Q.Explain the intermediate compound formation theory of catalysis with an example.

Step 1. The intermediate compound formation theory explains how a catalyst works, mainly in homogeneous catalysis, by proposing that the catalyst does not simply sit by and speed up the direct reaction — instead, it actively combines with one of the reactants to form a genuine, short-lived intermediate compound.

Step 2. For the overall reaction $A + B \rightarrow AB$, the catalysed pathway instead proceeds in two steps: first $A + C \rightarrow AC$ (C is the catalyst, AC the intermediate), then $AC + B \rightarrow AB + C$, regenerating the catalyst C at the end.

Step 3. Crucially, the activation energies of these two catalysed steps are each LOWER than the activation energy of the original, single uncatalysed step $A + B \rightarrow AB$ — it is exactly this lower-activation-energy pathway, opened up by the intermediate's formation and subsequent decomposition, that accelerates the overall reaction.

Step 4. A concrete worked example: the mechanism of the Friedel–Crafts alkylation of benzene with methyl chloride, catalysed by anhydrous $\text{AlCl}_3$. First, $\text{AlCl}_3$ reacts with $\text{CH}_3\text{Cl}$ to form the intermediate ion pair $[\text{CH}_3^+][\text{AlCl}_4^-]$: $\text{CH}_3\text{Cl} + \text{AlCl}_3 \rightarrow [\text{CH}_3^+][\text{AlCl}_4^-]$. This intermediate then reacts with benzene: $\text{C}_6\text{H}_6 + [\text{CH}_3^+][\text{AlCl}_4^-] \rightarrow \text{C}_6\text{H}_5\text{CH}_3 + \text{AlCl}_3 + \text{HCl}$, giving the final product toluene, releasing HCl, and regenerating the $\text{AlCl}_3$ catalyst unchanged for the next cycle. …

State the general two-step intermediate-formation mechanism (A+C→AC, AC+B→AB+C) and work through the Friedel–Crafts ex …

Omitting the crucial point that BOTH catalysed steps have lower activation energy than the single uncatalysed step — without this, the theor …