Question

Identify products \(P_1\) and \(P_2\) formed when Benzonitrile undergoes the following reactions:

(I) Benzonitrile \(\xrightarrow[(ii)\ H_3O^+]{(i)\ SnCl_2/HCl/ether} [X]\) (II) \([X] \xrightarrow[(ii)\ H_3O^+]{(i)\ HCN} [P_1]\) (III) \(2\) moles of \(X + 50% NaOH \rightarrow [P_2] +\) Salt

• A. \(P_1\) Benzoyl chloride and \(P_2\) Acetophenone
• B. \(P_1\) Acetophenone and \(P_2\) Benzoic acid
• C. \(P_1\) Phenol and \(P_2\) Sodium benzoate
• D. \(\alpha\)-Hydroxy phenylacetic acid and \(P_2\) Benzyl alcohol

Hint:

Aldehydes without \(\alpha\)-hydrogen undergo Cannizzaro reaction in concentrated alkali.

Answer 1

The Correct Option is D

Understanding the Concept: This question involves Stephen reduction, cyanohydrin formation and Cannizzaro reaction.
Step 1: Formation of compound \(X\).
Benzonitrile undergoes Stephen reduction: \[ C_6H_5CN \xrightarrow{SnCl_2/HCl} \text{imine salt} \] On hydrolysis: \[ \rightarrow C_6H_5CHO \] Thus: \[ X = \text{Benzaldehyde} \]
Step 2: Formation of \(P_1\).
Benzaldehyde reacts with HCN: \[ C_6H_5CHO + HCN \rightarrow C_6H_5CH(OH)CN \] Hydrolysis converts nitrile into carboxylic acid: \[ C_6H_5CH(OH)CN \xrightarrow{H_3O^+} C_6H_5CH(OH)COOH \] This compound is: \[ \alpha\text{-Hydroxy phenylacetic acid} \] also called Mandelic acid.
Step 3: Formation of \(P_2\).
Benzaldehyde lacks \(\alpha\)-hydrogen atoms. Therefore, with concentrated NaOH it undergoes Cannizzaro reaction: \[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \] Products formed:
Benzyl alcohol
Sodium benzoate
Hence: \[ P_2 = \text{Benzyl alcohol} \]