Consider the following reaction.
`C_(2)H_(5) I overset(Alc. KOH) to X overset(Br_(2)) to Y overset(KCN) to Z`
How many CN group are present in product Z?

Explanation

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

### Step 1: Identify the starting compound and the first reaction
The starting compound is ethyl iodide (C₂H₅I). When it reacts with alcoholic KOH, it undergoes an elimination reaction.

**Reaction:**
C₂H₅I + Alc. KOH → C₂H₄ (Ethylene) + HI

### Step 2: Formation of compound X
In this reaction, the alcoholic KOH acts as a base and removes a hydrogen atom (H⁺) from ethyl iodide, resulting in the formation of ethylene (C₂H₄).

**Product X:**
X = C₂H₄

### Step 3: Reaction of compound X with Br₂
Next, compound X (C₂H₄) reacts with bromine (Br₂). This is an addition reaction where bromine adds across the double bond.

**Reaction:**
C₂H₄ + Br₂ → C₂H₄Br₂ (1,2-dibromoethane)

### Step 4: Formation of compound Y
The product from the previous reaction is 1,2-dibromoethane.

**Product Y:**
Y = C₂H₄Br₂

### Step 5: Reaction of compound Y with KCN
Now, we take compound Y (C₂H₄Br₂) and react it with KCN. In this step, KCN acts as a nucleophile and will replace the bromine atoms (Br) with cyanide groups (CN).

**Reaction:**
C₂H₄Br₂ + 2 KCN → C₂H₄(CN)₂ + 2 KBr

### Step 6: Formation of compound Z
The final product Z will be 1,2-dicyanoethane (C₂H₄(CN)₂).

**Product Z:**
Z = C₂H₄(CN)₂

### Step 7: Count the CN groups in product Z
In the final product Z, there are two cyanide (CN) groups present.

### Final Answer:
The number of CN groups present in product Z is **2**.

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