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 problem of what compound is produced when excess aniline reacts with methyl iodide (CH3I), we can follow these steps:

### Step 1: Understand the Reactants
Aniline (C6H5NH2) is a primary amine. When it reacts with methyl iodide (CH3I), it can undergo alkylation.

### Step 2: First Alkylation Reaction
When aniline reacts with methyl iodide, the nitrogen atom in aniline attacks the methyl group (CH3) of methyl iodide. This is an SN2 reaction, where the iodide ion (I-) is displaced. The product of this reaction is N-methyl aniline (C6H5NH(CH3)) and hydrogen iodide (HI) is released as a byproduct.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{I} \rightarrow \text{C}_6\text{H}_5\text{NH(CH}_3\text{)} + \text{HI} \]

### Step 3: Second Alkylation Reaction
Since we have excess aniline, the newly formed N-methyl aniline can also react with another molecule of methyl iodide. The nitrogen atom in N-methyl aniline can again attack another methyl iodide, leading to the formation of N,N-dimethyl aniline (C6H5N(CH3)2) and another molecule of HI.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{NH(CH}_3\text{)} + \text{CH}_3\text{I} \rightarrow \text{C}_6\text{H}_5\text{N(CH}_3\text{)}_2 + \text{HI} \]

### Step 4: Further Alkylation
If there is still excess methyl iodide, the N,N-dimethyl aniline can react again to form N,N,N-trimethyl aniline (C6H5N(CH3)3) and release another HI.

**Reaction:**
\[ \text{C}_6\text{H}_5\text{N(CH}_3\text{)}_2 + \text{CH}_3\text{I} \rightarrow \text{C}_6\text{H}_5\text{N(CH}_3\text{)}_3 + \text{HI} \]

### Step 5: Final Product
However, if we consider that the question specifies "excess aniline," the final product will predominantly be N-methyl aniline, as the excess aniline will keep reacting with the methyl iodide until it is consumed.

### Conclusion
The main product formed when excess aniline reacts with methyl iodide is **N-methyl aniline** (C6H5NH(CH3)).

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