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 $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: