What is the major product [P3] formed when n-Hexane undergoes the given series of reactions?
n-Hexane $\xrightarrow[\text{20 atm, 773 K}]{\text{V}_2\text{O}_5}$ [P] $\xrightarrow[\Delta]{\text{C}_2\text{H}_5\text{Cl, Anhyd. AlCl}_3}$ [P2] $\xrightarrow[\text{333 K}]{\text{Conc. HNO}_3 + \text{H}_2\text{SO}_4}$ [P3]}
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Alkyl groups on a benzene ring are always ortho/para-directing. Because an ethyl group is bulky, the incoming nitro group prefers the less hindered para position, making the 1,4-substituted product (1-ethyl-4-nitrobenzene) the major product.
Understanding the Concept:
This sequence combines several key aromatic reactions:
Aromatization (reforming) of open-chain alkanes using transition metal oxide catalysts.
Friedel-Crafts Alkylation using an alkyl halide and a Lewis acid catalyst ($\text{Anhyd. AlCl}_3$).
Electrophilic aromatic substitution (nitration) governed by the directing effects of existing substituents.
Step 1: Determine the aromatization intermediate product [P].
When open-chain n-hexane is heated to high temperatures under pressure over vanadium pentoxide ($\text{V}_2\text{O}_5$), it undergoes dehydrogenation and cyclization to yield benzene [P]:
\[
\text{n-Hexane} \xrightarrow{\text{V}_2\text{O}_5} \text{Benzene [P]}
\]
Step 2: Track the Friedel-Crafts alkylation step to find [P2].
Reacting benzene [P] with ethyl chloride ($\text{C}_2\text{H}_5\text{Cl}$) in the presence of anhydrous $\text{AlCl}_3$ introduces an ethyl group onto the ring, producing ethylbenzene [P2]:
\[
\text{Benzene} + \text{C}_2\text{H}_5\text{Cl} \xrightarrow{\text{AlCl}_3} \text{Ethylbenzene [P2]}
\]
Step 3: Analyze the nitration directing effects to find final product [P3].
The ethyl group ($-\text{C}_2\text{H}_5$) is an electron-donating group due to inductive effects and hyperconjugation. This activates the ring and directs incoming electrophiles to the ortho and para positions. Because of steric hindrance at the ortho position, the para-substituted derivative forms as the major product. Nitration with a $\text{HNO}_3/\text{H}_2\text{SO}_4$ mixture yields 1-ethyl-4-nitrobenzene [P3].
