Question

What is the mass (in mg) of final product (C)? \[ \text{CH}_3\text{C}\text{NO}_2 \xrightarrow{\text{(i) Sn/HCl}} \text{CH}_3\text{C}\text{OOCOCH}_3 \xrightarrow{\text{(ii) pH neutral}} \text{CH}_3\text{C}\text{OH} \xrightarrow{\text{Br}_2/\text{Ac}} \text{Product (C)} \] Given: 137 mg of the starting compound.

Hint:

In chemical reactions, use stoichiometric relationships to calculate moles and convert them into masses based on molar mass.

Answer 1

Correct Answer: 228

Step 1: Moles of starting material.
Moles of starting material = \( \frac{137 \times 10^{-3}}{137} = 0.001 \) mole.
Step 2: Moles of the product.
Moles of product = 0.001 mole (since the reaction is 1:1).
Step 3: Mass of the product.
Mass of product = \( 0.001 \times 228 \, \text{g} = 0.228 \, \text{g} = 228 \, \text{mg} \).
Step 4: Conclusion.
Thus, the mass of the final product is 228 mg.