Cycloalkene (X) reacts with bromine. During the reaction, 1 mole of cycloalkane consumes 1 mole of Br$_2$ to form a product (Y). The product (Y) has C:Br ratio of 3:1. The percentage of bromine in product (Y) is:
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In reactions where a halogen reacts with an alkene, the percentage of halogen in the product can be determined using the molar mass and the ratio of carbon to halogen atoms in the product.
Step 1: Analyzing the reaction.
Cycloalkene (X) reacts with bromine (Br$_2$), and 1 mole of cycloalkene consumes 1 mole of Br$_2$. The product (Y) has a carbon (C) to bromine (Br) ratio of 3:1. This means for every 3 carbon atoms, 1 bromine atom is present in the product.
Step 2: Determining the molecular composition.
Let the molar mass of cycloalkene (X) be \(M_X\), and that of bromine (Br$_2$) be \(M_{Br_2}\). The molar mass of the product (Y) will be the sum of the molar masses of cycloalkene and bromine, adjusted for the given ratio. The molecular formula of the product (Y) can be expressed as \(C_3H_5Br\), based on the given C:Br ratio of 3:1.
Step 3: Calculating the percentage of bromine.
The molar mass of \(C_3H_5Br\) is:
\[
\text{Molar mass of } C_3H_5Br = (3 \times 12) + (5 \times 1) + (1 \times 80) = 36 + 5 + 80 = 121 \, \text{g/mol}
\]
The mass of bromine in the product is 80 g (since 1 mole of Br is 80 g). Therefore, the percentage of bromine in the product (Y) is:
\[
\text{Percentage of bromine} = \frac{80}{121} \times 100 = 66.11%
\]
Step 4: Conclusion.
Thus, the percentage of bromine in product (Y) is 66.11%, which corresponds to option (1).
