See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the reaction type: The substrate is a tosylate (OTs) heated (delta), indicating a pyrolytic elimination (E1 or E2). Tosylates are good leaving groups, and heating favors elimination. This is an E1 (or E1cb) elimination proceeding through a carbocation intermediate or a concerted pyrolytic pathway. Step 2 - Identify the substrate structure: The molecule is Ph-C(CH3)2-C^14(H)(OTs)-CH3. C^14 bears OTs and H, flanked by the quaternary carbon (bearing Ph, CH3, CH3) and a terminal CH3. Step 3 - Consider carbocation and rearrangement: If OTs leaves from C^14, a secondary carbocation forms at C^14. However, adjacent to C^14 is a tertiary carbon bearing Ph, CH3, CH3. A 1,2-hydride or 1,2-methyl shift could occur, but more importantly, the phenyl group can migrate (1,2-phenyl shift) from the adjacent tertiary carbon to C^14 to give a more stable tertiary (benzylic) carbocation. Alternatively, a 1,2-methyl shift from the adjacent carbon to C^14 can occur. Step 4 - 1,2-methyl shift: If a methyl group migrates from C(Ph)(CH3)2 to C^14 (which lost OTs), the carbocation moves to the carbon that originally bore Ph and two methyls. After methyl migration: the carbocation is at the original C(Ph)(CH3) position (now tertiary, benzylic). Then elimination of a proton from C^14 gives the alkene. After 1,2-methyl shift: carbocation at C bearing Ph and one CH3 (the other CH3 migrated to C^14). C^14 now bears: H, CH3 (migrated), CH3 (original), and is adjacent to the carbocation. Loss of H from the C^14 position gives: (CH3)2C=C^14(Ph)(CH3). This means the double bond is between the unlabeled carbon (bearing two methyls) and C^14 (bearing Ph and CH3). This matches option (c): (CH3)2C=C^14(Ph)(CH3). Step 5 - Zaitsev / more substituted alkene: Option (c) gives a trisubstituted alkene (two methyls + Ph/methyl = tetrasubstituted effectively), which is highly stable. The rearranged carbocation followed by elimination gives the most stable, most substituted alkene. Step 6 - Why other options fail: (a) Has C^14 bearing two methyls on the double bond carbon - this would require C^14 to gain an extra methyl, inconsistent with the shift. (b) Gives a terminal vinyl group at C^14 - less substituted, not the major product. (d) Has ^14CH3 as a substituent on the double bond, implying the label moved to a methyl group - this would require a different type of rearrangement and gives a less substituted product. Therefore, the correct answer is C.