Question

Write the mechanism of the bimolecular nucleophilic substitution reaction (S_N2) of an alkyl halide.

Hint:

Think one concerted step: the nucleophile attacks the carbon from the side opposite the halogen, giving a five-coordinate transition state and inversion of configuration.

Answer 1

Step 1: What S_N2 means. S_N2 stands for bimolecular nucleophilic substitution. "Bimolecular" means the slowest (rate-determining) step involves two species: the alkyl halide and the nucleophile. So the rate depends on the concentration of both.
\[ \text{Rate} = k[\text{R-X}][\text{Nu}^-] \]
Step 2: It is a one-step (concerted) reaction. Bond breaking (C-X) and bond making (C-Nu) happen at the same time in a single step. There is no intermediate.
Step 3: Backside attack. The nucleophile (Nu^-) attacks the carbon atom bearing the halogen from the side exactly opposite to the leaving halide group. This is because the halogen blocks the front side.
Step 4: Transition state. In the transition state the central carbon is bonded partially to both the incoming nucleophile and the outgoing halide. The three other groups on the carbon lie in a plane (the carbon is roughly \(sp^2\), giving a trigonal bipyramidal arrangement of five groups).
Step 5: Product and inversion. The halide leaves with the bonding electrons and the new C-Nu bond forms. The three other groups get pushed to the opposite side, so the configuration is turned inside out. This is called Walden inversion (inversion of configuration).
Example: \[ HO^- + CH_3Br \rightarrow CH_3OH + Br^- \]
The hydroxide ion attacks from behind, bromide leaves, and methanol is formed with inverted configuration.
\[\boxed{S_N2:\ \text{one step, backside attack, rate}=k[R\text{-}X][Nu^-],\ \text{inversion}}\]