`CH_(3) CH_(2) Br overset(AgCN) (to) A underset(H_(3)^(oplus) O) overset(NaOH, Delta ) (to) B`, B is :

Explanation

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

### Step 1: Identify the starting compound
The starting compound is **CH₃CH₂Br** (ethyl bromide). This is an alkyl halide.

### Step 2: Reaction with AgCN
When ethyl bromide (CH₃CH₂Br) reacts with **AgCN**, a nucleophilic substitution reaction occurs. The cyanide ion (CN⁻) from AgCN attacks the carbon atom bonded to the bromine atom, leading to the formation of a new compound.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{Br} + \text{AgCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{AgBr} \]

Here, the bromine atom is replaced by a cyanide group (CN), resulting in the formation of **ethyl cyanide (CH₃CH₂CN)**, which we will denote as **A**.

### Step 3: Hydrolysis of A
Next, we treat the compound A (ethyl cyanide) with **H₃O⁺** (acidic conditions) and **NaOH** (basic conditions) under heat (Δ). This process is known as hydrolysis.

During hydrolysis, the cyanide group (CN) is converted into an amine group (NH₂) through the formation of an intermediate carboxylic acid (propanoic acid) before finally forming the amine.

**Hydrolysis Reaction:**
1. The CN group is converted to a carboxylic acid:
\[ \text{CH}_3\text{CH}_2\text{CN} + 2\text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{NH}_3 \]

2. The carboxylic acid (propanoic acid) can then be converted to an amine:
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{H}_2\text{O} \]

### Step 4: Identify the final product B
After the hydrolysis of ethyl cyanide, we obtain **ethylamine (CH₃CH₂NH₂)** as the final product, which we will denote as **B**.

### Final Answer
Thus, the final product B is **CH₃CH₂NH₂** (ethylamine).

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