Consider the following sequence of reactions.
Total number of sp³ hybridised carbon atoms in the major product C formed is _________.

Solution

Chemistry ma’am · 4 steps

Diazotise, couple with phenol, then ethylate the OH: only the two ethyl groups give sp3 carbons.

1. Diazotisation. NaNO₂ and HCl at 0 to 5 °C turn the -NH₂ group of 4-ethoxyaniline into a diazonium group. So A is C₂H₅O-C₆H₄-N₂⁺Cl⁻.
2. Coupling with phenol. In NaOH, phenol becomes phenoxide, and the diazonium ion attacks its para position. Dilute HCl then turns -O⁻ back into -OH. So B is 4-ethoxy-4'-hydroxyazobenzene, C₂H₅O-C₆H₄-N=N-C₆H₄-OH. Count its atoms: C₁₄H₁₄N₂O₂, which matches the given formula.
3. Williamson ether synthesis. NaOH removes the phenolic H to give a phenoxide, which attacks ethyl bromide. The -OH becomes -OC₂H₅, adding C₂H₄ to B. This gives C₁₆H₁₈N₂O₂, as stated for C.

   C: C₂H₅O-C₆H₄-N=N-C₆H₄-OC₂H₅

4. So. All 12 ring carbons are sp². Each ethyl group has one CH₂ and one CH₃, both sp³. Two ethyl groups give 4 sp³ carbons.

The common slip: 2. Counting only the original ethoxy group gives 2. The ethyl bromide step adds a second -OC₂H₅ group, with two more sp3 carbons.

The idea to keep. Benzene ring carbons are always sp2, and the azo group adds no carbon at all. In azo dye questions, the sp3 carbons come only from the alkyl groups, so just count those.

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