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 $$mol^{-} H: 1, C: 12 , 0: 16, Br: 80$$)

We need to find the percentage of bromine in product Y.

To determine the cycloalkene, the cycloalkene X consumes one mole of $$Br_2$$ per mole of X (adds across the double bond), and in the product Y the C:Br ratio is 3:1 (by atoms). A cycloalkene $$C_nH_{2n-2}$$ (general formula) adds $$Br_2$$ to give $$C_nH_{2n-2}Br_2$$. For C:Br = 3:1, we need $$n:2 = 3:1$$, so $$n = 6$$. The cycloalkene is cyclohexene ($$C_6H_{10}$$), and the product is 1,2-dibromocyclohexane ($$C_6H_{10}Br_2$$).

Calculating the molecular mass of product Y gives $$M_Y = 6(12) + 10(1) + 2(80) = 72 + 10 + 160 = 242$$.

To calculate the percentage of bromine, we use $$\% Br = \frac{160}{242} \times 100 = 66.12\%$$.

The percentage of bromine is approximately $$66\%$$ (nearest integer). Therefore, the answer is 66.