As an organic chemistry student, one of the trickiest topics you will come across is the reaction set of substitution and elimination reactions. This is due to the average student mindset of learn, memorize, and apply. Many factors must be taken into account to fully understand these reactions. In this article I will help you understand how to carry out an SN2 reaction when faced with a poor hydroxyl leaving group.
Let’s start with a quick overview of the SN2 reaction mechanism. SN2 stands for bimolecular nucleophilic substitution. This reaction occurs when a strong nucleophile attacks a carbon forcing out the leaving group and attaching itself in its place.
This reaction relies on a number of factors as follows:
- SN2 requires a nucleophile strong enough to attack, preferably with a negative charge
- The leaving group or LG must be on a methyl, primary, or secondary carbon. SN2 cannot take place with a tertiary LG
- The LG must be willing to leave and form a weak stable molecule or ion in solution
- Polar aprotic solvents are preferred so as not to surround or ‘cage’ the strong nucleophile
This is easy to understand and follow when provided with a reaction sequence involving a 1-bromopropane reacting with sodium cyanide in dmso cream uses. We’re dealing with good LG on a primary carbon. Cyanide is strong enough to attack and the polar aprotic DMSO won’t get in the way.
But what about a reaction such as methanol reacting with HBr?
In this case we have an ideal methyl carbon involved, however OH- is a terrible leaving group. So terrible in fact, if the reaction were to take place, OH- would turn around and re-attack the moment it gets kicked off the molecule.
And what good is a leaving group that simply won’t stay gone?
This step requires a bit of help from the hydrobromic acid. We first must turn OH into a better leaving group by protonation. Methanol is allowed to attack the H in HBr. This protonates the alcohol and kicks off the negative bromide to reside as a decent nucleophile in solution.
While OH is a bad leaving group due to forming OH- in solution, OH2+ is a great leaving group. The OH2+ is now quite unhappy attached to the carbon atom. Oxygen has just 1 lone pair and 3 sigma bonds with a positive formal charge.