JEE Advanced 2023 — Paper 2 — CHEM-IV solution

JEE Advanced2023Paper 2CHEM-IV

PARAGRAPH II

A trinitro compound, $1, 3, 5$ -tris-(4-nitrophenyl) benzene, on complete reaction with an excess of Sn/HCl gives major product, which on treatment with an excess of NaNO $_{2}$ /HCl at $0^{\circ}$ C provides P as the product. P, upon treatment with excess of H $_{2}$ O at room temperature, gives the product Q. Bromination of Q in aqueous medium furnishes the product R. The compound P upon treatment with an excess of phenol under basic conditions gives the product S.

The molar mass difference between compounds Q and R is $474$ g mol $^{- 1}$ and between compounds P and S is $172.5$ g mol $^{- 1}$ .

Q2. The total number of carbon atoms and heteroatoms present in one molecule of S is _____.

[Use: Molar mass in g mol $^{- 1}$ : H $= 1$ , C $= 12$ , N $= 14$ , O $= 16$ , Br $= 80$ , Cl $= 35.5$ . Atoms other than C and H are considered as heteroatoms.]

Your answer

Solution

Under basic conditions, each diazonium group of P couples with phenol at the para position of phenol (azo coupling). All three diazonium groups react, giving an aromatic azo dye S:

$1, 3, 5$ -tris $(4 - ((4 -hydroxyphenyl) azo) phenyl)$ benzene.

Structurally, S has a central benzene ring (6 C) and three identical arms of the form $- C_{6} H_{4} - N = N - C_{6} H_{4} - O H$ .

Carbon count. Central ring contributes $6$ . Each arm has $2 \times 6 = 12$ carbons; three arms add $36$ . Total $C = 6 + 36 = 42$ .

Heteroatom count. Each arm has $2$ nitrogens (azo $- N = N -$ ) and $1$ oxygen (phenolic $- O H$ ), giving $3$ heteroatoms per arm. Three arms contribute $9$ heteroatoms.

Total of $C$ and heteroatoms $= 42 + 9 = 51$ .

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