Question #44483

In the Friedel-Crafts alkylation of benzene (used as solvent in large excess) with 1-chloropropane and anhydrous \ce{AlCl3} at $300\text{ K}$, the initial Lewis acid-base complex undergoes two competing pseudo-first-order processes:

\textbf{Path A:} Direct electrophilic aromatic substitution to form n-propylbenzene, with an activation energy $E_{a,\text{direct}}$. \textbf{Path B:} A 1,2-hydride shift to form a secondary carbocation complex, which rapidly reacts with benzene to form isopropylbenzene (Product $P$). The activation energy for the hydride shift is $E_{a,\text{shift}}$.

Given $E_{a,\text{direct}} - E_{a,\text{shift}} = 11.5\text{ kJ mol}^{-1}$ and assuming identical pre-exponential factors for both pathways, $P$ is isolated as the major product.

$P$ is oxidized with \ce{O2} at $400\text{ K}$ followed by acid hydrolysis to yield an aromatic compound $Q$ and an aliphatic compound $R$. $Q$ is then treated with \ce{CHCl3} and aqueous \ce{NaOH} at $340\text{ K}$, followed by acidification, to yield $S$ as the major product.

Identify the structure of $S$ and the approximate initial molar ratio of $P$ to n-propylbenzene. (Given: $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$, $\ln(10) = 2.30$)