See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type: Br2 in CH3OH (methanol) is a halogenation in nucleophilic solvent. Instead of dibromide formation (Br2/CCl4), the methanol acts as the nucleophile, giving a bromomethoxy (haloetherification) product: one carbon gets Br and the other gets OCH3. Step 2 - Identify the alkene and its substitution: The starting material is a 1,1-disubstituted cyclohexene (trisubstituted alkene). C1 bears H (wedge), H3C (dash), and a CH3 group, with the double bond extending to C2 of the ring. Step 3 - Mechanism: Br2 attacks the alkene pi bond to form a bromonium ion intermediate. The bromonium bridges the two carbons of the double bond. Methanol then attacks the more substituted carbon (Markovnikov selectivity, anti addition) from the back side (anti to bromine). Step 4 - Stereochemistry: The bromonium ion forms from one face. Anti addition means Br and OCH3 end up on opposite faces (trans/anti relationship). Given the stereochemistry of the starting material (H on wedge, H3C on dash at C1), the bromonium forms and methanol attacks anti to give a specific stereoisomer. Step 5 - Product: The product has OCH3 at the more substituted carbon (C1) and Br at C2 (less substituted), with anti stereochemistry. This corresponds to option (d): H and CH3/OCH3 on one carbon, CH3 and Br on the adjacent carbon with the appropriate anti relationship. Step 6 - Elimination of other options: (a) shows a dibromide-like arrangement without OCH3 at the correct position; (b) places OCH3 and Br on the same carbon which is not the methoxybromination product; (c) shows a vicinal dibromide (no OCH3), which would be the product in CCl4 not methanol. Option (d) correctly shows the bromomethoxy product with anti addition stereochemistry. Therefore, the correct answer is D.