Excess aniline undergoes alkylation with methyl iodide to yield which of the following ?

Options

A. N,N-dimethylaniline
B. N-methyl aniline
C. tetramethylammonium iodide salt
D. There is no reaction

Answer

To solve the question regarding the alkylation of excess aniline with methyl iodide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are aniline (C6H5NH2) and methyl iodide (CH3I).

2. **Understand the Reaction Mechanism**:
- Aniline has a lone pair of electrons on the nitrogen atom, which allows it to act as a nucleophile. Methyl iodide is an alkyl halide that can undergo nucleophilic substitution.

3. **Nucleophilic Attack**:
- In the presence of excess aniline, the nitrogen atom of aniline attacks the carbon atom of methyl iodide. This is an SN2 reaction where the iodide ion (I-) leaves as a leaving group.

4. **Formation of N-Methyl Aniline**:
- The product formed from this reaction is N-methyl aniline (C6H5NHCH3). The reaction can be represented as:
\[ \text{C6H5NH2} + \text{CH3I} \rightarrow \text{C6H5NHCH3} + \text{HI} \]

5. **Consider Further Alkylation**:
- Although N-methyl aniline can potentially react with another molecule of methyl iodide, the problem states that aniline is in excess while methyl iodide is limited. Therefore, N-methyl aniline will not further react with methyl iodide to form N,N-dimethyl aniline.

6. **Final Product**:
- Since further alkylation does not occur due to the limited amount of methyl iodide, the final product of the reaction is N-methyl aniline.

### Conclusion:
The product obtained from the reaction of excess aniline with methyl iodide is **N-methyl aniline**.