WBJEE 2024 Physics & Chemistry Paper

The correct answer is Option B: I > III > II. Here's why:The $\mathrm{S}_{\mathrm{N}} 1$ reaction proceeds through a two-step mechanism:Step 1: Formation of carbocation: This step involves the ionization of the alkyl halide to form a carbocation. The rate of this step depends on the stability of the carbocation formed. More stable carbocations form faster and thus lead to faster $\mathrm{S}_{\mathrm{N}} 1$ reaction.Step 2: Nucleophilic attack: This step involves the attack of the nucleophile on the carbocation to form the product.Let's analyze the stability of carbocations formed from each molecule:I. Allyl Chloride: The carbocation formed from allyl chloride is resonance stabilized:This resonance stabilization makes the allyl carbocation very stable. II. Chlorobenzene:Chlorobenzene does not easily form a carbocation because the benzene ring is highly stable and aromatic. The formation of a carbocation would disrupt this aromaticity, which is energetically unfavorable. Therefore, chlorobenzene is relatively unreactive towards the $\mathrm{S}_{\mathrm{N}} 1$ reaction.III. Ethyl Chloride:Ethyl chloride forms a simple ethyl carbocation, which is less stable than a resonance-stabilized carbocation but more stable than a primary carbocation found in many other alkyl halides. However, it is still less stable than the allyl carbocation.Therefore, the order of reactivity towards $\mathrm{S}_{\mathrm{N}} 1$ reaction is: Allyl chloride (I) > Ethyl chloride (III) > Chlorobenzene (II).This is because the resonance-stabilized allyl carbocation (I) is the most stable, followed by the ethyl carbocation (III), and finally, the chlorobenzene (II), which does not favor carbocation formation due to the loss of aromaticity.