See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Reaction 1 analysis: The starting material is a 2-methoxy-1,3-dioxolane bearing a 2-hydroxyethyl chain. Under acid (H+) conditions, the methoxy group at the acetal carbon is protonated and leaves as methanol, generating a stabilized oxocarbenium ion (a cyclic carbocation stabilized by two oxygen atoms of the dioxolane ring). This is a classic SN1-type process: the leaving group (OMe) departs first to form a stable carbocation intermediate, then the pendant hydroxyl group attacks intramolecularly to form the bicyclic product (the epoxide-fused dioxolane). The oxocarbenium ion is tertiary/acetal-stabilized, strongly favoring SN1. Therefore, Reaction 1 proceeds via SN1. Reaction 2 analysis: The substrate is a butyrolactone with a carbon bearing Br and two methyl groups (a neopentyl-like or tertiary-like center within the ring). However, despite the substitution, the carbon bearing Br is inside a rigid 5-membered lactone ring. Azide (N3-) is an excellent nucleophile. In this case, while the carbon is tertiary, the reaction with a good nucleophile like N3- in a lactone system proceeds via SN2 mechanism with inversion. The rigid ring constrains the geometry, and N3- is a strong enough nucleophile to displace Br directly. The product retains the lactone ring with N3 replacing Br. Therefore, Reaction 2 proceeds via SN2. Why other options fail: - (a) SN1, SN1: Reaction 2 is not SN1; N3- is a good nucleophile and the reaction is bimolecular. - (b) SN2, SN2: Reaction 1 is not SN2; it proceeds through a stable oxocarbenium ion (SN1). - (d) SN2, SN1: Reaction 1 is SN1, not SN2; and Reaction 2 is SN2, not SN1. The combination SN1 for Reaction 1 and SN2 for Reaction 2 corresponds to option (c). Therefore, the correct answer is C.