Identify (Y) in the following reaction series
`C_(2)H_(5)Ioverset("Alcoholic")underset("KOH")rarr (X)overset(Br_(2))rarr(Y)`

Options

A. `CH_(3)-CH_(2)-CN`
B. `underset(Br)underset(|)CH_(2)-underset(Br)underset(|)CH_(2)`
C. `underset(Br)underset(|)CH_(2)-underset(CN)underset(|)CH_(2)`
D. `underset(Br)underset(|)CH=underset(CN)underset("| ")"CH"`

Answer

To identify compound (Y) in the reaction series provided, we will follow the steps of the reaction carefully.

### Step-by-Step Solution:

1. **Starting Compound Identification**:
The starting compound is ethyl iodide, which is represented as \( C_2H_5I \).

2. **Reaction with Alcoholic KOH**:
When ethyl iodide reacts with alcoholic KOH, it undergoes an elimination reaction. Alcoholic KOH acts as a strong base, promoting the removal of a hydrogen atom and the iodine atom from the ethyl iodide.

The reaction can be represented as:
\[ C_2H_5I + KOH \rightarrow C_2H_4 + KI \]
Here, \( C_2H_4 \) (ethylene) is formed as compound (X).

3. **Formation of Compound (Y)**:
Next, the compound (X), which is ethylene \( C_2H_4 \), reacts with bromine \( Br_2 \). The reaction of ethylene with bromine is an electrophilic addition reaction. The double bond in ethylene acts as a nucleophile and attacks the bromine molecule.

The reaction can be represented as:
\[ C_2H_4 + Br_2 \rightarrow C_2H_4Br_2 \]
This results in the formation of 1,2-dibromoethane, which is compound (Y).

4. **Final Structure of Compound (Y)**:
The structure of compound (Y) is:
\[ CH_2Br - CHBr_2 \]
or simply \( C_2H_4Br_2 \).

### Conclusion:
Thus, the compound (Y) formed in the reaction series is 1,2-dibromoethane.