Heating of 2-chloro-1-phenylbutane with EtOK / EtOH gives X as the major product. Reaction X with Hg ( OAc ) _ 2 / H _ 2 O followed by NaB H _ 4 gives Y as the major product. Y is:

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Correct answer: A.

EtOK is potassium ethylene oxide and it is colourless gas and is liquid at low temperatures and is highly flammable i.e. that catches fire easily and undergoes alkylation reactions with many organic compounds and EtOH is the ethanol.

The heating of 2-chloro-1-phenylbutane with $EtOK / EtOH$ gives X as the major product. Reaction X with H $g{( OAc )}_{2} / {H}_{2}O$ followed by ${NaBH}_{4}$ gives Y as the major product. Y is 1-phenylbutanol.

Heating of 2-chloro-1-phenylbutane with EtOK / EtOH gives X as the major product. Reaction X with Hg ( OAc ) _ 2 / H _

Heating of 2-chloro-1-phenylbutane with EtOK / EtOH gives X as the...

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