Question

\(\mathrm{S}_{\mathrm{N}} 1\) reactions show first-order kinetics because: (A) the rate-limiting step is the first step to occur in the reaction. (B) the rate-limiting step involves only one molecule. (C) there is only one rate-limiting step. (D) the reaction involves only one molecule.

Short answer

Option B.

Step-by-step solution

- Understand the reaction

An \(\text{S}_{\text{N}}1\) reaction is a type of nucleophilic substitution where the rate-determining step is the dissociation of the leaving group to form a carbocation.

- Analyze the rate-determining step

The rate-determining step in \(\text{S}_{\text{N}}1\) reactions involves the formation of a carbocation intermediate from the original molecule.

- Determine rate dependence

In the \(\text{S}_{\text{N}}1\) mechanism's rate-determining step, the rate depends only on the concentration of the original molecule, not on the nucleophile.

- Evaluate the given options

Option (B) 'the rate-limiting step involves only one molecule' correctly describes the first-order kinetics because the rate is dependent on the concentration of only the substrate molecule.

Key concepts

Rate-Determining Step

In an \( \text{S}_{\text{N}}1 \) reaction, the rate-determining step is the slowest and most critical part of the reaction mechanism. It's like a bottleneck in a busy freeway where traffic slows down. Imagine trying to pass through a narrow bridge, while the rest of the road is wide and fast-moving. The overall progress of your journey would be limited by how quickly you can cross that bridge.
The same idea applies to chemical reactions. For \( \text{S}_{\text{N}}1 \) reactions, the rate-determining step is the stage where the leaving group separates from the substrate, forming a carbocation. This step is crucial because it largely determines how fast the reaction proceeds.
Understanding that the rate-determining step involves only the initial molecule helps to clarify why the concentration of other components (like the nucleophile) doesn't affect the overall rate.

Carbocation Formation

After the leaving group departs during the rate-determining step, a carbocation intermediate is formed. This is a positively charged ion that is highly reactive. Think of it as a fragile and unstable structure that is eager to react and stabilize itself.
Carbocations are central to \( \text{S}_{\text{N}}1 \) reactions because once formed, they can quickly react with nucleophiles to form the final product. However, their formation requires the substrate to lose a group, which takes energy and time, making this the slowest and most limiting step in the reaction.
Stability of carbocations plays a crucial role: tertiary carbocations (attached to three other carbon atoms) are more stable than secondary ones, which in turn are more stable than primary carbocations. The more stable the carbocation, the faster and more likely the \( \text{S}_{\text{N}}1 \) reaction will occur.

First-Order Kinetics

First-order kinetics means that the rate of the reaction depends on the concentration of just one reactant. For \( \text{S}_{\text{N}}1 \) reactions, this reactant is the substrate from which the leaving group departs to form the carbocation.
In this context, if you double the concentration of the substrate, the reaction rate doubles as well. This is different from second-order reactions, where the rate depends on the concentration of two reactants.
The mathematical expression for first-order kinetics is given by the rate law: \[ \text{Rate} = k [\text{substrate}] \]
Here, \( k \) is the rate constant and \( [\text{substrate}] \) represents the concentration of the substrate. This clear relationship is why \( \text{S}_{\text{N}}1 \) reactions are classified as first-order reactions, focusing only on one molecule's concentration to dictate the reaction speed.