`C_(2)H_(4) overset(Br_(2))rarr X overset(KCN)rarr Y, Y` is

Options

A. `CH_(3)CH_(2)CN`
B. `Nc-CH_(2)-CH_(2)-CN`
C. `Br-CH_(2)-CH_(2)CN`s
D. `Br-CH=CHCN`

Answer

To solve the question, we need to analyze the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is C₂H₄, which is ethylene (or ethene). Its structure can be represented as:
\[ \text{CH}_2= \text{CH}_2 \]

### Step 2: Reaction with Br₂
When ethylene reacts with bromine (Br₂), an addition reaction occurs. The double bond in ethylene opens up, and bromine atoms add across the double bond. This results in the formation of a vicinal dibromide.

The reaction can be represented as follows:
\[ \text{C}_2\text{H}_4 + \text{Br}_2 \rightarrow \text{BrCH}_2-\text{CHBr} \]

The product formed is 1,2-dibromoethane, which we will denote as X:
\[ \text{X} = \text{BrCH}_2-\text{CHBr} \]

### Step 3: Reaction of X with KCN
Next, we take the compound X (1,2-dibromoethane) and react it with potassium cyanide (KCN). In this reaction, a nucleophilic substitution occurs where the cyanide ion (CN⁻) attacks the carbon atoms that are bonded to the bromine atoms.

The reaction can be represented as:
\[ \text{BrCH}_2-\text{CHBr} + 2 \text{KCN} \rightarrow \text{CNCH}_2-\text{CHCN} + 2 \text{KBr} \]

### Step 4: Identify the final product Y
The final product Y will be a compound where both bromine atoms have been replaced by cyanide groups. The structure of Y can be represented as:
\[ \text{Y} = \text{CNCH}_2-\text{CHCN} \]

### Conclusion
Thus, the final product Y is 1,2-dicyanoethane.

### Final Answer
Y is 1,2-dicyanoethane.

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