Consider the following sequence of reactions:
The product ' B ' is

Solution

Chemistry ma’am · 4 steps

Aniline turns into a diazonium salt, which then couples at the para carbon of dimethylaniline.

1. Make A. Aniline with NaNO₂ and acid at 0 to 5 °C undergoes diazotisation. A is benzenediazonium chloride, C₆H₅N₂⁺Cl⁻. The cold keeps this salt from decomposing.
2. Find the electrophile. The terminal nitrogen of -N⁺≡N is a weak electrophile. It reacts only with very electron-rich rings, such as phenols and aromatic amines.
3. Pick the site. In N,N-dimethylaniline, the -N(CH₃)₂ group pushes electron density onto the ring at the ortho and para positions. The bulky diazonium ion attacks the less crowded para carbon, and one H leaves as H⁺.

   C₆H₅N₂⁺Cl⁻ + C₆H₅N(CH₃)₂ → C₆H₅N=NC₆H₄N(CH₃)₂ + HCl

4. So. B is 4-(N,N-dimethylamino)azobenzene, the yellow dye called butter yellow. Both nitrogens stay as -N=N⁻ joining the two rings, para to -N(CH₃)₂, which is option (a).

The idea to keep. In coupling, the diazonium ion keeps both nitrogens and adds as -N=N⁻ to the para carbon of an activated ring. Nothing is reduced and no N₂ is lost.

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