Question

Which of the following has highest reactivity for \(\text{S}_{\text{N}}1\) reactions?

• A. n-Butyl iodide
• B. sec-butyl iodide
• C. Isobutyl iodide
• D. tert-Butyl iodide

Hint:

For \(\text{S}_{\text{N}}1\) reactions, always check carbocation stability first: \[ 3^\circ > 2^\circ > 1^\circ \]

Answer 1

The Correct Option is D

Concept:
An \(\text{S}_{\text{N}}1\) reaction proceeds through formation of a carbocation intermediate. So the compound that forms the most stable carbocation will show the highest \(\text{S}_{\text{N}}1\) reactivity. ip

Step 1: Recall carbocation stability order.
The stability order of carbocations is: \[ 3^\circ > 2^\circ > 1^\circ > \text{methyl} \] ip

Step 2: Classify the given alkyl iodides.

• n-Butyl iodide is primary
• sec-Butyl iodide is secondary
• Isobutyl iodide is primary
• tert-Butyl iodide is tertiary e} ip

Step 3: Identify the most reactive compound in \(\text{S}_{\text{N}}1\).
Since tert-butyl iodide forms the most stable tertiary carbocation, it reacts fastest by the \(\text{S}_{\text{N}}1\) mechanism. ip Hence, the correct answer is:
\[ \boxed{(D)\ \text{tert-Butyl iodide}} \]