See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: NBS (N-bromosuccinimide) is a reagent used for allylic or benzylic free-radical bromination under light or peroxide conditions. It selectively brominates the C-H bond at the position most stabilized as a radical. Step 1 - Identify the substrate: The starting material is 1-methyl-1,2,3,4-tetrahydronaphthalene. The ring system is a fused bicyclic compound where one ring is benzene and the other is a cyclohexane ring. The methyl group is attached at C1, which is the benzylic position (adjacent to the aromatic ring). Step 2 - Identify abstractable hydrogens and radical stability: - C1 (benzylic, tertiary): bears the CH3 group and is part of the ring. However, C1 already has no H because it carries CH3 and is a tertiary carbon attached to the ring — wait, let us reconsider. C1 has: one bond to benzene ring carbon, one bond to CH3, one bond to C2 of cyclohexane ring, and one bond to C8a (the other ring junction). So C1 is a quaternary carbon if it has no H. Actually, in 1-methyl tetralin, C1 has: aromatic ring C, CH3, C2(CH2), and C8a — making it a quaternary carbon with no H available for abstraction. - The CH3 group at C1 is a benzylic methyl group. Abstraction of H from this CH3 gives a benzylic primary radical (PhCH2• type, stabilized by benzylic resonance). - C2: secondary, non-benzylic (one carbon removed), less stable radical. - C4: benzylic secondary position (adjacent to benzene ring on the other side), would give a benzylic secondary radical. Step 3 - Compare radical stabilities: The CH3 attached to C1 is directly benzylic. Abstraction from CH3 gives a primary benzylic radical. C4 is also benzylic (secondary benzylic radical). A secondary benzylic radical is more stable than a primary benzylic radical. However, option (a) shows bromination at the benzylic CH3 position (giving CH2Br), and option (c) shows bromination at C4. Step 4 - Reconsider option (a): Option (a) shows Br at C1 with the methyl still present AND Br on the ring carbon, indicating tertiary bromination at C1. But C1 has no H (it's quaternary). So option (a) cannot arise from H-abstraction at C1. Step 5 - Reconsider the structures: Looking more carefully at the answer choices and the given answer being (a): Option (a) depicts Br at C1 (the carbon bearing the methyl), meaning the methyl-bearing carbon gets brominated. This is only possible if C1 is tertiary (has one H). In 1-methyl tetralin, C1 is indeed tertiary — it connects to: the aromatic ring (C4a/C8a junction carbon), CH3, C2, and one H. So C1 is tertiary and benzylic. Step 6 - Radical stability at C1 vs CH2Br (side chain): A tertiary benzylic radical (at C1) is more stable than a primary benzylic radical (at CH3) or secondary benzylic radical (at C4). NBS selectively abstracts the H at the most stable radical position. Step 7 - Therefore, NBS abstracts the tertiary benzylic H at C1, generating a tertiary benzylic radical, which is then trapped by Br from NBS to give 1-bromo-1-methyl-1,2,3,4-tetrahydronaphthalene. Why other options fail: - (b) Bromination at the CH3 side chain gives a primary benzylic radical — less stable than tertiary benzylic, so not major. - (c) Bromination at C4 gives a secondary benzylic radical — less stable than tertiary benzylic at C1. - (d) Bromination at C2 gives a secondary non-benzylic radical — least stable, not preferred. Therefore, the correct answer is A.