`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 problem step by step, we will analyze the reactions mentioned in the question.

### Step 1: Reaction of CHCl₃ with MeCOMe (acetone)
- **Reactants**: CHCl₃ (chloroform) and MeCOMe (acetone).
- **Reaction**: When chloroform reacts with acetone in the presence of a base (OH⁻), the nucleophilic attack occurs. The base deprotonates the acetone, allowing the negatively charged oxygen to attack the carbon of chloroform, leading to the formation of a compound A.
- **Product A**: The product formed is a compound where the chloroform has been converted into a carbonyl compound with a trichloroethyl group. This can be represented as CH₃C(Cl)₂CCl₃.

### Step 2: Reaction of A with excess KOH
- **Reactant**: Compound A from Step 1.
- **Reaction**: In the presence of excess KOH, the chlorine atoms in compound A will be substituted by hydroxyl (OH) groups. This is a nucleophilic substitution reaction where the base replaces the chlorine atoms.
- **Product B**: The final product after this reaction will be a triol (with three hydroxyl groups) on the carbon skeleton. However, this triol is unstable and will dehydrate to form a carboxylic acid.
- **Final Product B**: CH₃C(OH)₂COOH (which can further dehydrate to form CH₃COOH).

### Step 3: Reaction of B with NaOH and CaO (Soda Lime)
- **Reactant**: Compound B from Step 2.
- **Reaction**: The reaction with NaOH and CaO is a decarboxylation reaction. Soda lime (NaOH + CaO) removes CO₂ from the carboxylic acid group.
- **Product C**: The product will be an alkane (specifically, an ethane derivative) after the removal of CO₂. This can be represented as CH₃CH₃ (ethane).

### Step 4: Reaction of C with KOBr
- **Reactant**: Compound C from Step 3.
- **Reaction**: The reaction with KOBr (potassium hypobromite) leads to a haloform reaction. In this reaction, a methyl ketone (if present) will yield bromoform (CHBr₃) and acetic acid (CH₃COOH).
- **Product D**: The product formed will be bromoform (CHBr₃) and acetic acid (CH₃COOH).

### Final Result
- **Product D**: The final product D is bromoform (CHBr₃) along with acetic acid (CH₃COOH).

### Summary of Products
- A: CH₃C(Cl)₂CCl₃
- B: CH₃C(OH)₂COOH
- C: CH₃CH₃
- D: CHBr₃ and CH₃COOH (acetic acid)