Consider the following sequence of reactions
`C_(2)H_(5)Broverset(AgCN)(to)Aoverset(H_(2)//Ni)(to)B`
The final product (2) is:

Options

A. `CH_(3)CH_(2)CH_(2)NH_(2)`
B. `CH_(3)CH_(2)NHC_(3)`
C. `(CH_(3)CH_(2))NH`
D. `CH_(3)CH_(2)NH_(2)`

Answer

To solve the given sequence of reactions, we will break down the reactions step by step.

### Step 1: Identify the initial reactant
The initial reactant is C₂H₅Br (ethyl bromide).

### Step 2: Reaction with AgCN
When ethyl bromide (C₂H₅Br) reacts with silver cyanide (AgCN), a nucleophilic substitution occurs. The bromine atom (Br) is replaced by the cyanide group (CN).

The reaction can be represented as follows:
\[ \text{C}_2\text{H}_5\text{Br} + \text{AgCN} \rightarrow \text{C}_2\text{H}_5\text{NC} \]

Here, we get ethyl isocyanide (or ethyl isocyanate), which can be represented as C₂H₅-NC.

### Step 3: Hydrogenation with H₂/Ni
The next step involves the hydrogenation of the isocyanide (C₂H₅-NC) using hydrogen gas (H₂) in the presence of a nickel catalyst (Ni). During this reaction, the isocyanide is reduced.

The reaction can be represented as follows:
\[ \text{C}_2\text{H}_5\text{NC} + \text{H}_2 \xrightarrow{\text{Ni}} \text{C}_2\text{H}_5\text{NH}_2 \]

Here, the isocyanide is converted to ethylamine (C₂H₅-NH₂).

### Final Product
The final product (B) after the sequence of reactions is:
\[ \text{C}_2\text{H}_5\text{NH}_2 \] (ethylamine)

### Summary
The final product (2) is ethylamine (C₂H₅NH₂).

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