Question

Which of the following halide shows highest reactivity towards S_N1 reaction?

• A. C₆H₅Cl
• B. C₆H₅CH₂CI
• C. CH₃-CH₂Cl
• D. CH₃-CH₂-CH₂-CH₂I

Answer 1

The Correct Option is B

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) : \(C_6H_5CH_2CI\)

Answer 2

In SN\(_1\) reactions, the rate-determining step is the dissociation of the leaving group to form a carbocation intermediate. The stability of the carbocation is the key factor influencing the reactivity in SN\(_1\) reactions.

C\(_6\)H\(_5\)Cl: The phenyl chloride undergoes SN\(_1\) very slowly due to the poor stability of the carbocation formed (the phenyl carbocation).
 
C\(_6\)H\(_5\)CH\(_2\)Cl: The benzyl carbocation formed here is highly stable due to the resonance stabilization provided by the phenyl group. Thus, this compound shows the highest reactivity towards SN\(_1\) reactions.
 
CH\(_3\)CH\(_2\)Cl: The ethyl carbocation formed here is less stable compared to a benzyl carbocation, making it less reactive than C\(_6\)H\(_5\)CH\(_2\)Cl.
 
CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)I: Although the iodide ion (I\(^-\)) is a very good leaving group, the stability of the butyl carbocation formed is lower than the benzyl carbocation, so it is less reactive than C\(_6\)H\(_5\)CH\(_2\)Cl.

Therefore, the highest reactivity is shown by C\(_6\)H\(_5\)CH\(_2\)Cl.