n-Hexane on heating to 773 K at 10-20 atmospheric pressure in the presence of oxides of vanadium supported over alumina, yields

Options

A. 1-hexene
B. 2-hexene
C. benzene
D. 2-methylpentane

Explanation

To solve the question regarding the reaction of n-hexane when heated in the presence of vanadium oxides supported over alumina, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactant:**
- The reactant is n-hexane, which has the molecular formula C6H14.

2. **Conditions of the Reaction:**
- The reaction occurs at a temperature of 773 K and under a pressure of 10 to 20 atmospheres.
- The catalyst used is oxides of vanadium (like V2O5) supported on alumina (Al2O3).

3. **Understanding the Reaction Type:**
- Under these conditions, n-hexane undergoes a process called aromatization. Aromatization is the transformation of aliphatic hydrocarbons into aromatic hydrocarbons.

4. **Products of the Reaction:**
- The primary product of the aromatization of n-hexane is benzene (C6H6).
- Additionally, during this reaction, hydrogen gas (H2) is produced. Specifically, for every molecule of n-hexane that is converted, 4 molecules of H2 are released.

5. **Writing the Balanced Reaction:**
- The balanced chemical equation for the reaction can be represented as follows:
\[ C6H14 \rightarrow C6H6 + 4H2 \]
- This indicates that one molecule of n-hexane yields one molecule of benzene and four molecules of hydrogen gas.

6. **Final Product Identification:**
- The final product of the reaction is benzene (C6H6).

### Conclusion:
The reaction of n-hexane when heated to 773 K at 10-20 atmospheric pressure in the presence of vanadium oxides supported over alumina yields benzene and hydrogen gas.