The SN\(_1\) reaction involves a two-step mechanism:
1. The first step is the formation of a carbocation after the leaving group departs.
2. The second step is the attack by the nucleophile. For the SN\(_1\) reaction, the reactivity depends on the stability of the carbocation formed. A more stable carbocation leads to higher reactivity.
In option (A) \( \text{C}_6\text{H}_5\text{Cl} \), the phenyl group is attached to the halide, and the carbocation formed would be unstable due to the lack of resonance stabilization.
In option (B) \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \), the phenyl group can provide some resonance stabilization to the carbocation formed, making it more stable and reactive towards the SN\(_1\) mechanism.
In options (C) and (D), the alkyl halides would form carbocations that are less stable compared to the phenylmethyl carbocation in option (B). Thus, \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \) shows the highest reactivity towards the SN\(_1\) reaction.
The correct option is (B) : \(C_6H_5CH_2CI\)
(b.)Calculate the potential for half-cell containing 0.01 M K\(_2\)Cr\(_2\)O\(_7\)(aq), 0.01 M Cr\(^{3+}\)(aq), and 1.0 x 10\(^{-4}\) M H\(^+\)(aq).