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).
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
