Question

Match List-I with List-II and select the correct answer from the given codes. \[ \begin{array}{|c|c|c|} \hline List-I (Reaction) & & List-II (Reagent/Catalyst) \hline A.\ \text{Cannizzaro reaction} & & 1.\ SnCl_2/HCl B.\ \text{Stephen's reaction} & & 2.\ NaOH C.\ \text{Clemmensen reduction} & & 3.\ Zn/Hg - conc.\ HCl D.\ \text{Rosenmund's method} & & 4.\ Pd/BaSO_4,\ \text{boiling xylene} \hline \end{array} \] Codes: {c c c c c} & A & B & C & D

• A. & 1 & 2 & 3 & 4
• B. & 2 & 1 & 3 & 4
• C. & 4 & 3 & 2 & 1
• D. & 1 & 4 & 2 & 3

Hint:

Common name reactions to remember:
Cannizzaro → NaOH
Stephen's → SnCl\(_2\)/HCl
Clemmensen → Zn(Hg)/HCl
Rosenmund → Pd/BaSO\(_4\)

Answer 1

The Correct Option is B

Concept: Each named reaction in organic chemistry is associated with specific reagents/catalysts.

Step 1: Match Cannizzaro reaction (A). Cannizzaro reaction involves the disproportionation of aldehydes without \(\alpha\)-hydrogen atoms in the presence of a strong base. \[ \text{Reagent: NaOH (or conc. alkali)} \quad \Rightarrow \quad (2) \]

Step 2: Match Stephen's reaction (B). Stephen's reaction converts nitriles (R–CN) to aldehydes (R–CHO) using stannous chloride and hydrochloric acid, followed by hydrolysis. \[ \text{Reagent: SnCl}_2/HCl \quad \Rightarrow \quad (1) \]

Step 3: Match Clemmensen reduction (C). Clemmensen reduction reduces carbonyl groups (C=O) of aldehydes or ketones to methylene (–CH\(_2\)–) groups using zinc amalgam and concentrated hydrochloric acid. \[ \text{Reagent: Zn/Hg - conc. HCl} \quad \Rightarrow \quad (3) \]

Step 4: Match Rosenmund's method (D). Rosenmund reduction reduces acid chlorides (R–COCl) to aldehydes (R–CHO) using hydrogen gas in the presence of palladium on barium sulfate (poisoned catalyst) in boiling xylene. \[ \text{Reagent: Pd/BaSO}_4,\ \text{boiling xylene} \quad \Rightarrow \quad (4) \]

Step 5: Final matching. \[ A \to 2,\quad B \to 1,\quad C \to 3,\quad D \to 4 \] This corresponds to option (B).