Acetone and chloroform reacts to produce

Options

A. `CH_(3)COOC_(2)H_(5)`
B. `CH_(3)CHO`
C. `(CH_(3))_(2)C(OH)C CI_(3)`
D. `HCHO`

Explanation

To determine the product formed from the reaction between acetone and chloroform, we can follow these steps:

### Step 1: Identify the Reactants
- Acetone (CH₃COCH₃) is a ketone with the structure:
\[ \text{CH}_3 - \text{C}(\text{O}) - \text{CH}_3 \]
- Chloroform (CHCl₃) is a haloalkane with the structure:
\[ \text{CHCl}_3 \]

### Step 2: Understand the Reaction Mechanism
- Chloroform contains a hydrogen atom that is acidic due to the presence of the electronegative chlorine atoms.
- In the presence of a base, the acidic hydrogen (H) from chloroform can be removed, resulting in the formation of a chloroform anion (CCl₃⁻).

### Step 3: Formation of the Enolate Ion
- The acetone can undergo deprotonation at one of its methyl groups in the presence of a strong base, forming an enolate ion:
\[ \text{CH}_3 - \text{C}(\text{O}) - \text{CH}_3 \rightarrow \text{CH}_2^{-} - \text{C}(\text{O}) - \text{CH}_3 \]

### Step 4: Nucleophilic Attack
- The enolate ion formed from acetone can act as a nucleophile and attack the carbon of the chloroform, resulting in the formation of a new carbon-carbon bond:
\[ \text{CH}_2^{-} - \text{C}(\text{O}) - \text{CH}_3 + \text{CCl}_3 \rightarrow \text{C}(\text{CCl}_3)(\text{OH}) - \text{C}(\text{O}) - \text{CH}_3 \]

### Step 5: Final Product Formation
- The final product will have a carbon atom bonded to a hydroxyl group (OH) and a trichloromethyl group (CCl₃) along with the remaining acetone structure:
\[ \text{C}(\text{CCl}_3)(\text{OH}) - \text{C}(\text{O}) - \text{CH}_3 \]

### Conclusion
The product formed from the reaction of acetone and chloroform is:
\[ \text{C}(\text{CCl}_3)(\text{OH}) - \text{C}(\text{O}) - \text{CH}_3 \]