Considering the above reaction, X and Y respectively are :

Solution

Chemistry ma’am · 4 steps

Cold diazotisation gives benzenediazonium chloride, and it couples para to the NMe₂ group.

1. Find X. Aniline with NaNO₂ and HCl at 273 to 278 K is diazotisation. The -NH₂ group becomes -N₂⁺, so X is benzenediazonium chloride, C₆H₅N₂⁺Cl⁻.

   C₆H₅NH₂ + NaNO₂ + 2HCl →(273-278 K) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

2. Why it stays. The low temperature is the whole point. Below about 278 K the diazonium salt is stable enough to use; it does not lose N₂ and give chlorobenzene or phenol.
3. Find Y. The diazonium ion is a weak electrophile, so it attacks only a strongly activated ring. The -N(CH₃)₂ group is ortho, para directing, and the two methyl groups crowd the ortho positions, so attack goes para.
4. So. X is C₆H₅N₂⁺Cl⁻ and Y is p-(dimethylamino)azobenzene, C₆H₅-N=N-C₆H₄-N(CH₃)₂ with -N=N⁻ para to NMe₂. That is option (b).

The idea to keep. Diazotisation needs 273 to 278 K to keep ArN₂⁺ alive. When it couples with phenols or anilines, the azo group goes para to the activating group unless that position is blocked.

Answer from the NTA final answer key. Solution by PrepWiser.