Azo Dye Formation with Beta Naphthol

To determine which of the given compounds forms a brilliant colored azo dye, we analyze the reaction conditions and the chemical nature of each compound:

1. Azo Dye Test Reaction Principle:
   The formation of an azo dye with $\beta$-naphthol (2-naphthol) involves two steps:

   • Diazotization: A primary aromatic amine ($\text{Ar-NH}_2$) reacts with nitrous acid ($\text{NaNO}_2 + \text{dil. HCl}$) at low temperatures ($0-5^\circ\text{C}$) to form a stable arenediazonium salt ($\text{Ar-N}_2^+\text{Cl}^-$):
     $\text{Ar-NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$
   • Azo Coupling: The arenediazonium ion acts as a weak electrophile and undergoes an electrophilic aromatic substitution reaction with an alkaline solution of $\beta$-naphthol to produce a brightly colored, conjugated azo compound (azo dye):
     $\text{Ar-N}_2^+\text{Cl}^- + \beta\text{-naphthol} \xrightarrow{\text{OH}^-} \text{Ar-N=N-(}\beta\text{-naphthol)} + \text{Cl}^- + \text{H}_2\text{O}$

2. Analysis of the Options:

   • Option (A) [$\text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2$]: This is a $3^\circ$ (tertiary) aromatic amine. It does not form a diazonium salt upon treatment with $\text{NaNO}_2/\text{HCl}$; instead, it undergoes electrophilic nitrosation at the para-position.
   • Option (B) [$\text{C}_6\text{H}_5\text{NHCH}_3$]: This is a $2^\circ$ (secondary) aromatic amine. It reacts with nitrous acid to give an oily $N$-nitrosamine ($\text{C}_6\text{H}_5\text{N(NO)CH}_3$) and does not form a diazonium salt.
   • Option (C) [$p\text{-CH}_3\text{-C}_6\text{H}_4\text{-NH}_2$]: This is a $1^\circ$ (primary) aromatic amine ($p$-toluidine). It smoothly undergoes diazotization to form a stable diazonium cation ($p\text{-CH}_3\text{-C}_6\text{H}_4\text{-N}_2^+$), which subsequently couples with $\beta$-naphthol in an alkaline medium to form a brilliant orange-red azo dye.
   • Option (D) [$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$]: This is a $1^\circ$ aliphatic/benzylic amine. Its diazonium salt is highly unstable and rapidly decomposes with the loss of $\text{N}_2$ gas to yield benzyl alcohol and other carbocation-derived products, thus failing to couple with $\beta$-naphthol.

Therefore, only compound (C) gives a positive azo dye test.

Correct Answer: C