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: Picture the reaction as a smooth handover happening in one motion. There is no ionic intermediate; the old C-X bond weakens while the new C-Nucleophile bond strengthens simultaneously, which is why it is called concerted and bimolecular.
Step 2: The electron-rich nucleophile is repelled by the halogen, so it comes in from directly behind the carbon-halogen bond (180 degrees away from X). This geometry is essential for the mechanism.
Step 3: Halfway along, the carbon is joined by half-bonds to both the nucleophile and the halide while its other three substituents are pushed into one flat plane, forming the unstable, high-energy transition state.
Step 4: When the reaction finishes, the halide departs and the three flat groups swing over to the opposite face. The spatial arrangement at carbon is therefore turned over, producing an inverted product (Walden inversion).
Step 5: Since the transition state involves both reactants, doubling either concentration doubles the rate, so \( \text{rate} = k[\text{R-X}][\text{Nu}] \). Bulky groups block the backside approach, so methyl and primary halides react fastest and tertiary halides are very slow by this path.