The given reaction involves the transformation of C6H5CN (benzene nitrile) through several steps. Let's break down each step to understand the formation of X, Y, and Z:
This is a typical Stephen reaction, where benzene nitrile (C6H5CN) is reduced by tin chloride (SnCl2) in the presence of hydrochloric acid (HCl). This reaction reduces the nitrile group to an imide group, forming benzylamine (C6H5CH2NH2) as X.
This step involves the treatment of the imide with concentrated KOH and H3O+. The Cannizzaro reaction is a disproportionation reaction of aldehydes (or compounds with an aldehyde group), where one molecule is reduced to alcohol and the other oxidized to an acid. Here, benzylamine undergoes this reaction, leading to the formation of Y (benzaldehyde) and Z (formic acid).
The reactions correspond to the following:
Correct Answer: Option B: Stephen reaction, Cannizzaro reaction
Reaction:
C\(_6\)H\(_5\)CN \(\xrightarrow{\text{SnCl}_2 + \text{HCl}}\) X \(\xrightarrow{\text{con. KOH}}\) Y + Z \(\xrightarrow{\text{H}_2\text{O}}\)
The formation of X, Y, and Z suggests the following reactions:
Step 1: The reaction of C\(_6\)H\(_5\)CN (Benzonitrile) with SnCl\(_2\) in the presence of HCl leads to the reduction of the nitrile group to an aldehyde group, forming X. This is known as the Stephen reaction.
Step 2: When X is treated with conc. KOH, it undergoes an aldol condensation, forming Y and Z. Y is likely to be a product of condensation, and Z could be the byproduct of the reaction. This is characteristic of the Cannizzaro reaction, which involves the disproportionation of aldehydes without an alpha-hydrogen.
Hence, the reactions correspond to the Stephen reaction followed by the Cannizzaro reaction.
Correct Answer: (D) Stephen reaction, Cannizzaro reaction.
Identify the products R and S in the reaction sequence given.