Question

Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with $Hg(OAc)_2/H_2O$ followed by $NaBH_4$ gives Y as the major product. Y is

• A. 1-phenylbutan-2-ol
• B. 2-phenylbutan-1-ol
• C. 1-phenylbutan-1-ol
• D. 2-phenylbutan-2-ol

Answer 1

Answer: (C)

1-phenylbutan-1-ol

Answer 2

The reaction begins with 2-chloro-1-phenylbutane (Ph-CH2-CH(Cl)-CH2-CH3). Treatment with a strong base like EtOK/EtOH at elevated temperatures promotes an E2 elimination reaction. The possible beta-hydrogens are on C1 (adjacent to the phenyl group) and C3.

1. Formation of X (Alkene):

   • Elimination of a beta-hydrogen from C1 and the chlorine from C2 yields Ph-CH=CH-CH2-CH3 (1-phenylbut-1-ene).
   • Elimination of a beta-hydrogen from C3 and the chlorine from C2 yields Ph-CH2-CH=CH-CH3 (1-phenylbut-2-ene). According to Zaitsev's rule, the more substituted alkene is favored. However, both are disubstituted. The alkene conjugated with the phenyl ring (1-phenylbut-1-ene) is more stable. Thus, X is predominantly 1-phenylbut-1-ene.

2. Formation of Y (Alcohol) via Oxymercuration-Demercuration: The reaction of alkene X with $Hg(OAc)_2/H_2O$ followed by $NaBH_4$ is oxymercuration-demercuration, a method for hydrating alkenes. This reaction follows Markovnikov's rule, where the hydroxyl group adds to the more substituted carbon of the double bond, and carbocation rearrangements do not occur.

   For X = Ph-CH=CH-CH2-CH3:

   • The double bond is between C1 (attached to Ph) and C2 (attached to CH2CH3).
   • C1, being directly attached to the phenyl ring, is considered more substituted in this context for Markovnikov addition.
   • Therefore, the -OH group adds to C1, and a hydrogen atom adds to C2.
   • The intermediate after oxymercuration is Ph-CH(OH)-CH(HgOAc)-CH2-CH3.
   • Demercuration with $NaBH_4$ replaces the -HgOAc group with a hydrogen atom.
   • The final product Y is Ph-CH(OH)-CH2-CH2-CH3, which is 1-phenylbutan-1-ol.

The major product Y is 1-phenylbutan-1-ol.