n-Heptane when heated to a temperature of about `800K` under
high pressure in the presence of `Cr_(2)O_(3)//Al_(2)O_(3)` catalyst gives-

Options

A. `1-`heptene
B. `2-` Methylhexane
C. Toluene
D. Xylene.

Explanation

To solve the question regarding the reaction of n-heptane when heated to about 800 K under high pressure in the presence of a Cr₂O₃/Al₂O₃ catalyst, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Conditions**:
- The reactant is n-heptane (C₇H₁₆).
- The reaction conditions include heating to approximately 800 K (around 527 °C) and applying high pressure in the presence of a catalyst (Cr₂O₃/Al₂O₃).

2. **Understand the Type of Reaction**:
- The reaction is an aromatization process. Aromatization involves converting aliphatic hydrocarbons (like n-heptane) into aromatic hydrocarbons (like toluene) through a series of steps.

3. **Outline the Steps of Aromatization**:
- **Cyclization**: The first step involves the cyclization of n-heptane, where it forms a cyclic compound. In this case, n-heptane can form methylcyclohexane (C₇H₁₂) by removing H₂.
- **Dehydrogenation**: The next step is the dehydrogenation of the cyclic compound. Methylcyclohexane loses hydrogen atoms to form toluene (C₇H₈).

4. **Final Product**:
- After the cyclization and dehydrogenation steps, the final product obtained from n-heptane under the given conditions is **toluene (C₇H₈)**.

### Summary of the Reaction:
- n-Heptane (C₇H₁₆) → Methylcyclohexane (C₇H₁₂) → Toluene (C₇H₈)

### Conclusion:
Thus, when n-heptane is heated to about 800 K under high pressure in the presence of a Cr₂O₃/Al₂O₃ catalyst, it yields **toluene**.

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