Question

Complete the equation :

\( \mathrm{C_6H_5{-}N_2^{+}Cl^{-}} + \mathrm{H{-}C_6H_4{-}NH_2} \rightarrow \)

Hint:

Benzenediazonium chloride couples at the para position of aniline to give the yellow dye p-aminoazobenzene and HCl.

Answer 1

Step 1: Understanding the Concept:
Benzenediazonium chloride is a weak electrophile. It attacks the para position of an activated ring such as aniline. This is called a coupling reaction.

Step 2: Identify the reactants:
The diazonium salt is \( \mathrm{C_6H_5N_2^+Cl^-} \).
The second reactant is aniline, \( \mathrm{C_6H_5NH_2} \). The \( \mathrm{-NH_2} \) group makes the ring rich in electrons.

Step 3: Write the product:
The diazonium ion joins the para carbon of aniline and loses \( \mathrm{HCl} \). The product has the \( \mathrm{-N=N-} \) azo link between two benzene rings.
\[ \mathrm{C_6H_5N_2^+Cl^- + C_6H_5NH_2 \xrightarrow{pH\ 4-5} C_6H_5{-}N{=}N{-}C_6H_4{-}NH_2\ (p) + HCl} \]
The product is p-aminoazobenzene (4-aminoazobenzene), a yellow azo dye.
The reaction runs in a weakly acidic medium, at pH 4 to 5.

Final Answer:
The coupling gives p-aminoazobenzene along with HCl. \[ \boxed{\mathrm{C_6H_5{-}N{=}N{-}C_6H_4{-}NH_2\ (p)\ +\ HCl}} \]