Ethylbenzene can be prepared by the Friedel-Crafts alkylation of benzene. While alkyl halides are typically used, alkenes can also serve as the alkylating agent in the presence of a Lewis acid catalyst such as anhydrous $\text{AlCl}_3$ (often with a trace of $\text{HCl}$ to generate the initial electrophile). Thus, the reaction of benzene ($\text{C}_6\text{H}_6$) with ethene ($\text{H}_2\text{C=CH}_2$) produces ethylbenzene.

• Option A: Reaction with propene ($\text{CH}_3\text{-CH=CH}_2$) would yield isopropylbenzene (cumene) .
• Option C: Alkanes like ethane ($\text{H}_3\text{C-CH}_3$) do not undergo this reaction.
• Option D: While alcohols can act as alkylating agents, they typically require a protic acid like $\text{H}_2\text{SO}_4$, as the water byproduct would complex with and deactivate a Lewis acid like $\text{AlCl}_3$.