See image — Amines Chemistry Question

💡 Solution & Explanation

Concept: Diazotization reaction requires a primary aromatic amine (-NH2 directly attached to the benzene ring). The reaction with NaNO2/HCl at 0-5°C converts ArNH2 into ArN2+Cl- (diazonium salt). Secondary aromatic amines (where the nitrogen bears one alkyl or aryl substituent and one H) do not form diazonium salts; instead they form N-nitroso compounds. Step 1 - Analyze each option: (a) m-Ethylaniline: The amino group (-NH2) is a primary amine directly on the benzene ring. The ethyl group is on the ring, not on nitrogen. This is a primary arylamine and will undergo diazotization. → Forms diazonium salt. (b) p-Aminoacetophenone: The amino group (-NH2) is a primary amine on the ring, para to the acetophenone group. This is a primary arylamine and will undergo diazotization. → Forms diazonium salt. (c) 4-Chloro-2-nitroaniline: The amino group (-NH2) is a primary amine on the ring, with chloro and nitro substituents on the ring. This is a primary arylamine and will undergo diazotization (though it may be sluggish due to electron-withdrawing groups, it still reacts). → Forms diazonium salt. (d) N-Ethyl-2-methylaniline: The nitrogen bears one ethyl group and one hydrogen, making it a secondary arylamine (ArNH-CH2CH3). Secondary arylamines do not form diazonium salts upon treatment with NaNO2/HCl; instead they form N-nitrosoamines (ArN(NO)-R). → Does NOT form a diazonium salt. Step 2 - Conclusion: Only option (d) N-Ethyl-2-methylaniline is a secondary arylamine and therefore cannot form a diazonium salt under these conditions. Therefore, the correct answer is D.