Match the compounds in List-I with the appropriate observations in List-II and choose the correct option.
- P → 1; Q → 5; R → 4; S → 2
- P → 2; Q → 5; R → 1; S → 3
- P → 5; Q → 2; R → 1; S → 4
- P → 2; Q → 1; R → 5; S → 3
Answer
(2) P → 2; Q → 5; R → 1; S → 3
Solution
(P) has a free α-NH2 (ninhydrin: purple) and a phenolic OH (neutral FeCl3: violet). So P → 2.
(Q) has no free NH2 (N-acetyl) and no phenol. Complete hydrolysis frees the amino acids, which give the ninhydrin test, but there is no phenol for the phthalein dye test. So Q → 5.
Q H3O+, [figure] + [figure] + CH3COOH + CH3OH
(R) Anilinium chloride gives free aniline, which couples with the phenyl diazonium ion to give p-aminoazobenzene (yellow dye). So R → 1.
[figure] + [figure] coupling [figure]
(S) The arylhydrazine condenses with the C=O of open-chain glucose to give the hydrazone. So S → 3.
[figure]
(4) needs both N and S in the compound (blood-red Fe(SCN)2+); none of the compounds has S.
Answer: (B)