Consider the above reaction, compound B is :

Solution

Chemistry ma’am · 4 steps

Sulphanilic acid gives a diazonium ion that couples para to -N(CH₃)₂: that is methyl orange.

1. Make the diazonium salt. NaNO₂ and HCl at 273 to 278 K turn the -NH₂ group of sulphanilic acid into -N₂⁺ Cl⁻. The -SO₃H group is not touched, so A is the 4-sulphobenzenediazonium salt.
2. Coupling keeps nitrogen. The diazonium ion is a weak electrophile. It attacks the very electron-rich ring of N,N-dimethylaniline. Both nitrogens stay in the product as an azo link, -N=N⁻.
3. Para position. -N(CH₃)₂ is a strong ortho, para director. The bulky ArN₂⁺ ion avoids the crowded ortho positions, so it attacks para to -N(CH₃)₂.

   HO₃S-C₆H₄-N₂⁺ + C₆H₅N(CH₃)₂ → HO₃S-C₆H₄-N=N-C₆H₄-N(CH₃)₂ + H⁺

4. So. B is p-(4-sulphophenylazo)-N,N-dimethylaniline, the acid form of the indicator methyl orange. This matches (c): -SO₃H and -N(CH₃)₂ both sit para to the azo link.

The idea to keep. Diazonium coupling keeps the -N=N⁻ bridge and puts it para to the activating -OH or -NR₂ group. Groups like -SO₃H elsewhere just come along unchanged.

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