Question

Which of the following statements is TRUE regarding the transition state of a bimolecular nucleophilic substitution ($S_N2$) reaction?

• A. The central carbon atom transitions to an $sp^3$ hybridization state with a tetrahedral geometry.
• B. The central carbon atom adopts an unstable pentacoordinate arrangement with approximate $sp^2$ hybridization.
• C. The reaction forms a stable, isolation-ready carbocation intermediate.
• D. Both the nucleophile and the leaving group carry full, non-interacting formal charges.

Hint:

The $S_N2$ mechanism involves a single concerted transition state (pentacoordinate carbon), while $S_N1$ proceeds via a distinct carbocation intermediate.

Answer 1

The Correct Option is B

Concept: The $\text{S}_\text{N}2$ pathway operates via a concerted, single-step mechanism. Unlike stepwise substitutions, it does not involve the formation of any distinct, low-energy intermediates such as carbocations. Instead, bond-making between the incoming nucleophile and the substrate occurs simultaneously with bond-breaking of the leaving group. This coordinated process passes through an unstable, high-energy transition state configuration.

Step 1: Analyzing the synchronous geometry and bond configuration of the reaction center.
During an $\text{S}_\text{N}2$ attack, the nucleophile approaches the substrate from the side directly opposite the leaving group ($180^\circ$ back-side attack) to minimize electrostatic repulsion. At the peak of the transition state, the bond to the incoming nucleophile ($\text{Nu}^{\delta-}$) is partially formed, while the bond to the departing leaving group ($\text{LG}^{\delta-}$) is partially broken. This arrangement forces the central carbon atom to coordinate temporarily with five separate groups at the same time.

Step 2: Determining the transient hybridization state and electronic geometry.
Because the partial bonds to the nucleophile and leaving group lie along a single linear axis, the central carbon uses its remaining unhybridized $p$-orbital to interact with them. The other three stable, non-reacting substituent bonds are pushed away and flatten out into a single perpendicular plane. This planar arrangement uses three $sp^2$ hybrid orbitals separated by $120^\circ$ angles. This creates an unstable pentacoordinate carbon center with a trigonal bipyramidal geometry in the transition state.