Question

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: 
\[ \text{H} = 1,\quad \text{C} = 12,\quad \text{O} = 16,\quad \text{Br} = 80 \]

Hint:

For alkene bromination, one mole of Br\(_2\) adds across one double bond, introducing two bromine atoms into the product.

Answer 1

Correct Answer: 67

To determine the percentage of bromine in the product (Y), start by analyzing the given information. The compound (X) is a cycloalkene, which reacts with one mole of bromine. The reaction suggests the formation of a dibromo compound. The C:Br ratio in the product (Y) is 3:1, meaning for every 3 carbon atoms, there is 1 bromine atom.

Assume the molecular formula of (Y) is \(C_xH_yBr_z\). Given the C:Br ratio of 3:1, set \(x=3k\), \(z=k\), for some integer \(k\). Bromination of a cycloalkene suggests adding Br₂ across a double bond, forming two C-Br bonds. This implies the simplest form is \(C_3H_4Br_2\) (because two bromine atoms are needed to break one double bond in a cycloalkene). However, given the ratio \(x:z = 3:1\), \(z=1\). So, the compound can be simplified to \(C_3H_5Br\).

Calculate the molar mass (M) of \(C_3H_5Br\):

\[ M = (3 \times 12) + (5 \times 1) + 80 = 36 + 5 + 80 = 121\]

Determine the percentage of bromine:

\[ \text{Percentage of Br} = \left(\frac{80}{121}\right) \times 100\]

Calculate:

\[ \frac{80}{121} \approx 0.6612 \quad \Rightarrow \quad 66.12\% \]

Round to the nearest integer: 66%

The percentage of bromine in product (Y) is 66%, which fits within the provided range.