Question

Observe the reaction sequence

Compound A forms sodium derivative with \(NaNH_2\). Find B and D.

• A. \(H_2/Ni\); \(CH_3CH_2CHO+HCHO\)
• B. \(H_2/Ni\); \(CH_3CHO+CH_3CHO\)
• C. \(H_2/Pd-C,quinoline\); \(CH_3CH_2CHO+HCHO\)
• D. \(H_2/Pd-C,quinoline\); \(CH_3CHO+CH_3CHO\)

Hint:

Terminal alkynes react with \(NaNH_2\) because acidic hydrogen is present. Partial hydrogenation uses poisoned palladium catalyst.

Answer 1

The Correct Option is C

Concept: Compound A forms sodium derivative with sodium amide. Only terminal alkynes show acidic hydrogen and react with \[ NaNH_2 \] Therefore A must be terminal alkyne. Partial hydrogenation of alkyne requires poisoned catalyst.

Step 1: Identify compound A. Given formula \[ C_4H_6 \] Possible terminal alkyne: \[ CH_3CH_2C\equiv CH \] This reacts with sodium amide.

Step 2: Choose correct hydrogenation catalyst. Partial hydrogenation requires Lindlar catalyst equivalent. Given option: \[ Pd-C,quinoline \] This converts alkyne into alkene. Thus B is \[ H_2/Pd-C,quinoline \]

Step 3: Perform ozonolysis. Alkene formed gives aldehydes after ozonolysis. Products: \[ CH_3CH_2CHO \] and \[ HCHO \] Thus final answer: \[ B=H_2/Pd-C,quinoline \] \[ D=CH_3CH_2CHO+HCHO \] Hence correct option \[ \boxed{C} \]