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 the compound Z in the given series of reactions, we will analyze each step carefully.

### Step 1: Reaction of Ethyl Iodide with Alcoholic KOH
- **Starting Compound**: Ethyl Iodide (C₂H₅I)
- **Reagent**: Alcoholic KOH
- **Type of Reaction**: Dehydrohalogenation (elimination reaction)

**Reaction**:
When ethyl iodide reacts with alcoholic KOH, a hydrogen atom from the carbon adjacent to the carbon bearing the iodine is removed along with the iodine atom. This results in the formation of a double bond between the two carbon atoms.

**Product (X)**:
\[ \text{C}_2\text{H}_5\text{I} \xrightarrow{\text{Alcoholic KOH}} \text{C}_2\text{H}_4 + \text{HI} \]
The product is Ethene (C₂H₄).

### Step 2: Reaction of Ethene with Bromine (Br₂)
- **Starting Compound**: Ethene (C₂H₄)
- **Reagent**: Bromine (Br₂)
- **Type of Reaction**: Electrophilic addition

**Reaction**:
When ethene reacts with bromine, bromine adds across the double bond, resulting in a dibromide.

**Product (Y)**:
\[ \text{C}_2\text{H}_4 + \text{Br}_2 \rightarrow \text{C}_2\text{H}_4\text{Br}_2 \]
The product is 1,2-Dibromoethane (C₂H₄Br₂).

### Step 3: Reaction of 1,2-Dibromoethane with KCN
- **Starting Compound**: 1,2-Dibromoethane (C₂H₄Br₂)
- **Reagent**: Potassium cyanide (KCN)
- **Type of Reaction**: Nucleophilic substitution

**Reaction**:
In this step, the cyanide ion (CN⁻) acts as a nucleophile and attacks the carbon atoms bonded to bromine. Since cyanide is an ambident nucleophile, it can attack from either carbon. However, it will preferentially attack the carbon atom due to the stability of the resulting bond.

**Product (Z)**:
When both bromine atoms are replaced by cyanide groups, the product formed is:
\[ \text{C}_2\text{H}_4\text{Br}_2 + 2 \text{KCN} \rightarrow \text{C}_2\text{H}_4\text{(CN)}_2 + 2 \text{KBr} \]
The product is 1,2-Dicyanoethane (C₂H₄(CN)₂).

### Final Answer:
The compound Z is **1,2-Dicyanoethane (C₂H₄(CN)₂)**.

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