See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The starting alcohol is 2-phenyl-3-methylcyclohexan-1-ol. The OH is at C1, Ph is at C2, and methyl is at C3 of the cyclohexane ring. Step 2 - Mechanism of acid-catalyzed dehydration: H3PO4 with heat promotes E1 elimination via protonation of OH to form water leaving group, generating a carbocation at C1, followed by loss of a proton from an adjacent carbon. Step 3 - Carbocation formation: Loss of water from C1 gives a secondary carbocation at C1. However, the adjacent C2 bears a phenyl group. A 1,2-hydride or 1,2-phenyl shift is not needed here; instead, the carbocation at C1 is stabilized by the adjacent phenyl at C2 through hyperconjugation/benzylic-type resonance if a shift occurs. Actually, the carbocation at C1 (secondary) can be stabilized by the phenyl at C2 via a 1,2-phenyl shift to give a tertiary benzylic carbocation at C1 with Ph migrated, but more directly, elimination (E1) occurs by loss of a proton from C2 or C6. Step 4 - Zaitsev's rule and thermodynamic control: Under thermodynamic (hot acid) conditions, the most stable alkene is formed. Elimination toward C2 (which bears Ph) would give a double bond between C1 and C2, placing Ph adjacent to the double bond (making it a styrene-type/conjugated system stabilized by Ph). This gives 1-phenyl-3-methylcyclohex-1-ene (double bond between C1-C2 with Ph at C2 now vinylic, and methyl at C3). Step 5 - Most stable product: The alkene with the double bond conjugated with (or allylic to) the phenyl ring is more stable. The double bond between C1-C2 with Ph at C2 gives an enyl-phenyl (styrenic) conjugation, making this the thermodynamically preferred product. This corresponds to option (c): a cyclohexene with Ph on the double bond carbon and methyl at the adjacent carbon. Step 6 - Why other options fail: - Option (a): Would place the double bond away from Ph, losing conjugation stabilization. - Option (b): Different regiochemistry, less stable. - Option (d): Another regioisomer without the conjugation benefit. The major product is the one where the double bond is between C1 and C2 (conjugated with Ph), retaining the methyl at C3, consistent with option (c). Therefore, the correct answer is C.