JEE Challenger
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ChemHardAdv 2025 • Paper 1

Calculation of Product Mass in Multi Step Organic Synthesis from Bromophenyl Acetal

The reaction sequence given below is carried out with 16 moles16\text{ moles} of X\mathbf{X}. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S\mathbf{S} produced is _______.

Use: Atomic mass (in amu): H=1,C=12,O=16,Br=80\text{Use: Atomic mass (in amu): } \text{H} = 1, \text{C} = 12, \text{O} = 16, \text{Br} = 80

Question Diagram 1
Official Numerical Answer175

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesHaloalkanes and HaloarenesStoichiometry

Step-by-Step Solution

To determine the amount of S\mathbf{S} produced, we analyze the reaction steps sequentially:

Step 1: Formation of P\mathbf{P}

  • Starting material X\mathbf{X}: 2-(2-bromophenyl)-1,3-dioxolane (C9H9BrO2\text{C}_9\text{H}_9\text{BrO}_2).
  • Reagents: (i) Na\text{Na}, dry ether (Wurtz-Fittig coupling) followed by (ii) H3O+\text{H}_3\text{O}^+ (acidic hydrolysis of the acetal protecting groups).
  • Reaction: Two molecules of X\mathbf{X} couple to form a biaryl intermediate, which upon hydrolysis of both acetals yields biphenyl-2,2'-dicarboxaldehyde (P\mathbf{P}). Formula of P:C14H10O2\text{Formula of } \mathbf{P}: \text{C}_{14}\text{H}_{10}\text{O}_2 Molar mass of P=(14×12)+(10×1)+(2×16)=168+10+32=210 g/mol\text{Molar mass of } \mathbf{P} = (14 \times 12) + (10 \times 1) + (2 \times 16) = 168 + 10 + 32 = 210\text{ g/mol}
  • Moles of P\mathbf{P} produced: Moles of P=16 moles of X2×100%=8 moles\text{Moles of } \mathbf{P} = \frac{16\text{ moles of } \mathbf{X}}{2} \times 100\% = 8\text{ moles}

Step 2: Formation of Q\mathbf{Q}

  • Reagents: (i) NaOH,Δ\text{NaOH}, \Delta (Intramolecular Cannizzaro reaction) followed by (ii) H3O+\text{H}_3\text{O}^+.
  • Product Q\mathbf{Q}: 2'-(hydroxymethyl)-[1,1'-biphenyl]-2-carboxylic acid (C14H12O3\text{C}_{14}\text{H}_{12}\text{O}_3).
  • Moles of Q\mathbf{Q} produced: Moles of Q=8 moles×50%=4 moles\text{Moles of } \mathbf{Q} = 8\text{ moles} \times 50\% = 4\text{ moles}

Step 3: Formation of R\mathbf{R}

  • Reagents: (i) NaOH, CaO\text{NaOH, CaO} (ii) Δ\Delta (Decarboxylation with soda lime).
  • Product R\mathbf{R}: [1,1'-biphenyl]-2-ylmethanol (C13H12O\text{C}_{13}\text{H}_{12}\text{O}).
  • Moles of R\mathbf{R} produced: Moles of R=4 moles×50%=2 moles\text{Moles of } \mathbf{R} = 4\text{ moles} \times 50\% = 2\text{ moles}

Step 4: Formation of T\mathbf{T}

  • Reagents: PBr3,(C2H5)2O\text{PBr}_3, (\text{C}_2\text{H}_5)_2\text{O} (Bromination of alcohol).
  • Product T\mathbf{T}: 2-(bromomethyl)-1,1'-biphenyl (C13H11Br\text{C}_{13}\text{H}_{11}\text{Br}).
  • Moles of T\mathbf{T} produced: Moles of T=2 moles×50%=1 mole\text{Moles of } \mathbf{T} = 2\text{ moles} \times 50\% = 1\text{ mole}

Step 5: Formation of S\mathbf{S}

  • Reagents: R+NaH,(C2H5)2O\mathbf{R} + \text{NaH}, (\text{C}_2\text{H}_5)_2\text{O} (Williamson ether synthesis).
  • Reaction: Nucleophilic substitution between the alkoxide of R\mathbf{R} and alkyl bromide T\mathbf{T} gives the ether S\mathbf{S}: S:bis([1,1′-biphenyl]−2-ylmethyl) ether (C26H22O)\mathbf{S}: \text{bis}([1,1'\text{-biphenyl}]-2\text{-ylmethyl})\text{ ether } (\text{C}_{26}\text{H}_{22}\text{O})
  • Moles of S\mathbf{S} produced: Moles of S=1 mole of T×50%=0.5 moles\text{Moles of } \mathbf{S} = 1\text{ mole of } \mathbf{T} \times 50\% = 0.5\text{ moles}

Calculation of Molar Mass and Mass of S\mathbf{S}:

  • Molar Mass of S\mathbf{S} (C26H22O\text{C}_{26}\text{H}_{22}\text{O}): MS=(26×12)+(22×1)+(1×16)=312+22+16=350 g/molM_{\mathbf{S}} = (26 \times 12) + (22 \times 1) + (1 \times 16) = 312 + 22 + 16 = 350\text{ g/mol}

  • Mass of S\mathbf{S}: Mass=Moles×Molar Mass=0.5 moles×350 g/mol=175 g\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.5\text{ moles} \times 350\text{ g/mol} = 175\text{ g}

Answer: 175

Calculation of Product Mass in Multi Step Organic Synthesis from Bromophenyl Acetal | Chemistry PYQ Solution - JEE Challenger