JEE Advanced 2023 Paper 2 · Q16 · Amines & Diazonium
PARAGRAPH II — A trinitro compound, 1,3,5-tris-(4-nitrophenyl)benzene, on complete reaction with an excess of Sn/HCl gives a major product, which on treatment with an excess of $NaNO_2/HCl$ at $0\,°C$ provides $P$ as the product. $P$, upon treatment with excess of $H_2O$ 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 $Q$ and $R$ is $474$ g·mol$^{-1}$ and between $P$ and $S$ is $172.5$ g·mol$^{-1}$. The number of heteroatoms present in one molecule of $R$ is ___. [Atoms other than C and H are heteroatoms]
Reveal answer + step-by-step solution
Correct answer:9
Solution
The trinitro compound → triamino → tris-diazonium $P$ (three $-N_2^+Cl^-$ groups). Hydrolysis: $P + 3H_2O \to Q$ = 1,3,5-tris-(4-hydroxyphenyl)benzene (three $-OH$ groups, $3$ heteroatoms = O). Bromination of $Q$: each phenol is highly activated; molar mass increase is $474$. Each Br adds $80 - 1 = 79$, so $474/79 = 6$ bromines added. Six bromine atoms attach (two per phenol ring at ortho positions). Total heteroatoms in $R$ = $3$ (O) + $6$ (Br) = $9$.
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