The major product formed when alkyl iodides is treated with excess of ammonia

Options

A. priamry amine
B. Secondary amine
C. Tertiary amine
D. quatermary ammonium salt

Answer

To determine the major product formed when alkyl iodides are treated with excess ammonia, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with an alkyl iodide, which can be represented as R-I, where R is the alkyl group and I is iodine.
- We also have excess ammonia (NH₃) as the nucleophile.

2. **Nucleophilic Attack**:
- Ammonia (NH₃) has a lone pair of electrons on the nitrogen atom, making it a good nucleophile.
- The nitrogen atom will attack the carbon atom of the alkyl iodide (R-I). This results in the displacement of the iodine atom (I⁻) as a leaving group.

3. **Formation of an Ammonium Ion**:
- When the nucleophilic attack occurs, the nitrogen atom from ammonia forms a bond with the carbon atom, leading to the formation of an ammonium ion (R-NH₃⁺).
- This ammonium ion is unstable due to the positive charge on nitrogen.

4. **Deprotonation**:
- The ammonium ion (R-NH₃⁺) can lose a proton (H⁺) to stabilize itself, resulting in the formation of a primary amine (R-NH₂).
- The reaction can be summarized as follows:
\[ R-I + NH₃ \rightarrow R-NH₃^+ \rightarrow R-NH₂ + H^+ \]

5. **Final Product**:
- The major product of this reaction is a primary amine (R-NH₂).
- Since we are using excess ammonia, the reaction does not proceed further to form secondary or tertiary amines or quaternary ammonium salts.

### Conclusion:
The major product formed when alkyl iodides are treated with excess ammonia is a primary amine (R-NH₂).