The given sequence involves the following steps:
Step 1: Friedel-Crafts Acylation
Benzene (\(\text{C}_6\text{H}_6\)) reacts with acetyl chloride (\(\text{CH}_3\text{COCl}\)) in the presence of \(\text{AlCl}_3\), an electrophilic catalyst. This forms acetophenone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) as compound \(A\):
\[\text{C}_6\text{H}_6 + \text{CH}_3\text{COCl} \xrightarrow[\text{AlCl}_3]{} \text{C}_6\text{H}_5\text{COCH}_3 \, (A).\]
Step 2: Clemmensen Reduction
Acetophenone (\(\text{C}_6\text{H}_5\text{COCH}_3\)) undergoes Clemmensen reduction using zinc amalgam (\(\text{Zn–Hg}\)) and hydrochloric acid (\(\text{HCl}\)). The carbonyl group (\(-\text{CO}\)) is reduced to a methylene group (\(-\text{CH}_2-\)), resulting in ethylbenzene (\(\text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\)) as compound \(B\):
\[\text{C}_6\text{H}_5\text{COCH}_3 \xrightarrow[\text{Zn–Hg, HCl}]{} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3 \, (B).\]
Step 3: Acidic Oxidation
Ethylbenzene reacts with \(\text{H}^+\) under oxidation conditions to form acetophenone again. This completes the reaction cycle.
Final Products:
\(A = \text{C}_6\text{H}_5\text{COCH}_3\) (Acetophenone).
\(B = \text{C}_6\text{H}_5\text{CH}_2\text{CH}_3\) (Ethylbenzene).
Conclusion: The correct identification of \(A\) and \(B\) is given in option \((3)\).
Final Answer: (3).
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to: