Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives $X$ as the major product. Reaction of X with $\ce{Hg(OAc)2}/\ce{H2O}$ followed by $\ce{NaBH4}$ gives Y as the major product. Y is:

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Hydrocarbons

Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives

X

as the major product. Reaction of X with

Hg (OAc) X_{2} / H X_{2} O

followed by

NaBH X_{4}

gives Y as the major product. Y is:

Answer: (A)

Heating 2-chloro-1-phenylbutane with the strong base

EtOK

drives an E2 elimination, and because a double bond conjugated with the ring is the most stable alkene, the major product

X

is 1-phenylbut-1-ene,

C X_{6} H X_{5} - CH = CH - CH X_{2} CH X_{3}

. Oxymercuration–demercuration then adds water with Markovnikov orientation and no rearrangement, placing the

- OH

on the benzylic carbon, which best supports the positive charge. So

Y

is 1-phenylbutan-1-ol,

C X_{6} H X_{5} - CH (OH) - CH X_{2} CH X_{2} CH X_{3}

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option