The reaction of chloroform with alcoholic `KOH` and p-toluidine form-

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Answer

To solve the question regarding the reaction of chloroform with alcoholic KOH and p-toluidine, we will follow these steps:

### Step 1: Identify the Reactants
- **Chloroform (CHCl3)**: A halogenated hydrocarbon.
- **Alcoholic KOH**: A strong base that can deprotonate compounds.
- **p-Toluidine**: An aromatic amine with the structure of a benzene ring with an amino group (-NH2) at the para position relative to a methyl group (-CH3).

### Step 2: Understand the Reaction Mechanism
The reaction involves the formation of isocyanide (also known as carbanamine) through a series of steps:
1. **Deprotonation**: The alcoholic KOH will deprotonate the amino group of p-toluidine, making it a stronger nucleophile.
2. **Formation of Isocyanide**: The deprotonated p-toluidine will then react with chloroform in the presence of KOH. Chloroform acts as a source of carbon and halide ions. The nucleophilic amine attacks the carbon atom in chloroform, leading to the formation of isocyanide.

### Step 3: Write the Balanced Reaction
The overall reaction can be summarized as:
\[ \text{p-Toluidine} + \text{CHCl}_3 + \text{KOH} \rightarrow \text{Isocyanide} + 3 \text{HCl} + 3 \text{H}_2\text{O} \]

### Step 4: Identify the Product
The main product formed from this reaction is **p-Tolyl isocyanide**. This compound is characterized by its foul smell and is used in the carbanamine test for primary amines.

### Conclusion
The final product of the reaction of chloroform with alcoholic KOH and p-toluidine is **p-Tolyl isocyanide**.

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