See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Br2 addition to an alkene proceeds via anti addition through a bromonium ion intermediate. The two bromine atoms add to opposite faces of the double bond (anti-stereochemistry). Step 2 - Starting material analysis: The starting material is (R)-3-bromocyclopentene. The double bond is between C3a and C4 (or C1–C2 of the cyclopentene numbering shown), and the existing bromine is at C3 (the allylic position) on a wedge bond (R configuration). Step 3 - Bromonium ion formation: Br2 attacks the double bond from either the top face or the bottom face, forming a bromonium ion intermediate. Because the starting material is enantiopure (R), attack from the two faces is diastereotopic — the two faces are not equivalent due to the existing stereocenter at C3. Step 4 - Attack from each face gives two different diastereomers: - Attack of Br2 from the top face followed by backside attack of Br⁻ from the bottom gives one diastereomer. - Attack of Br2 from the bottom face followed by backside attack of Br⁻ from the top gives the other diastereomer. These two products are Y and Z. Step 5 - Identifying the optically inactive product: Y is stated to be optically inactive (does not rotate plane-polarized light). Optical inactivity in a compound that still has stereocenters means it is a meso compound (internal plane of symmetry cancels chirality). A meso compound would result when the configuration at the two newly formed stereocenters is such that the molecule has an internal mirror plane. Step 6 - Evaluating option (c): Option (c) shows the cyclopentane ring with the original Br at C3 on a wedge, one new Br at C1 on a dashed wedge (going back), and one new Br at C2 on a bold wedge (coming forward). The anti addition places the two new bromines on opposite faces. With the existing C3-Br on wedge, the combination of stereocenters in option (c) produces a meso relationship — the molecule has an internal plane of symmetry passing through C3 and the midpoint of C1–C2, making the two halves mirror images of each other. This gives an achiral (meso) compound that is optically inactive. Step 7 - Why other options fail: - (a) shows both new Br atoms on the same face (both dashes) which would be syn addition — not the mechanism of Br2 addition. - (b) shows new Br atoms on the same face as each other (both on same side), inconsistent with anti addition or gives a chiral product. - (d) appears to show only monosubstitution or lacks the correct tribromide pattern. - (e) lacks the C3-Br or shows an incorrect stereochemical arrangement that would give an optically active product. Step 8 - Conclusion: The anti addition of Br2 to (R)-3-bromocyclopentene, when bromonium forms on the face that leads to a meso tribromide upon backside Br⁻ attack, gives the optically inactive (meso) product shown in option (c). Therefore, the correct answer is C.