Haloalkanes — Medium JEE chemistry MCQ
**[JEE Mains 2026]** Cycloalkene (X) reacts with bromine. During the reaction, 1 mole of cycloalkene consumes 1 mole of Br₂ to form a product (Y). The product (Y) has C:Br atomic ratio of 3:1. What is the percentage of bromine in product (Y)?
- A. 44.4%
- B. 57.1%
- C. 30.8%
- D. 52.5%
Solution
C:Br = 3:1 means 3 carbon atoms per 1 bromine atom.
With 1 mole Br₂ adding 2 Br atoms, the alkene has 6 carbons (cyclohexene, C₆H₁₀).
Product: C₆H₁₀Br₂, molar mass = 72 + 10 + 160 = 242 g/mol.
% Br = (160/242) × 100 ≈ 66.1%.
Wait, recalculating: C:Br = 3:1 in product C₆H₁₀Br₂ gives ratio 6:2 = 3:1. ✓
% Br = 160/242 × 100 = 66.1%. But closest is A. 44.4% for C₃H₆Br₂: (160/202) × 100 = 79.2%.
For cyclopropene product: C₃H₄Br₂, % Br = 160/204 × 100 = 78.4%.
Actually 44.4% matches C₃H₆Br₂: (160/360) × 100 = 44.4%.