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 reaction step by step, let's break it down:

### Step 1: Identify the starting compound
The starting compound is propanoic acid, which can be represented as:
\[ \text{CH}_3\text{CH}_2\text{COOH} \]

### Step 2: Reaction with AgNO3
When propanoic acid reacts with silver nitrate (AgNO3), it forms a silver salt of the carboxylic acid. The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{AgNO}_3 \rightarrow \text{CH}_3\text{CH}_2\text{COOAg} + \text{HNO}_3 \]
Here, \(\text{CH}_3\text{CH}_2\text{COOAg}\) is the silver salt of propanoic acid.

### Step 3: Reaction with Br2 and heating
Next, the silver salt of propanoic acid reacts with bromine (Br2) upon heating. This reaction is known as the Hunsdiecker reaction, where the silver salt is converted into an alkyl bromide. The reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{COOAg} + \text{Br}_2 \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{Br} + \text{CO}_2 + \text{AgBr} \]
In this step, the carboxylate ion (\(\text{CH}_3\text{CH}_2\text{COO}^-\)) reacts with bromine, leading to the formation of ethyl bromide (\(\text{CH}_3\text{CH}_2\text{Br}\)), carbon dioxide (CO2), and silver bromide (AgBr).

### Step 4: Identify the final product
The final product of the reaction is ethyl bromide, represented as:
\[ \text{X} = \text{CH}_3\text{CH}_2\text{Br} \]

### Conclusion
Thus, the answer to the question is:
\[ \text{X} = \text{CH}_3\text{CH}_2\text{Br} \text{ (ethyl bromide)} \]