Acetone `overset(NH_(2)OH)rarr X overset(LiAIH_(4))rarr Y`. In the above sequence Y is

Options

A. tertiary amine
B. secondary amine
C. primary amine
D. 2-amino propane.

Explanation

To solve the problem step by step, we will analyze the reactions involved in the sequence provided.

### Step 1: Identify the Reaction of Acetone with Hydroxylamine
Acetone (C₃H₆O) reacts with hydroxylamine (NH₂OH) to form an oxime. The reaction can be represented as follows:

1. **Starting Material**: Acetone (C₃H₆O)
2. **Reagent**: Hydroxylamine (NH₂OH)
3. **Reaction**: The carbonyl group (C=O) of acetone reacts with hydroxylamine, leading to the formation of acetoxime (X) and the release of water (H₂O).

The reaction can be summarized as:
\[ \text{C}_3\text{H}_6\text{O} + \text{NH}_2\text{OH} \rightarrow \text{C}_3\text{H}_7\text{NO} + \text{H}_2\text{O} \]
Where:
- Acetoxime (X) is formed:
\[ \text{CH}_3\text{C}(=N\text{OH})\text{CH}_3 \]

### Step 2: Identify the Reaction of Acetoxime with Lithium Aluminum Hydride
Next, we need to reduce the acetoxime (X) formed in the previous step using lithium aluminum hydride (LiAlH₄):

1. **Starting Material**: Acetoxime (C₃H₇NO)
2. **Reagent**: Lithium Aluminum Hydride (LiAlH₄)
3. **Reaction**: Lithium aluminum hydride is a strong reducing agent that will convert the oxime into a primary amine. The reaction will replace the oxime group (N=OH) with an amine group (NH₂).

The reaction can be summarized as:
\[ \text{C}_3\text{H}_7\text{NO} + \text{LiAlH}_4 \rightarrow \text{C}_3\text{H}_9\text{N} + \text{H}_2\text{O} \]
Where:
- The product (Y) formed is 2-aminopropane:
\[ \text{CH}_3\text{C}(NH_2)\text{CH}_3 \]

### Final Answer
Thus, the compound Y formed after the sequence of reactions is **2-aminopropane**.