`Ph-CH_(3)underset(Delta)overset(KMnO_(4))rarrAunderset(Delta)overset("Soda Lime")rarrB`
Compound `B` is :

Options

A. Toluene
B. Benzene
C. Cresol
D. Benzaldehyde

Answer

To solve the given question, we need to analyze the reactions step by step:

### Step 1: Identify the Starting Compound
The starting compound is toluene, which is represented as Ph-CH₃ (where Ph stands for the phenyl group).

### Step 2: Reaction with KMnO₄
When toluene (Ph-CH₃) is treated with KMnO₄ (potassium permanganate) under heating conditions, a strong oxidizing agent, it oxidizes the side chain (the CH₃ group) to a carboxylic acid.

- **Reaction**:
\[ \text{Ph-CH}_3 \xrightarrow{\Delta, \text{KMnO}_4} \text{Ph-COOH} \]
- **Product A**: The product formed is benzoic acid (Ph-COOH).

### Step 3: Reaction with Soda Lime
Next, the benzoic acid (Product A) is treated with soda lime (a mixture of sodium hydroxide (NaOH) and calcium oxide (CaO)) and heated. This reaction is known as decarboxylation.

- **Decarboxylation Reaction**:
\[ \text{Ph-COOH} \xrightarrow{\text{soda lime}, \Delta} \text{Ph-H} + \text{CO}_2 + \text{H}_2\text{O} \]
- **Product B**: The product formed after decarboxylation is benzene (Ph-H).

### Conclusion
Thus, the final product B after the reactions is benzene.

### Final Answer
Compound B is **benzene**.

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