In the given reaction `CH_3-CH_2-COOHoverset((i)AgNO_3)underset((ii)Br_2//Delta)rarr(X)` will be

Options

A. Ethyl bromide
B. Propyl bromide
C. Propyl propanote
D. All of these

Answer

To solve the given question step by step, we will analyze the reaction of propanoic acid (CH₃-CH₂-COOH) with silver nitrate (AgNO₃) followed by bromine (Br₂) under heating conditions.

### Step 1: Identify the Reactants
The reactant is propanoic acid, which has the structure:
\[ \text{CH}_3-\text{CH}_2-\text{COOH} \]

### Step 2: Reaction with Silver Nitrate (AgNO₃)
When propanoic acid reacts with silver nitrate, a decarboxylation reaction occurs. This means that the carboxylic acid group (-COOH) loses carbon dioxide (CO₂). The reaction can be summarized as follows:
\[ \text{CH}_3-\text{CH}_2-\text{COOH} + \text{AgNO}_3 \rightarrow \text{CH}_3-\text{CH}_3 + \text{CO}_2 + \text{Ag} \]

After this reaction, we are left with ethane (CH₃-CH₃).

### Step 3: Reaction with Bromine (Br₂)
Next, we take the product from the previous step (ethane, CH₃-CH₃) and react it with bromine (Br₂) under heating conditions. In this reaction, bromine will replace one of the hydrogen atoms in ethane to form ethyl bromide (CH₃-CH₂Br). The reaction can be summarized as follows:
\[ \text{CH}_3-\text{CH}_3 + \text{Br}_2 \rightarrow \text{CH}_3-\text{CH}_2\text{Br} + \text{HBr} \]

### Step 4: Identify the Final Product
The final product after both reactions is ethyl bromide (CH₃-CH₂Br).

### Conclusion
Thus, the product \( X \) in the given reaction is ethyl bromide.

### Final Answer
The answer is:
\[ X = \text{CH}_3-\text{CH}_2\text{Br} \text{ (Ethyl bromide)} \]

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