JEE Challenger
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ChemHardAdv 2023 • Paper 2

Total Carbon and Heteroatoms in Compound S Synthesis

Comprehension Passage

A trinitro compound, 1,3,5-tris-(4-nitrophenyl)benzene, on complete reaction with an excess of Sn/HCl\text{Sn/HCl} gives a major product, which on treatment with an excess of NaNO2/HCl\text{NaNO}_2/\text{HCl} at 0 ∘C0\,^\circ\text{C} provides P\textbf{P} as the product. P\textbf{P}, upon treatment with excess of H2O\text{H}_2\text{O} at room temperature, gives the product Q\textbf{Q}. Bromination of Q\textbf{Q} in aqueous medium furnishes the product R\textbf{R}. The compound P\textbf{P} upon treatment with an excess of phenol under basic conditions gives the product S\textbf{S}.

The molar mass difference between compounds Q\textbf{Q} and R\textbf{R} is 474 g mol−1474\text{ g mol}^{-1} and between compounds P\textbf{P} and S\textbf{S} is 172.5 g mol−1172.5\text{ g mol}^{-1}.

The total number of carbon atoms and heteroatoms present in one molecule of S\textbf{S} is ______ .

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

Official Numerical Answer51

Topics & Concepts

AminesAminesElectrophilic Aromatic SubstitutionOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the total number of carbon atoms and heteroatoms present in one molecule of compound S, we analyze the reaction sequence step-by-step:

1. Structure and Reaction Sequence

  • Starting Material: 1,3,5-tris-(4-nitrophenyl)benzene

    • Formula: C6H3(C6H4NO2)3=C24H15N3O6\text{C}_6\text{H}_3(\text{C}_6\text{H}_4\text{NO}_2)_3 = \text{C}_{24}\text{H}_{15}\text{N}_3\text{O}_6

  • Reduction and Diazotization (P\mathbf{P}):
    Complete reduction of the nitro groups with excess Sn/HCl\text{Sn/HCl} forms the corresponding triamine, which upon treatment with excess NaNO2/HCl\text{NaNO}_2/\text{HCl} at 0 ∘C0\,^\circ\text{C} gives the tris-diazonium chloride salt P.

    • Formula of P: C24H15N6Cl3\text{C}_{24}\text{H}_{15}\text{N}_6\text{Cl}_3
    • Molar Mass of P: Molar Mass(P)=24(12)+15(1)+6(14)+3(35.5)=493.5 g mol−1\text{Molar Mass}(\mathbf{P}) = 24(12) + 15(1) + 6(14) + 3(35.5) = 493.5 \text{ g mol}^{-1}

  • Diazo Coupling to form S\mathbf{S}:
    Treatment of P with excess phenol (C6H5OH\text{C}_6\text{H}_5\text{OH}) under basic conditions results in diazo coupling at the para-position of the phenol rings: C24H15N6Cl3+3 C6H5OH→OH−S+3 HCl\text{C}_{24}\text{H}_{15}\text{N}_6\text{Cl}_3 + 3\,\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \mathbf{S} + 3\,\text{HCl}

    • Structure of S: 1,3,5-tris(4−(4-hydroxyphenylazo)phenyl)\left(4-(4\text{-hydroxyphenylazo})phenyl\right)benzene
    • Formula of S: C6H3(C6H4−N=N−C6H4−OH)3=C42H30N6O3\text{C}_6\text{H}_3\left(\text{C}_6\text{H}_4-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}\right)_3 = \text{C}_{42}\text{H}_{30}\text{N}_6\text{O}_3
    • Molar Mass of S: Molar Mass(S)=42(12)+30(1)+6(14)+3(16)=666 g mol−1\text{Molar Mass}(\mathbf{S}) = 42(12) + 30(1) + 6(14) + 3(16) = 666 \text{ g mol}^{-1}

  • Verification of Molar Mass Difference: Molar Mass(S)−Molar Mass(P)=666−493.5=172.5 g mol−1\text{Molar Mass}(\mathbf{S}) - \text{Molar Mass}(\mathbf{P}) = 666 - 493.5 = 172.5 \text{ g mol}^{-1} This matches the given value of 172.5 g mol−1172.5 \text{ g mol}^{-1}.

2. Counting Carbon Atoms and Heteroatoms in S

  • Number of Carbon (C) atoms:
    Central benzene ring (6)+3×two benzene rings (3×12)=6+36=42 carbon atoms\text{Central benzene ring } (6) + 3 \times \text{two benzene rings } (3 \times 12) = 6 + 36 = 42 \text{ carbon atoms}

  • Number of Heteroatoms (N and O):

    • Nitrogen atoms (N\text{N}): 3×2=63 \times 2 = 6
    • Oxygen atoms (O\text{O}): 3×1=33 \times 1 = 3
      Total Heteroatoms=6+3=9\text{Total Heteroatoms} = 6 + 3 = 9

3. Total Count

Total number of carbon atoms and heteroatoms=42+9=51\text{Total number of carbon atoms and heteroatoms} = 42 + 9 = 51

Final Answer: 51

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