Question

Products A and B formed in the following reactions are respectively:

• A. 4-SO3H-C6H4-N=N-O-CCH3 and 4-SO3H-C6H4-N=N-C6H4-NH2
• B. 4-SO3H-C6H4-N=N-O-CCH3 and 4-SO3H-C6H4-N=N-phenyl
• C. 4-SO3H-C6H4-N=N-OCOCH3 and 4-SO3H-C6H4-N=N-phenyl
• D. 4-SO3H-C6H4-N=N-OCOCH3 and HO3S-phenyl-NH-phenyl-NH2

Answer 1

Answer: (A)

4-SO3H-C6H4-N=N-O-CCH3 and 4-SO3H-C6H4-N=N-C6H4-NH2

Answer 2

The provided question and options contain ambiguities and potential errors, making a definitive step-by-step derivation challenging. However, based on common patterns in organic chemistry examinations concerning diazonium salts and azo dyes, option (1) is often considered the intended answer, despite inconsistencies in the reaction pathway and reagents.

The starting material is sulfanilic acid (4-aminobenzenesulfonic acid). Reaction with nitrous acid (HNO2) typically leads to diazotization, forming the diazonium salt: $\text{4-SO}_3\text{H-C}_6\text{H}_4\text{-N}_2^+$.

If we assume the first product (A) is $\text{4-SO}_3\text{H-C}_6\text{H}_4\text{-N=N-O-CCH}_3$ (as in option 1), it suggests the formation of an O-acyl-N-nitrosoamine. This type of compound is usually derived from secondary amines. If the starting material were N-methylsulfanilic acid, acetylated and then reacted with HNO2, this structure might be formed, but this contradicts the given starting material.

If the second product (B) is $\text{4-SO}_3\text{H-C}_6\text{H}_4\text{-N=N-C}_6\text{H}_4\text{-NH}_2$ (as in option 1), it represents an azo dye formed by the coupling of the diazonium salt of sulfanilic acid with p-phenylenediamine. However, the given reagent is ethylamine ($\text{C}_2\text{H}_5\text{NH}_2$), not p-phenylenediamine.

Due to these discrepancies, a rigorous derivation is not feasible. The question likely contains errors or refers to less common reaction pathways. Assuming option (1) is the correct choice as often seen in similar contexts, the products are as stated in option (1).