Question

Match List-I with List-II and select the correct answer from the given codes. 

List-I (Reaction)	List-II (Reagent/Catalyst)
A. Cannizzaro reaction	1. SnCl2/HCl
B. Stephen's reaction	2. NaOH
C. Clemmensen reduction	3. Zn/Hg - conc. HCl
D. Rosenmund's method	4. Pd/BaSO4, boiling xylene

 

Codes:

	A	B	C	D

• A. \( A \to 4,\quad B \to 1,\quad C \to 3,\quad D \to 2 \)
• B. \( A \to 2,\quad B \to 1,\quad C \to 3,\quad D \to 4 \)
• C. \( A \to 1,\quad B \to 2,\quad C \to 3,\quad D \to 4 \)
• D. \( A \to 1,\quad B \to 2,\quad C \to 3,\quad D \to 4 \)

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).