Identify Z in the following series:
` C_(2) H_(5) I underset("KOH")overset("Alcoholic")(rarr) " X " overset(Br_(2))(rarr) " Y " overset("KCN")(rarr) Z `

Options

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

Answer

To identify Z in the given series of reactions, we will follow the steps outlined in the question.

### Step 1: Identify the starting compound
The starting compound is ethyl iodide, which has the formula \( C_2H_5I \) or \( CH_3CH_2I \).

### Step 2: Reaction with Alcoholic KOH
When ethyl iodide reacts with alcoholic potassium hydroxide (KOH), it undergoes a beta elimination reaction (dehydrohalogenation). In this reaction, the iodine atom (I) and a hydrogen atom from the beta carbon are eliminated, resulting in the formation of ethene (X).

**Reaction:**
\[ CH_3CH_2I \xrightarrow{KOH \, (alcoholic)} CH_2=CH_2 + HI \]

### Step 3: Reaction with Bromine
Next, the product ethene (X) reacts with bromine (Br2). This reaction results in the addition of bromine across the double bond, forming a dibromo derivative (Y).

**Reaction:**
\[ CH_2=CH_2 + Br_2 \rightarrow CH_2Br-CH_2Br \]

### Step 4: Reaction with Potassium Cyanide
The dibromo compound (Y) then reacts with potassium cyanide (KCN). In this reaction, the cyanide ion (CN-) acts as a nucleophile and attacks the electrophilic carbon atom bonded to bromine. Since there are two bromine atoms, both can be replaced by cyanide ions, leading to the formation of a dicyano derivative (Z).

**Reaction:**
\[ CH_2Br-CH_2Br + 2KCN \rightarrow CH_2(CN)-CH_2(CN) + 2KBr \]

### Conclusion
The final product Z is a dicyano derivative, specifically 1,2-ethanedicarbonitrile, which can be represented as \( CH_2(CN)-CH_2(CN) \) or \( C_2H_2N_2 \).

### Answer:
**Z is 1,2-ethanedicarbonitrile (dicyano derivative).**

---