N, N-Dimethylaniline is treated with aqueous $N a N O_{2} / H C l$ , the product formed is:

p-(N, N-dimethylamino) benzenediazonium chloride

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p-(N, N-dimethylamino) phenol

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p-nitroso - N, N-dimethylaniline

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p-nitro- N, N-dimethylaniline

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Solution

The correct option is C p-nitroso - N, N-dimethylaniline
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.

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Similar questions

H C l, N a N O_{2}

^{\circ} C

Benzaldehyde condenses with N, N-dimethylaniline in the presence of anhydrous $Z n C l_{2}$ to give

	Column I		Column II
(A)	Aniline	(i)	Can be made by Gabriel phthalimide reaction
(B)	N-Methylaniline	(ii)	Undergoes electrophilic substitution reaction with $H N O_{2}$
(C)	N, N-Dimethylaniline	(iii)	Forms yellow oily product with $H N O_{2}$
(D)	Benzylamine	(iv)	Gives azo dye test

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