Calculate Mass of Final Product in Multi Step Acetophenone Reaction

To find the total mass of the final product $\mathbf{D}$ produced, we need to trace the step-by-step chemical transformations from acetophenone, keeping track of the yield at each step and the molar mass of the resulting compounds.

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Step 1: Conversion of Acetophenone to Compound A

Acetophenone ($\text{C}_6\text{H}_5\text{COCH}_3$) undergoes a haloform reaction with sodium hypobromite ($\text{NaOBr}$), followed by acidic hydrolysis ($\text{H}_3\text{O}^+$), to yield benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$).

• Compound A: Benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$)
• Initial moles of acetophenone: $10\text{ moles}$
• Yield: $60\%$

$\text{Moles of } \mathbf{A} = 10\text{ moles} \times \frac{60}{100} = 6\text{ moles}$

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Step 2: Conversion of Compound A to Compound B

Benzoic acid ($\mathbf{A}$) reacts with ammonia ($\text{NH}_3$) upon heating ($\Delta$) to undergo amidation, forming benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$).

• Compound B: Benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$)
• Yield: $50\%$

$\text{Moles of } \mathbf{B} = 6\text{ moles} \times \frac{50}{100} = 3\text{ moles}$

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Step 3: Conversion of Compound B to Compound C

Benzamide ($\mathbf{B}$) undergoes Hofmann bromamide degradation upon treatment with $\text{Br}_2 / \text{KOH}$ to give aniline ($\text{C}_6\text{H}_5\text{NH}_2$).

• Compound C: Aniline ($\text{C}_6\text{H}_5\text{NH}_2$)
• Yield: $50\%$

$\text{Moles of } \mathbf{C} = 3\text{ moles} \times \frac{50}{100} = 1.5\text{ moles}$

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Step 4: Conversion of Compound C to Compound D

Aniline ($\mathbf{C}$) undergoes electrophilic aromatic substitution with $3\text{ equivalents}$ of bromine ($\text{Br}_2$) in acetic acid ($\text{AcOH}$) to form 2,4,6-tribromoaniline ($\text{C}_6\text{H}_4\text{Br}_3\text{N}$).

• Compound D: 2,4,6-tribromoaniline ($\text{C}_6\text{H}_4\text{Br}_3\text{N}$)
• Yield: $100\%$

$\text{Moles of } \mathbf{D} = 1.5\text{ moles} \times \frac{100}{100} = 1.5\text{ moles}$

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Calculation of Molar Mass of Compound D

The molecular formula of 2,4,6-tribromoaniline is $\text{C}_6\text{H}_4\text{Br}_3\text{N}$.

Using the given atomic weights ($\text{H} = 1$, $\text{C} = 12$, $\text{N} = 14$, $\text{Br} = 80$):

$\text{Molar Mass of } \mathbf{D} = (6 \times 12) + (4 \times 1) + (3 \times 80) + (1 \times 14)$ $\text{Molar Mass of } \mathbf{D} = 72 + 4 + 240 + 14 = 330\text{ g mol}^{-1}$

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Calculation of Mass of Compound D

$\text{Mass of } \mathbf{D} = \text{Moles of } \mathbf{D} \times \text{Molar Mass of } \mathbf{D}$ $\text{Mass of } \mathbf{D} = 1.5\text{ moles} \times 330\text{ g mol}^{-1} = 495\text{ g}$

Final Answer: The total mass of product $\mathbf{D}$ formed is $495$ (in grams).