`CH_(3)CH_(2)COOHunderset("Red P")overset(Br_(2))tooverset(NH_(3))toY`
Y in the above reaction is

Options

A. lactic acid
B. ethylamine
C. propylamine
D. alanine

Explanation

To solve the problem, we need to analyze the given reaction step by step. The starting compound is propanoic acid (CH₃CH₂COOH), and it undergoes a series of reactions involving bromine (Br₂) in the presence of red phosphorus and then ammonia (NH₃).

### Step-by-Step Solution:

1. **Identify the Starting Compound:**
- The starting compound is propanoic acid (CH₃CH₂COOH).

2. **First Reaction with Br₂ and Red Phosphorus:**
- When propanoic acid reacts with bromine (Br₂) in the presence of red phosphorus, the reaction leads to the formation of an alpha-bromo acid.
- The alpha hydrogen (the hydrogen on the carbon adjacent to the carboxylic acid group) is replaced by a bromine atom.
- The resulting compound is 2-bromopropanoic acid (CH₃CHBrCOOH).

3. **Second Reaction with Ammonia (NH₃):**
- The next step involves the reaction of 2-bromopropanoic acid with ammonia (NH₃).
- Ammonia acts as a nucleophile and attacks the carbon atom that is bonded to the bromine atom.
- As a result, the bromine atom leaves as a bromide ion (Br⁻), and the nitrogen from ammonia (NH₃) forms a bond with the carbon atom.

4. **Formation of the Final Product:**
- The product of this reaction is an amino acid. Specifically, after the nucleophilic attack and the loss of a proton (H⁺) from the nitrogen, the compound formed is alanine (CH₃CH(NH₂)COOH).
- Thus, the final product Y is alanine.

### Final Answer:
Y in the above reaction is **alanine (CH₃CH(NH₂)COOH)**.