In the following reaction
`RCH_(2)COOH overset(Br_(2)//P)rarr X overset("excess "NH_(3))rarr Y`
The major compounds X and Y are

Options

A. `RCH(Br)CONH_(2), RCH(NH_(2))COOH`
B. `RCH(Br)COOH,RCH(NH_(2))COOH`
C. `RCH_(2)COBr,RCH_(2)COONH_(4)`
D. `RCH(Br)COOH,RCH_(2)CONH_(2)`

Explanation

To solve the problem, we need to analyze the given reaction step by step.

### Step 1: Identify the starting compound
The starting compound is a carboxylic acid represented as RCH₂COOH. This compound contains an alpha hydrogen (the hydrogen atom on the carbon adjacent to the carboxylic acid group).

**Hint:** Identify the functional groups present in the starting compound to understand how they will react.

### Step 2: Understand the reaction with Br₂ and Red Phosphorus
When a carboxylic acid with alpha hydrogen is treated with bromine (Br₂) in the presence of red phosphorus (P), it undergoes halogenation at the alpha position. This reaction is known as the Hell-Volhard-Zelinsky (HVZ) reaction.

In this case, the alpha hydrogen is replaced by a bromine atom, resulting in the compound:
\[ RCHBrCOOH \]
This compound is our product X.

**Hint:** Remember that the HVZ reaction specifically targets the alpha hydrogen of carboxylic acids.

### Step 3: Reacting compound X with excess NH₃
Next, we treat the compound X (RCHBrCOOH) with excess ammonia (NH₃). The nitrogen in ammonia acts as a nucleophile and attacks the carbon atom that is bonded to the bromine atom. The bromine is a good leaving group, allowing the nitrogen to replace it.

The reaction can be represented as:
\[ RCHBrCOOH + NH₃ \rightarrow RCH(NH₂)COOH + HBr \]
This results in the formation of compound Y, which is an amino acid:
\[ RCH(NH₂)COOH \]

**Hint:** Recognize that nucleophiles can replace leaving groups in organic reactions.

### Conclusion
The major compounds formed in the reaction are:
- **X:** RCHBrCOOH (bromo acid)
- **Y:** RCH(NH₂)COOH (amino acid)

### Final Answer
- **X:** RCHBrCOOH
- **Y:** RCH(NH₂)COOH

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