Step 1 : Understanding the Question:
The topic of this question is Nucleophilic Substitution Reactions, specifically the $S_N1$ mechanism. The $S_N1$ (Substitution Nucleophilic Unimolecular) reaction occurs in two steps, where the first and slowest step is the formation of a carbocation intermediate. The question asks which of the given alkyl halides will react the fastest. Since the rate-determining step depends on the stability of the carbocation, we need to compare the stability of the cations formed by each halide.
Step 2 : Key Formulas and approach:
The approach follows the carbocation stability trend:
1. Identify the carbocation formed by removing the leaving group ($Cl^-$) from each option.
2. Rank them by stability: $Tertiary (3^\circ)>Secondary (2^\circ)>Primary (1^\circ)>Methyl$.
3. Stability is increased by $+I$ (inductive effect) and hyperconjugation from adjacent alkyl groups.
The halide that forms the most stable carbocation will have the lowest activation energy for the first step and thus the fastest overall rate.
Step 3 : Detailed Explanation:
Let's analyze the carbocations produced by each option:
- Option (A): $CH_3Cl \rightarrow CH_3^+$ (Methyl carbocation). This is the least stable possible carbocation.
- Option (C): $CH_3CH_2Cl \rightarrow CH_3CH_2^+$ (Primary carbocation). It is slightly stabilized by one ethyl group.
- Option (D): $(CH_3)_2CHCl \rightarrow (CH_3)_2CH^+$ (Secondary carbocation). It is stabilized by two methyl groups.
- Option (B): $(CH_3)_3CCl \rightarrow (CH_3)_3C^+$ (Tertiary carbocation).
The tertiary carbocation (tert-butyl cation) is exceptionally stable compared to the others. It has three electron-donating methyl groups that spread out the positive charge via the inductive effect.
Additionally, it has 9 $\alpha$-hydrogens available for hyperconjugation, which significantly lowers the energy of the intermediate.
In $S_N1$ reactions, the transition state for the rate-determining step resembles the carbocation. Therefore, the more stable the carbocation, the faster the reaction proceeds.
Tertiary alkyl halides like $(CH_3)_3CCl$ react almost exclusively via $S_N1$ in polar protic solvents, while primary halides like $CH_3Cl$ almost never do.
Step 4 : Final Answer:
The tertiary alkyl halide forms the most stable carbocation, leading to the fastest $S_N1$ reaction rate. The correct option is (B).