Isomeric amines with molecular formula C₈H₁₁N give the following tests
Isomer (P) ⇒ Can be prepared by Gabriel phthalimide synthesis
Isomer (Q) ⇒ Reacts with Hinsberg's reagent to give solid insoluble in NaOH
Isomer (R) ⇒ Reacts with HONO followed by β-naphthol in NaOH to give red dye.
Isomers (P), (Q) and (R) respectively are

Solution

Chemistry ma’am · 4 steps

Turn each test into an amine class: Gabriel is 1° aliphatic, Hinsberg solid is 2°, red dye is 1° aryl.

1. Gabriel synthesis. The phthalimide anion attacks an alkyl halide by S_N2, so this method gives only primary aliphatic amines. Aryl halides do not undergo this substitution, so anilines cannot be made this way.
2. Hinsberg test. A primary amine gives a sulfonamide that still has an acidic N–H, so it dissolves in NaOH. A secondary amine gives a sulfonamide with no N–H left, so it stays as a solid that does not dissolve in NaOH. A tertiary amine does not react.
3. Azo dye test. Only a primary aromatic amine forms a diazonium salt that is stable at 0–5 °C. That salt couples with alkaline β-naphthol to give the red-orange azo dye.

   Option	P: 1° aliphatic?	Q: 2° amine?	R: 1° aromatic?
   (a)	No, 2-ethylaniline is aromatic	No, 2-phenylethylamine is 1°	No, N-ethylaniline is 2°
   (b)	No, N-methylbenzylamine is 2°	No, 2-phenylethylamine is 1°	No, the NH₂ is on CH₂, not on the ring
   (c)	Yes, 2-phenylethylamine	Yes, an N-methylbenzylamine	Yes, 4-ethylaniline
   (d)	Yes, 2-phenylethylamine	No, N,N-dimethylbenzylamine is 3°	Yes, 2-ethylaniline

4. So. Only (c) passes all three tests. Its Q carries an extra ring methyl, so as drawn it is C₉H₁₃N and not C₈H₁₁N. That does not change the result, because the Hinsberg test depends only on Q having one N–H. The answer is (c).

The idea to keep. Decode each test before you look at structures. Gabriel needs S_N2 on an alkyl halide, Hinsberg insolubility needs no N–H left on the sulfonamide, and an azo dye needs the NH₂ right on the ring.

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