In the following reaction sequence, P, Q, S and T are the major products. [figure] The correct statement(s) about P, Q, S and T is(are)
- Q on treatment with ethanol generates an aromatic aldehyde.
- S gives positive phthalein dye test.
- P is a dinitro compound.
- T is a coloured compound.
Answer
(2) S gives positive phthalein dye test., (4) T is a coloured compound.
Solution
[figure] (CH3CO)2O, pyridine [figure]
Acetylation protects -NH2 (no oxidation, no anilinium ion, only one nitro group enters).
[figure] [288 K]conc. HNO3, conc. H2SO4 [figure] (P)
-NHCOCH3 is o/p-directing; the bulky group gives mainly the para product. P is a mononitro compound: (C) wrong.
[figure] H3O+ [figure]
Acid hydrolysis removes the acetyl group (4-nitroaniline).
[figure] [ 273-278 K]NaNO2/HCl [figure] (Q)
Diazotisation of the primary aromatic amine gives the diazonium salt.
[figure] CH3CH2OH [figure] + CH3CHO+N2+HCl
Ethanol reduces -N2+ to H (ethanol is oxidised to ethanal); no aromatic aldehyde is formed: (A) wrong.
[figure] O2 [figure] H3O+ [figure] + CH3COCH3
Cumene is oxidised to cumene hydroperoxide, which is cleaved by acid into phenol and acetone.
[figure] NaOH [figure] CO2 [figure] H3O+ [figure] (S)
Kolbe's reaction: phenoxide is carboxylated at the ortho position; acid gives salicylic acid. S has a free phenolic OH, so it gives the phthalein dye test: (B) correct.
[figure] [aq. NaOH]Q [figure] (T)
In alkaline medium the diazonium ion couples para to the phenolic OH; the azo compound (extended conjugation through –N=N–) is a coloured dye: (D) correct.
Answer: (B), (D)