Question

Which of the following alkyl halides undergoes \(\text{SN}^1\) reaction most readily?

• A. \((\text{CH}_3)_3\text{C-F}\)
• B. \((\text{CH}_3)_3\text{C-Cl}\)
• C. \((\text{CH}_3)_3\text{C-Br}\)
• D. \((\text{CH}_3)_3\text{C-I}\)

Hint:

All four share the same tert-butyl carbocation; the best leaving group goes fastest.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In \(\text{SN}^1\) the slow step is ionisation of the C-X bond to a carbocation. A better leaving group means faster ionisation.

Step 2: Detailed Explanation:
All four compounds give the same tert-butyl carbocation, so only the leaving group differs.
C-X bond strength falls in the order C-F > C-Cl > C-Br > C-I.
Leaving group ability rises as \(\text{F}^- < \text{Cl}^- < \text{Br}^- < \text{I}^-\), because the larger halide ion carries its charge better.
So \((\text{CH}_3)_3\text{C-I}\) ionises most readily.

Step 3: Final Answer:
Option (D), the tert-butyl iodide.

Final Answer:
Iodide is the best leaving group in SN1. \[ \boxed{\text{(D) }(\text{CH}_3)_3\text{C-I}} \]