Question

The product B formed in the following reaction sequence is: \[ {C}_6{H}_5{CN} \xrightarrow{{HCl}} (A) \xrightarrow{{AgCN}} (B) \]

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

For nucleophilic substitution reactions, always identify the intermediate and use the correct reagents for further transformations.

Answer 1

The Correct Option is C

To identify the product B formed in the given reaction sequence, we need to analyze each step:

1. First Step: Hydrolysis of Benzonitrile
   

   The starting compound is benzonitrile, \({C}_6{H}_5{CN}\). When benzonitrile undergoes hydrolysis in the presence of hydrochloric acid (HCl), it converts to the corresponding amide, benzamide (\({C}_6{H}_5{CONH}_2\)). This transformation is depicted in the reaction below:

   \[ {C}_6{H}_5{CN} + 2H_2O + HCl \rightarrow {C}_6{H}_5{CONH}_2 + NH_4Cl \]

2. Second Step: Reaction with Silver Cyanide
   

   The benzamide product (A) is then reacted with silver cyanide (AgCN). In this step, a nucleophilic substitution takes place where the amide group is converted to the isocyanate group, forming phenyl isocyanate (\({C}_6{H}_5{NCO}\)). The reaction can be summarized as follows:

   \[ {C}_6{H}_5{CONH}_2 + AgCN \rightarrow {C}_6{H}_5{NCO} + Ag \]

The final product B is phenyl isocyanate (\({C}_6{H}_5{NCO}\)).