`CH_(3)CH_(2)CH_(2)CONH_(2)` is boiled with aqueous NaOH, then the reaction mixture is acidified with HCl. The products obtained are

Options

A. `CH_(3)CH_(2)CH_(2)COO^(-)+NH_(3)`
B. `CH_(3)CH_(2)CH_(2)COONa+NH_(3)`
C. `CH_3CH_(2)CH_(2)COOH+NH_(4)Cl`
D. `CH_(3)CH_(2)CH_(2)COO^(-)+NH_(4)Cl`

Answer

To solve the problem step by step, we will analyze the reaction of the compound `CH₃CH₂CH₂CONH₂` (butanamide) when boiled with aqueous NaOH and then acidified with HCl.

### Step 1: Identify the Reactants
The given compound is butanamide, which can be represented as:
- **Butanamide**: `CH₃CH₂CH₂CONH₂`

### Step 2: Reaction with Aqueous NaOH
When butanamide is boiled with aqueous NaOH, a hydrolysis reaction occurs. The hydroxide ions (OH⁻) from NaOH attack the carbonyl carbon of the amide group, leading to the formation of a carboxylate ion and ammonia.

- The reaction can be represented as:
\[ CH₃CH₂CH₂CONH₂ + NaOH \rightarrow CH₃CH₂CH₂COO^{-}Na^{+} + NH₄^{+} \]

### Step 3: Formation of Carboxylate Salt
After the reaction with NaOH, we obtain a carboxylate salt (sodium butanoate) and ammonium ion:
- **Carboxylate Salt**: `CH₃CH₂CH₂COO⁻Na⁺`
- **Ammonium Ion**: `NH₄⁺`

### Step 4: Acidification with HCl
Next, the mixture is acidified with HCl. The H⁺ ions from HCl will react with the carboxylate ion to form the corresponding carboxylic acid.

- The reaction can be represented as:
\[ CH₃CH₂CH₂COO^{-}Na^{+} + HCl \rightarrow CH₃CH₂CH₂COOH + NaCl \]

### Step 5: Products Obtained
After the acidification, the products obtained are:
1. **Butanoic Acid**: `CH₃CH₂CH₂COOH`
2. **Ammonium Chloride**: `NH₄Cl`

### Final Answer
The products obtained from the reaction are:
- Butanoic Acid: `CH₃CH₂CH₂COOH`
- Ammonium Chloride: `NH₄Cl`

### Summary of Products
- **Main Product**: Butanoic Acid (`CH₃CH₂CH₂COOH`)
- **By-product**: Ammonium Chloride (`NH₄Cl`)