The given question involves understanding the mechanism and stability of SN2 reactions in organic chemistry, specifically comparing the reactivities between benzyl bromide (C6H5CH2Br) and ethyl bromide (CH3CH2Br).
SN2 Reaction Overview:
Explanation:
Therefore, the correct answer is: Both (A) and (R) are correct and (R) is the correct explanation of (A).
Explanation:
1. Assertion (A): Correct. - In \( C_6H_5CH_2Br \), the \( CH_2-Br \) bond is connected to a benzyl group. The phenyl ring allows for stabilization of the transition state via resonance, facilitating the \( S_N2 \) reaction. This makes the reaction proceed more readily compared to \( CH_3CH_2Br \), where no such stabilization exists.
2. Reason (R): Correct. - The unhybridized \( p \)-orbital formed during the trigonal bipyramidal transition state interacts with the conjugated system of the phenyl ring, providing extra stabilization.
3. Conclusion: Both (A) and (R) are correct, and (R) is the correct explanation for (A).
Final Answer is option (3).
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is