`Ph-CH_(2)-CH=CH-CH_(3) overset((i)Br_(2))underset((ii)Alc.KOH)to`

Options

A. `Ph-CH=CH-CH=CH_(2)`
B. `Ph-CH_(2)-underset(OH)underset(|)(CH)-underset(OH)underset(|)(CH)-CH_(3)`
C. `Ph-CH_(2)-C-=C-CH_(3)`
D. `Ph-C-=C-CH_(2)-CH_(3)`

Answer

To solve the given problem step by step, we will analyze the reaction of the compound with bromine (Br2) followed by treatment with alcoholic KOH.

### Step 1: Identify the starting compound
The starting compound is:
\[ \text{Ph-CH}_2\text{-CH=CH-CH}_3 \]
This compound has a phenyl group (Ph) attached to a carbon chain that contains a double bond.

### Step 2: Reaction with Br2
When the compound is treated with bromine (Br2), an addition reaction occurs across the double bond. The double bond between the second and third carbon atoms will react with Br2, resulting in the addition of bromine atoms to these carbon atoms.

The reaction can be represented as follows:
\[ \text{Ph-CH}_2\text{-CH=CH-CH}_3 + \text{Br}_2 \rightarrow \text{Ph-CH}_2\text{-CHBr-CHBr-CH}_3 \]

The product after this step is:
\[ \text{Ph-CH}_2\text{-CHBr-CHBr-CH}_3 \]

### Step 3: Reaction with alcoholic KOH
Next, the product from Step 2 is treated with alcoholic KOH. In this step, an elimination reaction occurs where two molecules of HBr are eliminated. The elimination will occur from the carbon atoms that are bonded to bromine and adjacent hydrogen atoms.

The elimination can be represented as follows:
1. One bromine atom and one hydrogen atom from adjacent carbons are removed to form a double bond.
2. The other bromine atom and its adjacent hydrogen are also removed to form a triple bond.

The final product after this elimination reaction is:
\[ \text{Ph-CH}_2\text{-C}\equiv\text{C-CH}_3 \]

### Final Product
Thus, the final product of the reaction is:
\[ \text{Ph-CH}_2\text{-C}\equiv\text{C-CH}_3 \]

### Conclusion
The correct answer is option 3: \( \text{Ph-CH}_2\text{-C}\equiv\text{C-CH}_3 \).

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