JEE Challenger
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ChemHardAdv 2022 • Paper 1

Amount of Final Product Formed from Acetylene Sequence with Yields

If the reaction sequence given below is carried out with 15 moles15\text{ moles} of acetylene, the amount of the product D\mathbf{D} formed (in g\text{g}) is ____________.

The yields of A\mathbf{A}, B\mathbf{B}, C\mathbf{C} and D\mathbf{D} are given in parentheses. [Given: Atomic mass of H=1,C=12,O=16,Cl=35\text{H} = 1, \text{C} = 12, \text{O} = 16, \text{Cl} = 35]

Question Diagram 1
Official Numerical Answer136

Topics & Concepts

HydrocarbonsAlkanes, Alkenes, AlkynesStoichiometry

Step-by-Step Solution

To determine the mass of final product D\mathbf{D} formed from 15 moles15\text{ moles} of acetylene, we trace the mole conversions and percentage yields through each reaction step:

  1. Trimerization of 15 moles15\text{ moles} of acetylene yields benzene (A\mathbf{A}): Moles of A=153×0.80=4 moles\text{Moles of } \mathbf{A} = \frac{15}{3} \times 0.80 = 4\text{ moles}

  2. Friedel-Crafts alkylation of benzene with 1-chloropropane gives cumene (B\mathbf{B}): Moles of B=4×0.50=2 moles\text{Moles of } \mathbf{B} = 4 \times 0.50 = 2\text{ moles}

  3. Cumene process yields phenol (C\mathbf{C}): Moles of C=2×0.50=1 mole\text{Moles of } \mathbf{C} = 2 \times 0.50 = 1\text{ mole}

  4. Acetylation of phenol with acetyl chloride gives phenyl acetate (D\mathbf{D}, molar mass =136 g/mol= 136\text{ g/mol}): Moles of D=1×1.00=1 mole\text{Moles of } \mathbf{D} = 1 \times 1.00 = 1\text{ mole}

Mass of D=1 mole×136 g/mol=136 g\text{Mass of } \mathbf{D} = 1\text{ mole} \times 136\text{ g/mol} = 136\text{ g}

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