Question

If the reaction sequence given below is carried out with 15 moles of acetylene, the amount of the product D formed (in g) is _____.

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

Answer 1

Correct Answer: 136

Step 1: Identify the products and calculate molar masses.

• Acetylene ($\text{C}_2\text{H}_2$): $2 \times 12 + 2 \times 1 = 24 + 2 = 26 \text{ g/mol}$
• Reaction 1: $3 \text{ HC} \equiv \text{CH} \rightarrow \text{C}_6\text{H}_6$ (Benzene, A) 
  Molar mass of A (Benzene, $\text{C}_6\text{H}_6$): $6 \times 12 + 6 \times 1 = 72 + 6 = 78 \text{ g/mol}$
• Reaction 2: Benzene (A) reacts with 1-chloropropane. This is a Friedel-Crafts alkylation. Due to carbocation rearrangement (primary to secondary), isopropylbenzene (Cumene) is formed. 
  A ($\text{C}_6\text{H}_6$) + $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{CH(CH}_3)_2$ (Cumene, B) 
  Molar mass of B (Cumene, $\text{C}_9\text{H}_{12}$): $9 \times 12 + 12 \times 1 = 108 + 12 = 120 \text{ g/mol}$
• Reaction 3: Cumene process. Cumene is oxidized to cumene hydroperoxide, which then rearranges to form Phenol and Acetone. 
  B ($\text{C}_6\text{H}_5\text{CH(CH}_3)_2$) $\rightarrow \text{C}_6\text{H}_5\text{OH}$ (Phenol, C) + $\text{CH}_3\text{COCH}_3$ 
  Molar mass of C (Phenol, $\text{C}_6\text{H}_5\text{OH}$): $6 \times 12 + 6 \times 1 + 1 \times 16 = 72 + 6 + 16 = 94 \text{ g/mol}$
• Reaction 4: Phenol reacts with acetyl chloride in the presence of pyridine (acylation). 
  C ($\text{C}_6\text{H}_5\text{OH}$) + $\text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{OCOCH}_3$ (Phenyl acetate, D) 
  Molar mass of D (Phenyl acetate, $\text{C}_8\text{H}_8\text{O}_2$): $8 \times 12 + 8 \times 1 + 2 \times 16 = 96 + 8 + 32 = 136 \text{ g/mol}$

Step 2: Calculate moles of each product formed considering the yield.

Initial moles of acetylene = 15 moles.

• Formation of A (Benzene): 
  Reaction: $3 \text{ HC} \equiv \text{CH} \rightarrow \text{A (C}_6\text{H}_6)$ 
  From stoichiometry, 3 moles of acetylene produce 1 mole of A. 
  Theoretical moles of A from 15 moles of acetylene = $15 \text{ moles acetylene} \times \frac{1 \text{ mole A}}{3 \text{ moles acetylene}} = 5 \text{ moles A}$. 
  Yield of A = 80% 
  Actual moles of A formed = $5 \text{ moles} \times 0.80 = 4 \text{ moles A}$.
• Formation of B (Cumene): 
  Reaction: $\text{A (C}_6\text{H}_6) + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \rightarrow \text{B (C}_6\text{H}_5\text{CH(CH}_3)_2)$ 
  From stoichiometry, 1 mole of A produces 1 mole of B. 
  Starting moles of A = 4 moles. 
  Theoretical moles of B from 4 moles of A = $4 \text{ moles A} \times \frac{1 \text{ mole B}}{1 \text{ mole A}} = 4 \text{ moles B}$. 
  Yield of B = 50% 
  Actual moles of B formed = $4 \text{ moles} \times 0.50 = 2 \text{ moles B}$.
• Formation of C (Phenol): 
  Reaction: $\text{B (C}_6\text{H}_5\text{CH(CH}_3)_2) \rightarrow \text{C (C}_6\text{H}_5\text{OH}) + \text{CH}_3\text{COCH}_3$ 
  From stoichiometry, 1 mole of B produces 1 mole of C. 
  Starting moles of B = 2 moles. 
  Theoretical moles of C from 2 moles of B = $2 \text{ moles B} \times \frac{1 \text{ mole C}}{1 \text{ mole B}} = 2 \text{ moles C}$. 
  Yield of C = 50% 
  Actual moles of C formed = $2 \text{ moles} \times 0.50 = 1 \text{ mole C}$.
• Formation of D (Phenyl acetate): 
  Reaction: $\text{C (C}_6\text{H}_5\text{OH}) + \text{CH}_3\text{COCl} \rightarrow \text{D (C}_6\text{H}_5\text{OCOCH}_3)$ 
  From stoichiometry, 1 mole of C produces 1 mole of D. 
  Starting moles of C = 1 mole. 
  Theoretical moles of D from 1 mole of C = $1 \text{ mole C} \times \frac{1 \text{ mole D}}{1 \text{ mole C}} = 1 \text{ mole D}$. 
  Yield of D = 100% 
  Actual moles of D formed = $1 \text{ mole} \times 1.00 = 1 \text{ mole D}$.

Step 3: Calculate the mass of product D formed.

Moles of D formed = 1 mole.

Molar mass of D (Phenyl acetate, $\text{C}_8\text{H}_8\text{O}_2$) = 136 g/mol.

Mass of D formed = Moles of D $\times$ Molar mass of D

Mass of D = $1 \text{ mole} \times 136 \text{ g/mol} = 136 \text{ g}$.

Therefore, the amount of product D formed is 136 g.