Question

Find the product (C) in the reaction given below:
${{C}_{6}}{{H}_{5}}N{{H}_{2}}\text{ }\xrightarrow[NaN{{O}_{2}}\text{ + HCl}]{0\text{ }-\text{ }{{5}^{o}}C}\text{ (A) }\xrightarrow[C{{u}_{2}}{{(CN)}_{2}}]{KCN}\text{ (B) }\xrightarrow[{}]{{{H}^{+}}/\text{ }{{\text{H}}_{2}}O}\text{ (C)}$
A.${{C}_{6}}{{H}_{5}}C{{H}_{2}}N{{H}_{2}}$
B.${{C}_{6}}{{H}_{5}}COOH$
C.${{C}_{6}}{{H}_{5}}COC{{H}_{3}}$
D.All of the above

Answer 1

Hint: Aniline is an aromatic compound with an amino group ($N{{H}_{2}}$) attached to it. The first reaction of aniline forms an intermediate compound (A) which is a diazonium salt. Compound (A) further reacts to form a cyano compound (B). Compound B dissolves in water to form an acid (C).

Complete answer:
The initial reactant of the equation is Aniline (${{C}_{6}}{{H}_{5}}N{{H}_{2}}$). It reacts with Sodium nitrite ($NaN{{O}_{2}}$) and Hydrochloric acid ($HCl$) in cold conditions (0-5o C) to form Benzene diazonium chloride (${{C}_{6}}{{H}_{5}}N_{2}^{+}C{{l}^{-}}$). The product formed is a salt of Benzene and behaves as an unstable ion. Their reaction is as follows:
${{C}_{6}}{{H}_{5}}N{{H}_{2}}\text{ }\xrightarrow[NaN{{O}_{2}}\text{ + HCl}]{0\text{ }-\text{ }{{5}^{o}}C}\text{ }{{C}_{6}}{{H}_{5}}N_{2}^{+}C{{l}^{-}}$
Product (A) is ${{C}_{6}}{{H}_{5}}N_{2}^{+}C{{l}^{-}}$
Now, Benzene diazonium chloride reacts with Potassium cyanide and Copper cyanide to form Cyano Benzene (${{C}_{6}}{{H}_{5}}CN$)with the liberation of Nitrogen gas. This reaction facilitates the removal of the Amino group from the Benzene ring and introduces the cyanide group. The reaction is a substitution reaction. The reaction is as follows:
${{C}_{6}}{{H}_{5}}{{N}^{+}}C{{l}^{-}}\text{ }\xrightarrow[Cu{{(CN)}_{2}}]{KCN}\text{ }{{C}_{6}}{{H}_{5}}CN\text{ + }{{\text{N}}_{2}}$
Product (B) is ${{C}_{6}}{{H}_{5}}CN$
Finally, cyano benzene hydrolyses in water to form, Benzoic acid (${{C}_{6}}{{H}_{5}}COOH$). The cyanide group dissolves in water to form a carboxylic acid. The reaction is as follows:
${{C}_{6}}{{H}_{5}}CN\text{ }\xrightarrow{{{H}_{2}}O\text{ / }{{\text{H}}^{+}}}\text{ }{{C}_{6}}{{H}_{5}}COOH$
Product (C) is ${{C}_{6}}{{H}_{5}}COOH$
The correct answer is Option B: ${{C}_{6}}{{H}_{5}}COOH$.

Note:
The above reaction is a mix of three different types of reactions. In the first reaction remember that it is the chemical reaction of Aniline to form Diazonium salt. The second reaction with KCN is Sandmeyer’s reaction. In the third reaction, the cyanide group dissolves in water to form an acid with the functional group COOH.