`CH_(3)CH_(2)CO OHunderset("Red P")overset(Br_(2))to[X]overset(NH_(3)(alc.))to[Y]`
[Y] in the above reactions is

Options

A. Lactic acid
B. n-propyl amine
C. Ethylamine
D. Alanine

Answer

To solve the given problem, we will follow the steps outlined in the video transcript and provide a clear, step-by-step explanation of the reactions involved.

### Step-by-Step Solution:

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

2. **First Reaction with Bromine and Red Phosphorus**:
The first reaction involves the treatment of propanoic acid with bromine (\( \text{Br}_2 \)) in the presence of red phosphorus. This is known as the Hell-Volhard-Zelinsky (HVZ) reaction. In this reaction, bromine adds to the alpha carbon of the carboxylic acid.

The reaction can be summarized as follows:
\[ \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{Br}_2, \text{Red P}} \text{CH}_3\text{CHBr}\text{COOH} \]
Here, bromine (\( \text{Br} \)) is added to the alpha carbon, resulting in the formation of a bromo acid, which we will denote as compound \( [X] \):
\[ [X] = \text{CH}_3\text{CHBr}\text{COOH} \]

3. **Second Reaction with Ammonia**:
The next step involves treating the bromo acid \( [X] \) with ammonia (\( \text{NH}_3 \)) in alcoholic conditions. In this reaction, ammonia will substitute the bromine atom.

The reaction can be represented as:
\[ \text{CH}_3\text{CHBr}\text{COOH} \xrightarrow{\text{NH}_3(\text{alc})} \text{CH}_3\text{CH(NH}_2)\text{COOH} + \text{HBr} \]
Here, the bromine atom is replaced by an amino group (\( \text{NH}_2 \)), and hydrogen bromide (\( \text{HBr} \)) is eliminated.

4. **Final Product**:
The final product \( [Y] \) is:
\[ [Y] = \text{CH}_3\text{CH(NH}_2)\text{COOH} \]
This compound is known as alanine, which is an amino acid.

### Conclusion:
Thus, the final product \( [Y] \) in the above reactions is **alanine**.