JEE Advanced 2023 · Paper 2 · Chemistry Q16

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ChemistryNumerical+3Question 50 of 51

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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 $\ce{NaNO2}$ /HCl at 0 °C provides P as the product. P, upon treatment with excess of $\ce{H2O}$ 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 $\mathrm{g\,mol^{-1}}$ and between compounds P and S is 172.5 $\mathrm{g\,mol^{-1}}$ .

The number of heteroatoms present in one molecule of R is _____.

[Use: Molar mass (in $\mathrm{g\,mol^{-1}}$ ): 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

Answer and worked solution

Official answer: 9

The sequence.

Sn/HCl reduces the three $\ce{NO2}$ groups to $\ce{NH2}$ .
Excess $\ce{NaNO2}$ /HCl at 0 °C gives the tris-diazonium chloride P.
Warm water replaces each $\ce{N2+}$ by OH. Q is 1,3,5-tris(4-hydroxyphenyl)benzene.

Bromination. In water, each phenol ring is brominated at both positions ortho to OH; the para position is taken. That is 6 Br in all.

Check. Swapping 6 H for 6 Br adds $6\times79=474$ g/mol, which matches the given difference.

Heteroatoms in R. 3 O + 6 Br $=9$ .

Topics:Diazonium salts Phenols Nitro compounds

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