A organic compound (P ) upon reacting with `NH_(3)` gives (Q). On heating, (Q) gives (R). (R) in presence of KOH reacts with `Br_(2)` to give ethylamine. (P) is :

Options

A. `CH_(3)-underset(CH_(3))underset(|)(CH)-COOH`
B. `CH_(3)CH_(2)COOH`
C. `CH_(3)COOH`
D. `CH_(3)CH_(2)CH_(2)COOH`

Explanation

To solve the problem, we need to identify the organic compound (P) that reacts with ammonia (NH₃) to form compound (Q), which upon heating gives compound (R). Finally, (R) reacts with KOH and Br₂ to yield ethylamine.

### Step-by-Step Solution:

1. **Identify the Final Product (Ethylamine)**:
- The final product we need to obtain is ethylamine (C₂H₅NH₂).

2. **Understanding the Reaction of (R) with KOH and Br₂**:
- The reaction of an amide with KOH and Br₂ is known as the Hofmann degradation reaction. This reaction reduces the carbon chain of the amide by one carbon atom, yielding a primary amine.
- Since we are getting ethylamine (C₂H₅NH₂) from (R), it implies that (R) must be an amide with three carbons (C₃H₇NO).

3. **Determine (R)**:
- If (R) leads to ethylamine, we can deduce that (R) is propanamide (C₂H₅C(O)NH₂), which has three carbons.

4. **Relate (Q) to (R)**:
- Since (R) is propanamide, we can see that (Q) must be the corresponding ammonium salt formed from the reaction of propanoic acid with ammonia.
- Therefore, (Q) is propanamide (C₂H₅C(O)NH₂) formed by the reaction of propanoic acid with ammonia.

5. **Identify (P)**:
- To find (P), we recognize that (P) must be the carboxylic acid that reacts with ammonia to form (Q). Since (Q) is propanamide, (P) must be propanoic acid (C₂H₅COOH).

### Conclusion:
The organic compound (P) is propanoic acid (C₂H₅COOH).