Question

In $S_{N}2$ solvolysis of $RX$ in which the solvent is a nucleophiIe, what is the order and molecularity of the reaction?

• A. First order; unimolecular
• B. First order; bimolecular
• C. Second order; bimolecular
• D. Pseudo first-order; bimolecular

Answer 1

Answer: (D)

Pseudo first-order; bimolecular

$S_{N}2$ solvolysis reaction is bimolecular and the rate law is
Rate = $K\left[RX\right]\left[Solvent\right]$
Since the concentration of the solvent is very large and exceeds the amount of $RX$. therefore, its concentration is constant. Hence the reaction will follow pseudo first-order rate law.