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JEE Advanced 2023 · Paper 2, Chemistry question 17

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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 NaNOX2\ce{NaNO2}/HCl at 0 °C provides P as the product. P, upon treatment with excess of HX2O\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 g mol−1\mathrm{g\,mol^{-1}} and between compounds P and S is 172.5 g mol−1\mathrm{g\,mol^{-1}}.

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

[Use: Molar mass (in g mol−1\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]

Answer and worked solution

Official answer: 51

The coupling. P couples with phenol in base at each diazonium group, giving the tris-azo compound S:
Ar−NX2X+ClX−+CX6HX5OH→Ar−N=N−CX6HX4OH.\ce{Ar-N2+Cl- + C6H5OH -> Ar-N=N-C6H4OH}.

Check. Each coupling replaces Cl (35.5) by CX6HX4OH\ce{C6H4OH} (93), adding 57.5. Three couplings add 3×57.5=172.53\times57.5=172.5 g/mol, which matches the given difference.

Atom count in S.

  • Carbons: core CX6HX3(CX6HX4)X3\ce{C6H3(C6H4)3} has 24 C; three CX6HX4OH\ce{C6H4OH} groups add 18. Total 42 C.
  • Heteroatoms: 6 N + 3 O =9=9.

Carbon atoms + heteroatoms =42+9=51=42+9=51.

Topics:Diazonium saltsPhenols
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