See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: When Br2 reacts with an alkene in the presence of water (step 1: Br2, step 2: H2O), the reaction proceeds via bromonium ion formation followed by nucleophilic attack by water to give a bromohydrin. The bromonium ion forms on the less hindered face, and water attacks anti to bromine at the more substituted carbon (Markovnikov-like regioselectivity for nucleophile). Step 1 – Identify the alkene: The starting material is 1-ethylcyclohex-1-ene (a cyclohexene ring with an ethyl group at C1, the more substituted carbon of the double bond). Step 2 – Bromonium ion formation: Br2 attacks the double bond to form a cyclic bromonium ion. The bromonium bridges C1 and C2 of the ring. Step 3 – Regioselectivity of water attack: Water (the nucleophile in step 2) attacks the more substituted carbon (C1, which bears the ethyl group) from the side anti to the bromonium (anti addition). This places OH at C1 (the more substituted carbon) and Br at C2 (the less substituted carbon), consistent with Markovnikov regioselectivity for the nucleophile. Step 4 – Stereochemistry: Anti addition means OH and Br are trans to each other across the ring. In option (b), C1 has CH2CH3 (wedge) and OH (wedge) — both on the same face — while C2 has Br (dash) and H (down/bold), showing OH at C1 and Br at C2 in an anti relationship consistent with anti addition through the bromonium. Step 5 – Why other options fail: - Option (a): Shows dibromide addition product (Br2 addition without water), not a bromohydrin. Incorrect product for conditions with H2O. - Option (c): Shows an allylic or vinylic bromination product with the double bond intact, which would require different conditions (e.g., NBS or HBr). Incorrect. - Option (d): Similar to (b) but the stereochemistry at C2 shows H on dash and Br bold-down in a configuration that does not correctly represent anti addition stereochemistry, or the relative orientation differs from the expected anti-addition product. Option (b) correctly shows the bromohydrin product with OH at the more substituted carbon (C1, Markovnikov), Br at C2, and anti stereochemistry. Therefore, the correct answer is B.