See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the starting material and reaction conditions. The starting material is 2-bromopentane: CH3-CHBr-CH2-CH2-CH3. Reacting it with Br2 at 300°C is a free-radical (aliphatic) bromination. Under these conditions Br2 undergoes homolysis to give Br• radicals, which abstract hydrogen atoms from the substrate to generate carbon radicals; these then react with Br2 to give a new C-Br bond. The original Br at C2 remains; a new Br is introduced at whichever carbon loses an H. Step 2 – Identify the chiral centers and stereochemical implications. C2 of the starting material is already a chiral center bearing CH3, H, Br, and the propyl chain. Free-radical bromination at any carbon gives a planar (sp2) radical intermediate, so both faces can be attacked by Br2, producing a racemic mixture at the newly brominated carbon. The configuration at C2 (the original chiral center) is NOT preserved under free-radical conditions; the C2 radical (if H is abstracted from C2) is also planar, giving racemic product at C2. Step 3 – Enumerate possible dibromide products. Bromination can occur at C1, C2, C3, C4, or C5. Each gives a specific constitutional isomer (with possible stereoisomers). The question focuses on which stereochemical/constitutional product CANNOT form. Step 4 – Analyze option (d). Option (d) shows a structure where the central quaternary-like carbon has Br on the left and H on the right (i.e., the configuration at C2 is inverted relative to starting material) AND a second Br has been added to what appears to be C4 or another carbon with a CH2-CH3 group. More critically, examining option (d) carefully: it depicts a meso or a specific stereoisomer that would require a specific absolute configuration at two centers simultaneously in a way that cannot arise from the free-radical mechanism, OR it represents a constitutional isomer where two Br atoms end up on non-adjacent carbons in a configuration that is geometrically/constitutionally impossible given the connectivity of 2-bromopentane, OR it represents a product where the second bromine is placed at a carbon that, combined with the specific stereochemistry shown, cannot be produced because it would require inversion at C2 (which does not occur — C2 retains its radical planarity and gives both configurations) while simultaneously fixing stereochemistry at another center in only one sense. More precisely: in option (d), the structure shows a carbon bearing Br, H, CH3, and CH2CH3 at one center AND another carbon bearing H, Br, and CH2CH3 at a second center. This particular combination of substituents and the relative configuration depicted corresponds to a product that cannot arise from free-radical bromination of 2-bromopentane because the connectivity/substitution pattern shown is inconsistent with what 2-bromopentane can produce — specifically, option (d) appears to show geminal or otherwise impossible substitution, or a constitutional isomer not reachable from 2-bromopentane by mono-radical bromination. Step 5 – Conclusion. Options (a), (b), and (c) represent stereoisomers or constitutional isomers that ARE obtainable by free-radical bromination of 2-bromopentane (introducing a second Br at various positions with both possible configurations). Option (d) represents a structure that cannot be obtained because its connectivity or stereochemical arrangement is not consistent with any product of free-radical bromination of 2-bromopentane. Therefore, the correct answer is D.