Three isomers A, B and C (mol. formula C₈H₁₁N) give the following results : A and C→(Diazotization)P+Q→((i) Hydrolysis)R (product of A); +; S (product of C)
R has lower boiling point than S
B →(C₆H₅SO₂Cl) alkali-insoluble product
A, B and C, respectively are :

Solution

Chemistry ma’am · 4 steps

Turn A and C into hydroxybenzoic acids, then let hydrogen bonding tell you which one is ortho.

1. Diazotize then oxidise. Diazotization turns each -NH₂ on the ring into -N₂⁺. Warm hydrolysis replaces it with -OH. Then KMnO₄/H⁺ oxidises the side-chain ethyl group to -COOH, so A and C each end up as a hydroxybenzoic acid.
2. Use boiling points. In ortho-hydroxybenzoic acid (salicylic acid) the -OH and -COOH sit side by side and form an intramolecular hydrogen bond, so neighbouring molecules attract each other less. The para isomer can only hydrogen bond between molecules, so its boiling point is higher. R is lower boiling, so A must be the ortho ethylaniline and C cannot be ortho.
3. Hinsberg test on B. With C₆H₅SO₂Cl a primary amine gives a sulfonamide that still has an acidic N–H, so it dissolves in alkali. A secondary amine gives a sulfonamide with no N–H, which is alkali-insoluble. So B must be a secondary amine.

   Option	A (gives R)	B	C (gives S)	Fits?
   (a)	2-ethylaniline	3-methylbenzylamine, primary	3-ethylaniline	No: B gives an alkali-soluble product
   (b)	3-ethylaniline	N-methylbenzylamine, secondary	2-ethylaniline	No: S would be the ortho acid, lower boiling
   (c)	4-ethylaniline	N-ethylaniline, secondary	2-ethylaniline	No: R would be para, higher boiling
   (d)	2-ethylaniline	N-methylbenzylamine, secondary	4-ethylaniline	Yes

4. So. A is 2-ethylaniline, which gives salicylic acid (R, lower boiling). B is C₆H₅CH₂NHCH₃, a secondary amine. C is 4-ethylaniline, which gives 4-hydroxybenzoic acid (S, higher boiling). The answer is (d).

The idea to keep. Ortho isomers with -OH next to -COOH or -NO₂ form intramolecular hydrogen bonds and boil lower. In the Hinsberg test, an alkali-insoluble sulfonamide means a secondary amine.

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