Question

Benzonitrile on reduction with stannous chloride in presence of hydrochloric acid followed by acid hydrolysis forms,

• A. Benzal chloride
• B. Benzoyl chloride
• C. Benzophenone
• D. Benzaldehyde

Hint:

Stephen Reduction: Nitrile + $\text{SnCl}_2$/$\text{HCl} \rightarrow$ Imine $\xrightarrow{\text{H}_3\text{O}^+}$ Aldehyde.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The reaction described in the question is a specific organic name reaction known as the Stephen aldehyde synthesis (or Stephen reduction).
It is used to convert alkyl or aryl cyanides (nitriles) into the corresponding aldehydes.
Step 2: Key Formula or Approach:
The general reaction sequence is:
1. \( \text{R-C}\equiv\text{N} + \text{SnCl}_2 + \text{HCl} \rightarrow \text{R-CH=NH} \cdot \text{HCl} \) (Imine hydrochloride intermediate)
2. \( \text{R-CH=NH} \cdot \text{HCl} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{R-CHO} + \text{NH}_4\text{Cl} \)
Step 3: Detailed Explanation:
The reactant given is Benzonitrile, which has the formula \( \text{C}_6\text{H}_5\text{CN} \).
Step 1: Reduction using Stannous Chloride and HCl.
Benzonitrile reacts with \( \text{SnCl}_2 \) and \( \text{HCl} \) (which provide nascent hydrogen) to form an aldimine stannichloride, which is effectively an imine hydrochloride intermediate.
\[ \text{C}_6\text{H}_5\text{-C}\equiv\text{N} + 2[\text{H}] + \text{HCl} \xrightarrow{\text{SnCl}_2/\text{HCl}} \text{C}_6\text{H}_5\text{-CH=NH} \cdot \text{HCl} \]
Step 2: Acid Hydrolysis.
The intermediate aldimine complex is then subjected to acidic hydrolysis.
The imine group (\( \text{=NH} \)) is hydrolyzed, replacing the nitrogen atom with an oxygen atom to form a carbonyl group.
\[ \text{C}_6\text{H}_5\text{-CH=NH} \cdot \text{HCl} + \text{H}_2\text{O} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_6\text{H}_5\text{-CHO} + \text{NH}_4\text{Cl} \]
The final product, \( \text{C}_6\text{H}_5\text{CHO} \), is Benzaldehyde.
Step 4: Final Answer:
The product formed is Benzaldehyde.