What product will form when N , N - dimethyl aniline reacts with NaNO2 and dilute HCl at low temperature ?

Question

What product will form when N , N - dimethyl aniline reacts with NaNO2 and dilute HCl at low temperature ? (A)  p -Nitroso- N , N - dimethylaniline (B) Methyl- n - hexylamine (C)  m - Benzenediazonium chloride (D)  N - Nitroso- N - methylaniline

Tardigrade Admin · Accepted Answer

In tertiary aromatic amines, there is no N−H bond. Therefore, nitrosoation on Nitrogen atom and salt formation is not possible, instead the aromatic ring is nitrosoated at the para position forming a p-nitroso derivative. This derivative on reaction with alkali gives parrot green colouration.

Q. What product will form when $N$ , $N$ - dimethyl aniline reacts with $N a N O_{2}$ and dilute $H Cl$ at low temperature ?

$p$ -Nitroso- $N$ , $N$ - dimethylaniline

Methyl- $n$ - hexylamine

$m$ - Benzenediazonium chloride

$N$ - Nitroso- $N$ - methylaniline

In tertiary aromatic amines, there is no N−H bond. Therefore, nitrosoation on Nitrogen atom and salt formation is not possible, instead the aromatic ring is nitrosoated at the para position forming a p-nitroso derivative. This derivative on reaction with alkali gives parrot green colouration.

Q. What product will form when N , N - dimethyl aniline reacts with NaNO2​ and dilute HCl at low temperature ?

p -Nitroso- N , N - dimethylaniline

Methyl- n - hexylamine

m - Benzenediazonium chloride

N - Nitroso- N - methylaniline

In tertiary aromatic amines, there is no N−H bond. Therefore, nitrosoation on Nitrogen atom and salt formation is not possible, instead the aromatic ring is nitrosoated at the para position forming a p-nitroso derivative. This derivative on reaction with alkali gives parrot green colouration.

Q. What product will form when $N$ , $N$ - dimethyl aniline reacts with $N a N O_{2}$ and dilute $H Cl$ at low temperature ?

$p$ -Nitroso- $N$ , $N$ - dimethylaniline

Methyl- $n$ - hexylamine

$m$ - Benzenediazonium chloride

$N$ - Nitroso- $N$ - methylaniline