See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: SN1 reactivity depends on the stability of the carbocation formed after departure of the leaving group (I-). The more stable the carbocation intermediate, the faster the SN1 reaction. Step 2 - Identifying the carbocation formed: In all four compounds, ionization of the C-I bond generates a carbocation at the benzylic/allylic carbon (the CH2 position becomes CH2+, which is actually the carbon adjacent to the double bond, i.e., an allylic/benzylic cation). Step 3 - Analyzing each option: (a) The carbocation formed is allylic, stabilized by the cross-conjugated cyclopentadienyl system. The cyclopentadienyl cation (formed if resonance extends into the ring) is actually antiaromatic (4 pi electrons), which destabilizes the cation. So despite multiple double bonds, the antiaromatic character of the cyclopentadienyl cation makes this less favorable. (b) The carbocation is allylic, stabilized by conjugation with one endocyclic double bond and one exocyclic methylene. This gives a somewhat stabilized allylic cation but not exceptionally so. (c) Similar to (b) but with a saturated ring and one exocyclic methylene. The allylic stabilization is present but limited. (d) The carbocation formed is a benzylic cation - the CH2-I loses I- to give a benzylic carbocation (ArCH2+) where Ar is a dimethyl-substituted benzene (aromatic ring). Benzylic carbocations are highly stabilized by resonance delocalization into the aromatic pi system. The aromatic ring provides extensive resonance stabilization, making this the most stable carbocation among all four options. Additionally, the two methyl groups on the aromatic ring donate electron density inductively/mesomerically, further stabilizing the positive charge. Step 4 - Why (d) wins: Benzylic carbocation stabilization through an intact aromatic ring (resonance with 6 pi electron aromatic system) is superior to simple allylic stabilization in (b) and (c), and far superior to the antiaromatic destabilization in (a). The dimethyl-substituted benzene ring in (d) provides maximum stabilization of the benzylic cation. Step 5 - Conclusion: Compound (d) forms the most stable carbocation intermediate and therefore undergoes SN1 most readily. Therefore, the correct answer is D.