Identification of Alkane Compound X From Reactions

To determine the identity of compound $(\text{X})$, let us analyze each of the given chemical reactions:

1. Isomerization: When straight-chain alkanes ($n$-alkanes) are heated in the presence of anhydrous $\text{AlCl}_3$ and $\text{HCl}$ gas, they undergo isomerization to form branched-chain alkanes. Heptane ($\text{C}_7\text{H}_{16}$) isomerizes to yield branched isomers, such as $2,4$-dimethylpentane (which also has 7 carbon atoms): $\text{CH}_3-(\text{CH}_2)_5-\text{CH}_3 \xrightarrow[\Delta]{\text{anh. AlCl}_3 + \text{HCl}} \text{CH}_3-\underset{\underset{\text{CH}_3}{\mid}}{\text{CH}}-\text{CH}_2-\underset{\underset{\text{CH}_3}{\mid}}{\text{CH}}-\text{CH}_3 \quad (\text{2,4-dimethylpentane})$

2. Aromatization: $n$-Alkanes containing 6 to 8 carbon atoms undergo aromatization (dehydrocyclization) when heated at high temperatures ($773\text{ K}$) and pressures ($10\text{--}20\text{ atm}$) in the presence of catalyst oxides such as $\text{Cr}_2\text{O}_3$, $\text{V}_2\text{O}_5$, or $\text{Mo}_2\text{O}_3$ supported on alumina. $n$-Heptane yields toluene: $\text{CH}_3-(\text{CH}_2)_5-\text{CH}_3 \xrightarrow{\text{Cr}_2\text{O}_3 / \text{Al}_2\text{O}_3, \Delta} \text{C}_6\text{H}_5\text{CH}_3 + 4\text{H}_2$

3. Cyclisation: During reforming, $n$-heptane can undergo catalytic cyclisation to form a six-membered ring derivative, methylcyclohexane: $\text{CH}_3-(\text{CH}_2)_5-\text{CH}_3 \xrightarrow{\text{Catalyst}, \Delta} \text{Methylcyclohexane} + \text{H}_2$

Therefore, the compound $(\text{X})$ is Heptane.

Correct Option: C (Heptane)