The cycloalkene ( X ) on bromination consumes one mole of bromine per mole of ( X ) and gives the product ( Y ) in which C : Br ratio is 3:1. The percentage of bromine in the product ( Y ) is \_\_\_\_%. (Nearest integer) (Given : molar mass in g mol ^ -1 H : 1, C : 12, O : 16, Br : 80)
A. A
A cycloalkene (X) undergoes bromination with 1 mole of Br₂, consuming 2 moles of Br atoms (addition to the double bond). The product (Y) has C:Br ratio of 3:1. With 2 Br atoms present, the number of C atoms is 2 × 3 = 6. Therefore, product Y has composition C₆H₁₂Br₂ with molar mass = 6(12) + 12(1) + 2(80) = 72 + 12 + 160 = 244 g/mol. Percentage of Br = (2 × 80 / 244) × 100 = (160 / 244) × 100 = 65.57%, which rounds to 66% (nearest integer).
Related: Chemistry — Hydrocarbons · All PYQ Banks