The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.
P → 5; Q → 4; R → 2; S → 1
(P) Stephen reaction:
Stephen reaction reduces nitriles (RCN) to aldehydes. The precursor is usually benzonitrile, which is derived from benzoic acid.
\[ \Rightarrow P \rightarrow \boxed{2} \quad \text{(Benzoic acid)} \] (Q) Sandmeyer reaction:
Used to substitute an amino group on an aromatic ring (from aniline) via diazotization. Requires a nitro compound as a precursor.
\[ \Rightarrow Q \rightarrow \boxed{3} \quad \text{(Nitrobenzene)} \] (R) Hoffmann bromamide degradation reaction:
Converts amides to amines with one fewer carbon. The amine product is Toluene.
\[ \Rightarrow R \rightarrow \boxed{4} \quad \text{(Toluene)} \] (S) Cannizzaro reaction:
Occurs with aldehydes having no alpha-H (like benzaldehyde), which can be derived from oxidation of toluene.
\[ \Rightarrow S \rightarrow \boxed{1} \quad \text{(Toluene)} \]
Final Answer: \( \boxed{\text{B}} \)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is