Which would be obtained by boiling CHCl3 with dilute KOH?

Options

A. `CH_(3)`COOK
B. HCOOk
C. `K_(2)C_(2)O_(4)`
D. `CH_(3)`OH

Answer

To determine the product obtained by boiling chloroform (CHCl3) with dilute potassium hydroxide (KOH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
We start with chloroform (CHCl3) and dilute KOH. Chloroform contains three chlorine atoms that can act as leaving groups.

2. **Reaction with KOH**:
When chloroform is boiled with dilute KOH, the KOH acts as a base. The hydroxide ions (OH⁻) from KOH can replace the chlorine atoms in chloroform.

3. **Substitution Reaction**:
The reaction proceeds through a nucleophilic substitution mechanism where the three chlorine atoms are replaced by hydroxyl groups (OH). This results in the formation of a compound with three hydroxyl groups attached to the carbon atom.

\[ \text{CHCl}_3 + 3 \text{KOH} \rightarrow \text{CH(OH)}_3 + 3 \text{KCl} \]

4. **Formation of Formic Acid**:
The product formed is trichloroethanol (CH(OH)₃). However, since we are boiling with water present (dilute KOH), dehydration can occur, leading to the formation of formic acid (HCOOH) after the removal of water.

5. **Final Product**:
The final product after the reaction and subsequent dehydration is formic acid (HCOOH). In the presence of KOH, formic acid can further react to form potassium formate (HCOOK).

\[ \text{HCOOH} + \text{KOH} \rightarrow \text{HCOOK} + \text{H}_2\text{O} \]

6. **Conclusion**:
Therefore, the product obtained by boiling CHCl3 with dilute KOH is potassium formate (HCOOK).

### Final Answer:
The product obtained is potassium formate (HCOOK).

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