Q3. Add the two equations of Route 1 to show what happens overall.
C6H5N=N–NHC6H5 + HCl → C6H5N2+Cl− + C6H5NH2, then C6H5N2+Cl− + C6H5NH2 → C6H5N=NC6H4NH2 + HCl. The diazonium salt, the aniline and the HCl cancel, leaving diazoaminobenzene → p-aminoazobenzene: HCl is a catalyst, and the two compounds are isomers (reasoned from the Lab Manual’s scheme, page 111).
Q4. Diazoaminobenzene and aniline yellow are isomers. How do their structures differ, and how can a melting point show that Route 1 has worked?
Both are C12H11N3. In diazoaminobenzene, C6H5N=N–NHC6H5, the –N=N– group is bonded to the nitrogen of the second aniline unit; in aniline yellow, C6H5N=NC6H4NH2, it is bonded to the para carbon of the second ring, which keeps a free –NH2 group (reasoned). Beyond the NCERT text, diazoaminobenzene melts at 95–96 °C (Wikipedia, “1,3-Diphenyltriazene”) and aniline yellow at 123–126 °C (Wikipedia, “Aniline Yellow”), so a dry, recrystallised product that melts near 123–126 °C, not near 95 °C, supports formation of the expected product, but melting behaviour alone does not conclusively establish identity or exclude impurities (reasoned).
Q5. How many moles of NaNO2 are in the Lab Manual’s 1.6 g, and what mass of aniline would that diazotise, mole for mole? How does that compare with the 2 mL of aniline used, taking aniline’s density as 1.02 g mL−1? (H = 1, C = 12, N = 14, O = 16, Na = 23.)
M(NaNO2) = 23 + 14 + 32 = 69 g mol−1: 1.6/69 = 0.0232 mol. M(C6H5NH2) = 72 + 7 + 14 = 93 g mol−1: 0.0232 × 93 = 2.16 g of aniline. The 2 mL of aniline is 2 × 1.02 = 2.04 g, or 2.04/93 = 0.0219 mol, so the nitrite is only about 6% more than mole for mole, close to the equation, as the Lab Manual’s note on azo dye syntheses asks (page 111) (reasoned; the density is from PubChem, citing the US National Institute for Occupational Safety and Health).
Q6. In Route 1, how many moles of aniline hydrochloride are used per mole of diazoaminobenzene? (Cl = 35.5.)
Aniline hydrochloride, C6H5NH3+Cl−: M = 72 + 8 + 14 + 35.5 = 129.5 g mol−1, so 1.5/129.5 = 0.0116 mol. Diazoaminobenzene: 3/197 = 0.0152 mol (Example 1). The ratio is 0.0116/0.0152 = 0.76. If each HCl were used up, at most 76% of the diazoaminobenzene could react; because the coupling step gives the HCl back, all of it can, so the aniline hydrochloride acts as a catalyst (reasoned).