The \(S_N1\) reaction mechanism proceeds through the formation of a carbocation intermediate. The stability of the carbocation plays a critical role in determining whether the reaction will proceed.
- Halide (A): \( \text{H}_2\text{C} = \text{CH} - \text{CH}_2\text{Cl} \) forms an allylic carbocation upon ionization, which is stabilized by resonance. Therefore, it is likely to undergo an \(S_N1\) reaction.
- Halide (B): \( \text{CH}_3 - \text{CH} = \text{CH} - \text{Cl} \) forms a carbocation that is not stabilized by resonance or inductive effects. This makes it unlikely to undergo an \(S_N1\) reaction, as the carbocation formed would be highly unstable.
- Halide (C): This compound forms a benzylic carbocation upon ionization, which is highly stabilized due to resonance with the aromatic ring, making it suitable for an \(S_N1\) reaction.
- Halide (D): \( \text{H}_3\text{C} - \text{C(Cl)H}_2 \) forms a tertiary carbocation, which is stable and favorable for the \(S_N1\) reaction due to hyperconjugation and inductive effects.
Therefore, the only halide that will not show an \(S_N1\) reaction is (B).
The Correct answer is: (B) only
Let \( S = \left\{ m \in \mathbb{Z} : A^m + A^m = 3I - A^{-6} \right\} \), where
\[ A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \end{bmatrix} \]Then \( n(S) \) is equal to ______.