Q. 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 gmol−1 H : 1, C : 12, O : 16, Br : 80)
Complete Step-by-Step Calculation
Step 1: Nature of Reaction
Cycloalkene + Br₂ → Vicinal dibromide
One mole of Br₂ adds across one double bond.
Step 2: Understanding C : Br Ratio
Given in product (Y), C : Br = 3 : 1
This means:
For every 3 carbon atoms, there is 1 bromine atom.
Step 3: Since bromination adds 2 Br atoms
Let number of carbon atoms = C
Then number of Br atoms = C / 3
But bromination adds exactly 2 Br atoms
So:
C / 3 = 2
Therefore,
C = 6
Step 4: Molecular Formula of Product
Cycloalkene with 6 carbons = C₆H₁₀
After bromination → C₆H₁₀Br₂
Step 5: Molar Mass Calculation
C₆ = 6 × 12 = 72
H₁₀ = 10 × 1 = 10
Br₂ = 2 × 80 = 160
Total molar mass = 72 + 10 + 160 = 242 g/mol
Step 6: Percentage of Bromine
Mass of bromine = 160
% Br = (160 / 242) × 100
= 66.11%
Nearest integer = 66%