See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Electrophilic addition of bromine to alkenes proceeds via a bromonium ion intermediate. When the reaction is carried out in a nucleophilic solvent such as methyl alcohol (CH3OH), the solvent itself can act as a nucleophile and open the bromonium ion intermediate. Step 1: When Br2 reacts with ethylene, the first step is the formation of a cyclic bromonium ion intermediate (a three-membered ring with Br bridging the two carbons). This is the 'ion formed initially.' Step 2: This bromonium ion intermediate is an electrophile that can be attacked by any available nucleophile in solution. In methyl alcohol solution, two nucleophiles are present: Br- (the bromide ion generated when Br2 donated Br+ to the alkene) and CH3OH (the solvent methyl alcohol). Step 3: If Br- attacks the bromonium ion, the product is 1,2-dibromoethane (BrCH2CH2Br). If CH3OH attacks the bromonium ion, the product is BrCH2CH2OCH3 (after deprotonation). Both pathways are available, which is why both products are observed. Step 4: This directly matches option (b): 'the ion formed initially may react with Br- or CH3OH.' Why other options fail: - Option (a): Solvation of bromine by methyl alcohol does not directly explain why a different product (BrCH2CH2OCH3) forms; solvation is not the mechanistic reason. - Option (c): This is an ionic (electrophilic addition) mechanism, not a free radical reaction. Free radical reactions give different regiochemistry and products. - Option (d): Markovnikov's rule applies to addition of HX to unsymmetrical alkenes; ethylene is symmetrical and this reaction proceeds via a bromonium ion, not carbocation, so Markovnikov's rule is not the relevant explanation here. Therefore, the correct answer is A.