When chloroform is treated with amine and KOH , we get:

Options

A. Rose odour smell
B. Sour almond like smell
C. Offensive odour
D. Sour oil of winter green like smell

Answer

To determine the product formed when chloroform (CHCl3) is treated with an amine (RNH2) and potassium hydroxide (KOH), we can follow these steps:

### Step 1: Identify the Reactants
We have chloroform (CHCl3) and a primary amine (RNH2) in the presence of a strong base, KOH.

### Step 2: Understand the Reaction Mechanism
In this reaction, the primary amine will react with chloroform in the presence of KOH. The base (KOH) will deprotonate the amine, making it a stronger nucleophile.

### Step 3: Formation of Isocyanide
1. The deprotonated amine (RNH) will attack one of the chlorine atoms in chloroform (CHCl3).
2. This forms a carbonium ion intermediate, which will lead to the elimination of HCl and the formation of an isocyanide (R-N≡C).

### Step 4: Final Products
The overall reaction can be summarized as:
- 3 moles of KOH react with 1 mole of chloroform and 1 mole of primary amine to produce:
- Isocyanide (R-N≡C)
- Water (H2O)
- Potassium chloride (KCl)

### Step 5: Conclusion
The final product of the reaction between chloroform, a primary amine, and KOH is isocyanide (R-N≡C), along with water and potassium chloride.

### Summary of the Reaction:
\[ \text{RNH}_2 + \text{CHCl}_3 + 3 \text{KOH} \rightarrow \text{R-N}\equiv\text{C} + 3 \text{H}_2\text{O} + 3 \text{KCl} \]

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