In bromination of Propyne, with Bromine 1, 1, 2, 2-tetrabromopropane is obtained in 27% yield. The amount of 1,1,2,2 tetrabromopropane obtained from 1 g of Bromine in this reaction is ____ × 10⁻¹ g. (Nearest integer) (Molar Mass : Bromine = 80 g/mol)
Answer: 3
Solution
The bromination of propyne (CH₃-C≡CH) with Br₂ produces 1,1,2,2-tetrabromopropane (CH₃-CBr₂-CHBr₂).
From the reaction stoichiometry, 1 mole of propyne reacts with 2 moles of Br₂ to give 1 mole of product.
Given: 1 g of Br₂, Molar mass of Br₂ = 160 g/mol, Molar mass of product = 360 g/mol, Yield = 27%
Moles of Br₂ = 1/160 mol
Moles of product formed = (1/160) × (1/2) = 1/320 mol
Theoretical mass of product = (1/320) × 360 = 1.125 g
Actual mass (27% yield) = 1.125 × 0.27 = 0.30375 g = 3.0375 × 10⁻¹ g
