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

Options

A.
B.
C.
D.

Explanation

To solve the problem step by step, we will analyze the reactions involved and identify the major products formed.

### Step 1: Identify the starting compound
The starting compound is 2-chloro-1-phenylbutane. This compound has a chlorine atom attached to the second carbon of the butane chain, which is also connected to a phenyl group.

### Step 2: Reaction with EtOK/EtOH
When 2-chloro-1-phenylbutane is heated with potassium ethoxide (EtOK) in ethanol (EtOH), it undergoes an elimination reaction (E2 mechanism). In this reaction, the chlorine atom (a good leaving group) is removed along with a hydrogen atom from an adjacent carbon, leading to the formation of a double bond.

- **Elimination Reaction**:
- The chlorine atom leaves, and a hydrogen atom is removed from the adjacent carbon (C-3).
- This results in the formation of an alkene.

The major product (X) of this elimination reaction is **1-phenylbut-2-ene**.

### Step 3: Reaction of X with Hg(OAc)₂/H₂O
Next, we take the alkene (1-phenylbut-2-ene) and react it with mercuric acetate (Hg(OAc)₂) in the presence of water (H₂O). This reaction is known as oxymercuration, which adds a hydroxyl group (OH) across the double bond.

- **Oxymercuration**:
- The alkene reacts with mercuric acetate to form a mercurial intermediate.
- Water then attacks the more substituted carbon of the double bond, leading to the formation of an alcohol.

After the reaction, the product (Y) is **1-phenylbutan-2-ol**, which is an alcohol.

### Step 4: Final product
Thus, the final product Y is **1-phenylbutan-2-ol**.

### Summary of Products
- **X (major product from elimination)**: 1-phenylbut-2-ene
- **Y (major product from oxymercuration)**: 1-phenylbutan-2-ol

### Final Answer
Y is **1-phenylbutan-2-ol**.

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