See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The starting material is acenaphthylene, which consists of a naphthalene unit fused with a five-membered ring that contains one C=C double bond (the 1,2-double bond of the five-membered ring). Its molecular formula is C12H8. Step 2 - Identify the reaction: Addition of 1 equivalent of HBr to acenaphthylene gives C13H17Br. Wait - acenaphthylene is C12H8, and HBr adds across the double bond to give C12H9Br (C12H8 + HBr = C12H9Br, which is C12H9Br, not C13H17Br). Re-examining: the product is C13H17Br. The starting material shown has molecular formula consistent with acenaphthene or a partially saturated analog. Looking more carefully, the starting material appears to be a tricyclic structure - it looks like acenaphthylene (C12H8) but the product C13H17Br suggests a larger starting material. Actually, counting the structure: it appears to be a fluorene-like or phenalene-like structure. The structure shown is acenaphthylene fused with an additional ring - this is actually a structure with formula C13H16 (acenaphthene with an extra CH2, or fluoranthene partially reduced). Given C13H16 + HBr = C13H17Br, the starting material is C13H16, which is consistent with the tricyclic structure shown (acenaphthylene with one additional CH2 giving a larger five-membered ring, i.e., fluorene-type or acenaphthene extended). Step 3 - Electrophilic addition of HBr: HBr adds to the C=C double bond in the five-membered ring following Markovnikov's rule. The proton (H+) adds to the carbon that gives the more stable carbocation. In this fused ring system, the carbocation formed at the benzylic position (adjacent to the aromatic naphthalene system) is more stable because it is stabilized by resonance with the aromatic pi system. Step 4 - Regiochemistry (Markovnikov): The proton adds to the less substituted or less stable carbocation-forming carbon, and Br- then attacks the more stable (more substituted/benzylic) carbocation. Both carbons of the double bond in the five-membered ring are benzylic, but the geometry and substitution dictate which carbocation is more stable. Since no rearrangement occurs, Br ends up at the more stable carbocation position. Step 5 - Stereochemistry: The bromide ion attacks the carbocation from the less hindered face. In the fused ring system, the less hindered face leads to the product shown in option (d), where Br is on a plain (equatorial/anti) bond at the terminal carbon of the five-membered ring. Step 6 - Why other options fail: - (a) shows Br on a wedge bond - this would be the more hindered face attack, not the major product. - (b) shows Br at a different regiochemical position inconsistent with Markovnikov addition. - (c) shows Br on a dash bond - this corresponds to syn addition from the more hindered face, not the major product. - (d) shows Br in the correct regiochemical position with correct stereochemistry (attack from less hindered face), consistent with Markovnikov's rule and no carbocation rearrangement. Therefore, the correct answer is D.