Question:
Benzonitrile (A) + X \(\rightarrow\) B; A + Y \(\rightarrow\) C. B + C \(\xrightarrow{dil.\ NaOH}\) \(\alpha,\beta\)-unsaturated carbonyl compound. What are X and Y?
Benzonitrile (A) + X \(\rightarrow\) B; A + Y \(\rightarrow\) C. B + C \(\xrightarrow{dil.\ NaOH}\) \(\alpha,\beta\)-unsaturated carbonyl compound. What are X and Y?
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Stephen reduction converts nitriles into aldehydes without affecting the aromatic ring.
Updated On: Jun 8, 2026
- DIBAL-H, \(H_2O\); \((CH_3)_2Cd\)
- \(SnCl_2 + HCl,\ H_3O^+\); \(CH_3MgBr,\ H_2O\)
- DIBAL-H, \(H_2,\ Ni\)
- \(SnCl_2 + HCl,\ H_2O\); \((CH_3)_2Cd\)
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The Correct Option is D
Solution and Explanation
Concept:
Aldol condensation requires an aldehyde and a ketone.
Step 1: Formation of B from benzonitrile.
Stephen reduction converts nitriles into aldehydes: \[ C_6H_5CN \xrightarrow{SnCl_2/HCl} C_6H_5CHO \] Thus: \[ B = \text{benzaldehyde} \]
Step 2: Formation of C from benzonitrile.
Reaction with dimethyl cadmium gives ketone: \[ C_6H_5CN \xrightarrow{(CH_3)_2Cd} C_6H_5COCH_3 \] Thus: \[ C = \text{acetophenone} \]
Step 3: Aldol condensation.
Benzaldehyde and acetophenone undergo Claisen--Schmidt condensation to form an \(\alpha,\beta\)-unsaturated carbonyl compound. Hence: \[ \boxed{(D)} \]
Step 1: Formation of B from benzonitrile.
Stephen reduction converts nitriles into aldehydes: \[ C_6H_5CN \xrightarrow{SnCl_2/HCl} C_6H_5CHO \] Thus: \[ B = \text{benzaldehyde} \]
Step 2: Formation of C from benzonitrile.
Reaction with dimethyl cadmium gives ketone: \[ C_6H_5CN \xrightarrow{(CH_3)_2Cd} C_6H_5COCH_3 \] Thus: \[ C = \text{acetophenone} \]
Step 3: Aldol condensation.
Benzaldehyde and acetophenone undergo Claisen--Schmidt condensation to form an \(\alpha,\beta\)-unsaturated carbonyl compound. Hence: \[ \boxed{(D)} \]
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