See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Concept: Acid-catalyzed hydration of alkenes follows Markovnikov's rule, and carbocation intermediates can undergo rearrangements (hydride or methyl shifts) to form more stable carbocations. Step 1: Draw 3-phenylbut-1-ene. The structure is CH2=CH-CH(Ph)-CH3, where the double bond is between C1 and C2, and C3 bears a phenyl group and a methyl group. Step 2: Protonation of the double bond according to Markovnikov's rule. The proton adds to C1 (the terminal carbon), placing the positive charge on C2: CH3-CH+(C2)-CH(Ph)-CH3. This gives a secondary carbocation at C2. Step 3: Evaluate possibility of rearrangement. The carbocation at C2 is secondary. Adjacent C3 bears a phenyl group. A 1,2-hydride shift from C3 to C2 would move the positive charge to C3, generating a carbocation at C3 that is stabilized by the adjacent phenyl group (benzylic carbocation). A benzylic secondary carbocation is more stable than a simple secondary carbocation due to resonance delocalization into the aromatic ring. Step 4: The rearranged carbocation is at C3 (benzylic position): CH3-CH2-C+(Ph)-CH3. This is a tertiary benzylic carbocation (C3 is attached to CH3, CH2CH3, and Ph), making it especially stable. Step 5: Water attacks the carbocation at C3, followed by deprotonation, giving the alcohol at C3 of the original chain... Let me re-examine. After the 1,2-hydride shift from C3 to C2: the carbocation is now at C3: structure becomes CH3-CH2(C2 now has H from shift)-C+(C3)(Ph)-CH3. C3 is bonded to Ph, CH3, and C2 (which is now CH2CH3 after receiving the hydride). So C3 carbocation: Ph-C+(CH3)-CH2CH3. This is a tertiary benzylic carbocation. Step 6: Nucleophilic attack of water on C3 gives: Ph-C(OH)(CH3)-CH2CH3, which is 2-phenylbutan-2-ol. Why other options fail: - (a) 3-Phenylbutan-1-ol: would require anti-Markovnikov addition, not favored. - (b) 3-Phenylbutan-2-ol: would be the product without rearrangement (water attacks C2 secondary carbocation), but this is less stable than the rearranged product. - (d) 2-Phenylbutan-1-ol: does not correspond to any logical carbocation intermediate. The rearranged product 2-phenylbutan-2-ol is the major product due to formation of the more stable tertiary benzylic carbocation. Therefore, the correct answer is C.