See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic addition of Br2 to an alkene in the presence of a nucleophilic solvent (here CH3OH acting like water) proceeds via a bromonium ion intermediate, which is then attacked by the nucleophile. Step 1 – Identify the alkene and the bromonium ion formation: Methyl vinyl ether is H2C=CH-OCH3. When Br2 approaches, it forms a cyclic bromonium ion across the double bond. The bromonium ion is unsymmetrical because the OCH3 group on C2 stabilizes positive charge at C2 (oxygen lone pairs donate electron density). Therefore, the bromonium ion has more carbocation character at C2 (the carbon bearing OCH3). Step 2 – Nucleophilic attack by CH3OH: CH3OH, acting as water would, attacks the more electrophilic carbon (C2, adjacent to OCH3) in an anti fashion. However, because C2 bears the OCH3 group and has significant carbocation character, attack occurs preferentially at C2. After attack by CH3OH and loss of a proton, C2 gains an OCH3 from solvent. C1 retains the Br from ring opening. Step 3 – Product formation: C1: -CH2Br (Br from bromonium) C2: -CH(OCH3)(OCH3) (original OCH3 plus new OCH3 from methanol) This gives BrCH2-CH(OCH3)2, which matches option (b): carbon 1 has Br and two H's, carbon 2 has two OCH3 groups. Step 4 – Why other options fail: (a) Shows OCH on one carbon and H3CO on another — incorrect connectivity and groups. (c) Shows geminal dibromide-like or wrong substitution pattern with two OCH3 on C1 — incorrect regiochemistry. (d) Shows an ester (bromoacetate) — would require oxidation, not simple electrophilic addition. Therefore, the correct answer is B.