Question

The reaction involves two steps:

1. Diazotization of aniline: Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) reacts with sodium nitrite ($\text{NaNO}_2$) and hydrochloric acid ($\text{HCl}$) at $0-5^\circ\text{C}$ to form benzenediazonium chloride ($[\text{C}_6\text{H}_5\text{N}_2]^+ \text{Cl}^-$). $\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} [\text{C}_6\text{H}_5\text{N}_2]^+ \text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$
2. Azo coupling: The benzenediazonium chloride acts as an electrophile and undergoes an electrophilic aromatic substitution reaction (azo coupling) with N,N-dimethylaniline ($\text{C}_6\text{H}_5\text{NMe}_2$). N,N-dimethylaniline is a strongly activated aromatic ring due to the electron-donating $-\text{NMe}_2$ group, which directs the coupling to the para position. $[\text{C}_6\text{H}_5\text{N}_2]^+ \text{Cl}^- + \text{C}_6\text{H}_5\text{NMe}_2 \rightarrow \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4\text{NMe}_2 (\text{para}) + \text{HCl}$

The product P is N,N-dimethyl-4-(phenyldiazenyl)aniline, also known as methyl yellow or dimethyl yellow.

The product is an azo compound formed by the coupling of a phenyl group and a para-substituted N,N-dimethylaniline group via an azo linkage ($-\text{N}=\text{N}-$).

• A. N,N-dimethyl-4-(phenyldiazenyl)aniline
• B. Aniline
• C. Benzenediazonium chloride
• D. N,N-dimethylaniline

Answer 1

Answer: (A)

N,N-dimethyl-4-(phenyldiazenyl)aniline

Answer 2

The reaction describes the synthesis of an azo compound through diazotization of aniline followed by azo coupling with N,N-dimethylaniline. The product of this reaction is N,N-dimethyl-4-(phenyldiazenyl)aniline, commonly known as methyl yellow or dimethyl yellow.