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
List-I | List-II | ||
(A) | [Co(NH3)5(NO2)]Cl2 | (I) | Solvate isomerism |
(B) | [Co(NH3)5(SO4)]Br | (II) | Linkage isomerism |
(C) | [Co(NH3)6] [Cr(CN)6] | (III) | Ionization isomerism |
(D) | [Co(H2O)6]Cl3 | (IV) | Coordination isomerism |
List-I | List-II | ||
(A) | 1 mol of H2O to O2 | (I) | 3F |
(B) | 1 mol of MnO-4 to Mn2+ | (II) | 2F |
(C) | 1.5 mol of Ca from molten CaCl2 | (III) | 1F |
(D) | 1 mol of FeO to Fe2O3 | (IV) | 5F |
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32