JEE Advanced 2025 Paper 1 · Q16 · Amines & Diazonium

Question

Match the compounds in List-I (above) with the appropriate observations in List-II and choose the correct option.

List-I (compounds): (P) Tyrosine–Gly–OMe dipeptide (free α-NH$_2$, phenolic–OH); (Q) N-acetyl phenylglycyl-Val-OMe dipeptide (no free α-NH$_2$, no phenolic–OH); (R) Aniline hydrochloride (PhNH$_3^+$Cl$^-$); (S) 2,4,6-trimethylphenylhydrazine (mesityl-NHNH$_2$).

List-II (observations):
(1) Reaction with phenyl diazonium salt gives yellow dye.
(2) Reaction with ninhydrin gives purple color and it also reacts with FeCl$_3$ to give violet color.
(3) Reaction with glucose will give corresponding hydrazone.
(4) Lassaigne extract of the compound treated with dilute HCl followed by addition of aqueous FeCl$_3$ gives blood red color.
(5) After complete hydrolysis, it will give ninhydrin test and it 	extbf{DOES NOT} give positive phthalein dye test.

SolveKar AI · Accepted Answer

(P) The Tyr–Gly–OMe dipeptide has a free α-NH$_2$ (gives Ruhemann's purple with ninhydrin) AND a phenolic –OH on the tyrosyl ring (gives violet colour with FeCl$_3$). → (2).

(Q) N-acetyl Phg–Val–OMe has both NH groups acylated/peptide-bound; α-NH$_2$ is liberated only after complete hydrolysis. The hydrolysate (Phg + Val + AcOH) gives a positive ninhydrin test, and since neither residue has a phenolic–OH there is no phthalein dye — matching observation (5).

(R) Aniline hydrochloride contains an –NH$_3^+$Cl$^-$ on an aromatic ring. On freebasing (or directly via the aryl-amine after deprotonation), it couples with phenyldiazonium chloride to give a yellow azo dye (p-aminoazobenzene). → (1).

(S) Mesitylhydrazine (Ar–NHNH$_2$) condenses with the carbonyl of glucose to give the corresponding phenylhydrazone (osazone-precursor). → (3).

Match: P→2, Q→5, R→1, S→3 → option (B).

JEE Advanced 2025 Paper 1 · Q16 · Amines & Diazonium

Solution