Calculate Mass of Product R in Multi Step Hydrocarbon Reaction Scheme

To determine the mass of product $\mathbf{R}$ ($\mathbf{x}\text{ g}$), we analyze the step-by-step chemical transformations and stoichiometry:

Step 1: Identification of Compound P and calculation of its moles

Hydrolysis of magnesium dicarbide ($\text{Mg}_2\text{C}_3$) yields propyne ($\mathbf{P}$): $\text{Mg}_2\text{C}_3 + 4\text{H}_2\text{O} \rightarrow 2\text{Mg(OH)}_2 + \text{CH}_3\text{-C}\equiv\text{CH}$

• Compound $\mathbf{P}$: Propyne ($\text{C}_3\text{H}_4$)
• Molar mass of $\mathbf{P}$ ($\text{C}_3\text{H}_4$): $\text{Molar Mass} = (3 \times 12) + (4 \times 1) = 40\text{ g mol}^{-1}$
• Given mass of $\mathbf{P}$: $4.0\text{ g}$
• Moles of $\mathbf{P}$: $n_{\mathbf{P}} = \frac{4.0\text{ g}}{40\text{ g mol}^{-1}} = 0.1\text{ mol}$

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Step 2: Formation of Compound Q

Propyne ($\mathbf{P}$) undergoes terminal alkyne deprotonation using $\text{NaNH}_2$, followed by alkylation with methyl iodide ($\text{CH}_3\text{I}$) to form but-2-yne ($\mathbf{Q}$): $\text{CH}_3\text{-C}\equiv\text{CH} \xrightarrow{\text{NaNH}_2} \text{CH}_3\text{-C}\equiv\text{C}^-\text{Na}^+ \xrightarrow{\text{CH}_3\text{I}} \text{CH}_3\text{-C}\equiv\text{C-CH}_3$

• Compound $\mathbf{Q}$: But-2-yne ($\text{C}_4\text{H}_6$)
• Reaction yield: $75\%$
• Moles of $\mathbf{Q}$ formed: $n_{\mathbf{Q}} = n_{\mathbf{P}} \times \frac{75}{100} = 0.1 \times 0.75 = 0.075\text{ mol}$

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Step 3: Trimerization of Q to form Compound R

Passing but-2-yne ($\mathbf{Q}$) through a red-hot iron tube at $873\text{ K}$ causes cyclic trimerization to form hexamethylbenzene ($\mathbf{R}$): $3\,\text{CH}_3\text{-C}\equiv\text{C-CH}_3 \xrightarrow[\text{873 K}]{\text{red hot iron tube}} \text{C}_6(\text{CH}_3)_6$

• Compound $\mathbf{R}$: Hexamethylbenzene ($\text{C}_{12}\text{H}_{18}$)
• Stoichiometry: $3\text{ moles of } \mathbf{Q} \rightarrow 1\text{ mole of } \mathbf{R}$
• Theoretical moles of $\mathbf{R}$: $n_{\mathbf{R, theoretical}} = \frac{n_{\mathbf{Q}}}{3} = \frac{0.075}{3} = 0.025\text{ mol}$
• Reaction yield: $40\%$
• Actual moles of $\mathbf{R}$: $n_{\mathbf{R, actual}} = 0.025 \times \frac{40}{100} = 0.010\text{ mol}$

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Step 4: Calculation of the Mass of Compound R ($\mathbf{x}$)

• Molar mass of $\mathbf{R}$ ($\text{C}_{12}\text{H}_{18}$): $\text{Molar Mass} = (12 \times 12) + (18 \times 1) = 144 + 18 = 162\text{ g mol}^{-1}$
• Mass of $\mathbf{R}$ ($\mathbf{x}\text{ g}$): $\mathbf{x} = n_{\mathbf{R, actual}} \times \text{Molar Mass of } \mathbf{R} = 0.010\text{ mol} \times 162\text{ g mol}^{-1} = 1.62\text{ g}$

Final Answer: The value of $\mathbf{x}$ is 1.62