JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

The cycloalkane (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 $^{- 1}$ H : 1, C : 12, O : 16, Br : 80)

Your answer

Solution

One mole of $B r_{2}$ delivers two Br atoms. A C:Br atomic ratio of $3 : 1$ in Y means the molecule has $3 n$ carbons and $n$ bromines. With 2 Br per molecule, $n = 2$ and so the carbon count is 6.

So Y is $C_{6} H_{10} B r_{2}$ . Molar mass: $6 (12) + 10 (1) + 2 (80) = 72 + 10 + 160 = 242$ g/mol.

$% B r = \frac{160}{242} \times 100 = 66.11% \approx 66%$ .

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