Question

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

Hint:

Concerted one-step backside attack by the nucleophile, a pentacoordinate transition state, inversion of configuration, and second-order rate \( = k[\text{RX}][\text{Nu}] \).

Answer 1

Step 1 (What S_N2 means): S_N2 stands for substitution, nucleophilic, bimolecular. It is a single-step (concerted) reaction in which the nucleophile forms a bond to carbon at the same time as the leaving group (halide) breaks away. Both the alkyl halide and the nucleophile take part in the rate-determining step.
Step 2 (Backside attack): The nucleophile (for example OH^-) approaches the carbon bearing the halogen from the side exactly opposite to the leaving group. This is because the C-X bond has a region of low electron density on the far side of carbon.
Step 3 (Transition state): At the midpoint a transition state forms in which carbon is partially bonded to both the incoming nucleophile and the outgoing halide. The three other groups on carbon lie in a plane, giving a trigonal bipyramidal (pentacoordinate) arrangement.
Step 4 (Product and inversion): As the nucleophile completes its bond, the halide leaves with the bonding pair, and the three groups flip through to the other side, like an umbrella turning inside out. This gives inversion of configuration (Walden inversion) at the reacting carbon.
Step 5 (Kinetics and reactivity): Because both species appear in the slow step, the rate law is \( \text{rate} = k[\text{alkyl halide}][\text{nucleophile}] \), i.e. second order. Less hindered carbons react faster, so the order of reactivity is primary > secondary > tertiary alkyl halide.
Example: CH₃Br + OH^- → CH₃OH + Br^-.