`CHCl_(3)+MeCOMe overset(OH^(-))rarr A overset("excess KOH")rarr B overset(NaOH - CaO)rarr C overset("KOBr")rarr D`

Options

A. `MeCH_(2)OH`
B. `MeCOOH`
C. `Me_(2)CHOH`
D. `CH_(2)Cl_(2)`

Answer

To solve the question step by step, we will analyze the reactions sequentially:

### Step 1: Reaction of CHCl₃ with MeCOMe in the presence of OH⁻
- **Starting Materials**: CHCl₃ (chloroform) and MeCOMe (acetone).
- **Reaction**: When CHCl₃ is treated with a strong base like OH⁻, it undergoes deprotonation and forms a carbanion (CCl₃⁻). This carbanion can then attack the carbonyl carbon of acetone.
- **Product A**: The resulting product A will be a compound where the CCl₃ group is attached to the acetone structure. The product can be represented as:
\[ \text{A} = \text{CH}_3\text{C(OH)(CCl}_3)\text{H} \]

### Step 2: Reaction of A with excess KOH
- **Starting Material**: Product A.
- **Reaction**: Excess KOH will replace the chlorine atoms with hydroxyl groups (OH) due to nucleophilic substitution.
- **Product B**: The product after this reaction will have three hydroxyl groups, which is unstable, leading to the loss of a water molecule. The final product B will be:
\[ \text{B} = \text{CH}_3\text{C(OH)}_3 \]
This can further dehydrate to form a carboxylic acid, specifically acetic acid (CH₃COOH).

### Step 3: Reaction of B with NaOH and CaO (Soda Lime)
- **Starting Material**: Product B (which is a carboxylic acid).
- **Reaction**: The combination of NaOH and CaO (soda lime) leads to decarboxylation, where the carboxyl group (-COOH) is removed as CO₂.
- **Product C**: After decarboxylation, we will obtain:
\[ \text{C} = \text{CH}_3\text{CH}_3 \]
This is propane.

### Step 4: Reaction of C with KOBr
- **Starting Material**: Product C (propane).
- **Reaction**: The presence of KOBr will lead to a haloform reaction. In this reaction, the presence of the methyl group adjacent to a carbonyl will lead to the formation of a haloform (bromoform in this case).
- **Product D**: The final product D will be:
\[ \text{D} = \text{CH}_3\text{CBr}_3 \]
This is bromoform.

### Final Summary
The final product D after all the reactions is bromoform (CHBr₃).

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