Question

What is the percentage of carbon in the product ‘X’ formed in the following reaction?

• A. 85.6 %
• B. 80.6 %
• C. 90.6 %
• D. 70.6 %

Hint:

Friedel's Crafts alkylation of benzene with \(\text{CH}_3\text{Cl}\) produces toluene (\(\text{C}_7\text{H}_8\)). - The calculated \(\%\text{C}\) in \(\text{C}_7\text{H}_8\) is around \(91\%\); the given option nearest to this is \(90.6\%\).

Answer 1

The Correct Option is C

Step 1: Identify the Product \(X\) - The reaction of benzene with methyl chloride \(\bigl(\text{CH}_3\text{Cl}\bigr)\) in the presence of \(\text{AlCl}_3\) is a Friedel–Crafts alkylation. - The product \(X\) is toluene \(\bigl(\text{C}_6\text{H}_5\text{CH}_3\bigr)\), which has the molecular formula \(\text{C}_7\text{H}_8\). 

Step 2: Calculate the Percentage of Carbon in \(\text{C}_7\text{H}_8\) \[ \text{Molecular mass of C}_7\text{H}_8 = (7 \times 12) + (8 \times 1) = 84 + 8 = 92 \] \[ \text{Mass of carbon} = 7 \times 12 = 84 \] \[ \%\text{ Carbon} = \frac{84}{92} \times 100 \approx 91.3\% \] Among the given options, \(90.6\%\) is the closest. Hence, the percentage of carbon in the product \(X\) (toluene) is approximately \(90.6\%\).