An organic compound 'A' contains nitrogen and chlorine. It dissolves readily in water to give a solution that turns litmus red. Titration of compound 'A' with standard base indicates that the molecular weight of 'A' is 131± 2. When a sample of 'A' is treated with aq. NaOH, a liquid separates which contains N but not Cl. Treatment of the obtained liquid with nitrous acid followed by phenol gives orange precipitate. The compound 'A' is :

Solution

Chemistry ma’am · 3 steps

An orange azo dye with phenol needs an aromatic primary amine, so A must be its hydrochloride salt.

1. Read the clues. A dissolves in water and turns litmus red, so it is an acidic salt like RNH₃⁺ Cl⁻. Its chlorine is ionic, because NaOH frees a liquid amine that keeps the N but loses the Cl as NaCl.

   C₆H₅NH₃⁺Cl⁻ + NaOH → C₆H₅NH₂ + NaCl + H₂O

2. Azo dye test. Nitrous acid followed by phenol gives an orange precipitate of p-hydroxyazobenzene. Only an aromatic primary amine, with NH₂ directly on a benzene ring, gives a diazonium salt stable enough to couple. An NH₂ on an sp³ carbon loses N₂ at once and gives no dye.

   Option	Molar mass	Acidic salt?	Amine freed by NaOH	Orange dye?
   (a)	C₆H₁₀NCl, 131.5	yes	NH₂ on sp³ C of a non-aromatic ring	no
   (b)	C₆H₆ClN, 127.5	no, free amine	Cl stays on the ring	not applicable
   (c)	C₆H₁₀NCl, 131.5	yes	NH₂ on a CH₂, aliphatic	no
   (d)	C₆H₈NCl, 129.5	yes	aniline, no Cl	yes, orange

3. So. A is anilinium chloride, option (d). Its molar mass of 129.5 lies within 131 ± 2, and the aniline it releases diazotises and couples with phenol to give the orange azo dye.

The idea to keep. Diazotisation and coupling to give a coloured azo dye is the test for an aromatic primary amine. If the NH₂ sits on an sp³ carbon, the diazonium salt breaks down to N₂ and no dye forms.

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