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]

The correct answer is 136.00.

Let us analyze the reaction sequence step-by-step to find the structure of each intermediate and the final product D, as well as calculate the amount of product D formed starting with 15 moles of acetylene (HC≡CH).

Step 1: Conversion of Acetylene to Product A

Acetylene undergoes trimerization when passed through a red-hot iron tube:

$3\text{ HC≡CH}\to {\text{C}}_6{\text{H}}_6\text{ (Benzene, A)}$

From the stoichiometry, 3 moles of acetylene yield 1 mole of benzene.
Initial moles of acetylene = 15 moles.
Theoretical moles of A (Benzene) = $\frac{15}{3}=5\text{ moles}$.
Given percentage yield for step A = 80%.
Actual moles of A formed = $5\times 0.80=4.0\text{ moles}$.

Step 2: Friedel-Crafts Alkylation of A to form B

Benzene (A) reacts with 1-chloropropane (H₃C-CH₂-CH₂-Cl) in the presence of anhydrous AlCl₃. The primary propyl carbocation rearranges via a 1,2-hydride shift to form the more stable secondary isopropyl carbocation, yielding cumene (isopropylbenzene) as product B.

$\text{Benzene (A)}+{\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{Cl}\stackrel{{\text{AlCl}}_3}{\to }\text{Cumene (B)}$

1 mole of benzene forms 1 mole of cumene.
Given percentage yield for step B = 50%.
Actual moles of B (Cumene) formed = $4.0\times 0.50=2.0\text{ moles}$.

Step 3: Cumene Hydroperoxide Rearrangement to form C

Cumene (B) undergoes oxidation with O₂ followed by acid hydrolysis (H₃O⁺) to yield phenol (C) and acetone as a side product:

$\text{Cumene (B)}\underset{2.{\text{ H}}_3{\text{O}}^{+}}{\overset{1.{\text{ O}}_2}{\to }}\text{Phenol (C)}+{\text{CH}}_3{\text{COCH}}_3$

1 mole of cumene gives 1 mole of phenol.
Given percentage yield for step C = 50%.
Actual moles of C (Phenol) formed = $2.0\times 0.50=1.0\text{ mole}$.

Step 4: Acetylation of Phenol to form D

Phenol (C) reacts with acetyl chloride (CH₃COCl) in the presence of pyridine to form phenyl acetate (D):

${\text{C}}_6{\text{H}}_5\text{OH (C)}+{\text{CH}}_3\text{COCl}\stackrel{\text{Pyridine}}{\to }{\text{C}}_6{\text{H}}_5{\text{OCOCH}}_3\text{ (Phenyl Acetate, D)}$

1 mole of phenol gives 1 mole of phenyl acetate.
Given percentage yield for step D = 100%.
Actual moles of D (Phenyl Acetate) formed = $1.0\times 1.00=1.0\text{ mole}$.

Calculation of Molar Mass and Amount of Product D:

Molecular formula of Product D (Phenyl Acetate, C₈H₈O₂):
Molar mass of D = $\left(8\times 12\right)+\left(8\times 1\right)+\left(2\times 16\right)=96+8+32=136\text{ g/mol}$.

Amount of product D formed = $\text{Moles of D}\times \text{Molar mass of D}=1.0\text{ mol}\times 136\text{ g/mol}=136.00\text{ g}$.