Step 1: Understanding the Concept:
The \(\text{S}_{\text{N}}1\) (Substitution Nucleophilic Unimolecular) reaction proceeds via a two-step mechanism involving the formation of a carbocation intermediate.
The rate-determining step is the formation of this carbocation. Step 2: Key Formula or Approach:
The reactivity for \(\text{S}_{\text{N}}1\) reactions depends directly on the stability of the resulting carbocation.
The stability order for alkyl carbocations is: \(\text{Tertiary } (3^\circ)>\text{Secondary } (2^\circ)>\text{Primary } (1^\circ)\). Step 3: Detailed Explanation:
Let's analyze the carbocations formed by the given iodides:
(A) n-Butyl iodide: Forms a \(1^\circ\) carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+\)). Very low stability.
(B) sec-Butyl iodide: Forms a \(2^\circ\) carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_2\text{CH}_3\)). Moderate stability.
(C) Isobutyl iodide: Forms a \(1^\circ\) carbocation (\((\text{CH}_3)_2\text{CHCH}_2^+\)). Low stability.
(D) tert-Butyl iodide: Forms a \(3^\circ\) carbocation (\((\text{CH}_3)_3\text{C}^+\)). High stability due to 9 alpha-hydrogens (hyperconjugation) and the +I effect of three methyl groups.
Since the tert-butyl carbocation is the most stable, tert-butyl iodide will react the fastest via the \(\text{S}_{\text{N}}1\) mechanism. Step 4: Final Answer:
tert-Butyl iodide has the highest reactivity for \(\text{S}_{\text{N}}1\) reactions.