicate an Sn2 mechanism?
To depict a mechanism, we often use arrow-pushing diagrams as a way to describe how a reaction proceeds step-by-step, that is, bonds being broken and bonds being formed. We depict the flow of electrons using arrows, where one arrow represents the formation/breaking of one bond composed of 2 electrons. In this logic, electrons flow from electron rich species such as a nucleophile to electron poor species such as an electrophile.
Let’s look at an example, the reaction of hydroxide with methyl iodide as shown below.
In this reaction we first need to identify any competent nucleophiles. Here, the hydroxide is a strong nucleophile, as determined by its lone pairs reflected in its negative charge. As an electron rich species, the hydroxide ion will therefore attack methyl iodide. The methyl iodide contains a strong dipole moment in which the iodine withdraws electron density away from the carbon center. Following electronegativity trends, we also know this means iodine anion will be stabilized and therefore represents a good leaving group.
We therefore draw electrons from the hydroxide ion O to attack the carbon center of the methyl iodide electrophile. This displaces electron density back onto the iodine. In the transition state, a new bond forms between the hydroxide and carbon, and the old carbon-iodine bond is partially broken. This is until the energy of activation needed to reach the transition state is accessed, and then the new O-C bond will form, and the old C-I bond will be broken.
Sn2 reactions really have only one step: the simultaneous bond breaking and bond formation step, making it the “rate determining step.” This step is called the rate determining step (RDS) because it is the slowest. Therefore, the rate determining step is determine by the formation of the product.
Let’s take a look at one more example, the reaction of a simple enolate with molecular bromine.
In this case, it is important to remember that not all nucleophiles are lone pair electrophiles. In this case, a pi-bond is able to act as an electron rich species for the Sn2 reaction. To react with molecular bromine, which is a good electrophile given the low bond dissociation energy of a bromine-bromine bond, the pi-bond alpha carbon will attack one bromine atom of the molecular bromine. This alpha carbon is the best nucleophile since drawing the resonance form of the enolate reveals an electron rich carbanion, which is a better nucleophile than sp2 hybridized oxygen. The result is a new bromide ion, and an alpha-substituted ketone known as an alpha-halo ketone.
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