Alkyl halide on heating with alc. `NH_3` in a sealed tube results

Options

A. `1^@` amino
B. `2^@` amino
C. `3^@` amino
D. all of these

Explanation

To solve the question regarding the reaction of alkyl halides with alcoholic ammonia in a sealed tube, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- We have an alkyl halide (RX), where R is the alkyl group and X is the halogen (like Cl, Br, or I).
- We are reacting this with alcoholic ammonia (NH₃ in alcohol).

2. **Understand the Mechanism:**
- Ammonia (NH₃) acts as a nucleophile due to the presence of a lone pair of electrons on the nitrogen atom.
- The alkyl halide (RX) has a polar bond due to the electronegativity difference between R and X, making the carbon atom partially positive (electrophilic center).

3. **Nucleophilic Attack:**
- The lone pair of electrons from ammonia attacks the electrophilic carbon in the alkyl halide (RX).
- This results in the formation of an ammonium ion (RNH₃⁺) and the halide ion (X⁻) is released.

4. **Deprotonation:**
- The ammonium ion (RNH₃⁺) can lose a proton (H⁺) to form a primary amine (RNH₂).
- The reaction can be represented as:
\[ RX + NH₃ \rightarrow RNH₃^+ + X^- \]
\[ RNH₃^+ \rightarrow RNH₂ + H^+ \]

5. **Further Reactions:**
- The primary amine (RNH₂) can further react with more alkyl halide (RX) in a similar manner:
- The nitrogen in RNH₂ can act as a nucleophile and attack another alkyl halide, forming a secondary amine (R₂NH).
- This process can continue, leading to the formation of tertiary amines (R₃N) and quaternary ammonium salts (R₄N⁺X⁻).

6. **Final Products:**
- The final products of the reaction include:
- Primary amine (RNH₂)
- Secondary amine (R₂NH)
- Tertiary amine (R₃N)
- Quaternary ammonium salt (R₄N⁺X⁻)

### Conclusion:
The reaction of alkyl halides with alcoholic ammonia in a sealed tube results in the formation of primary, secondary, and tertiary amines, as well as quaternary ammonium salts.